Intermediate Algebra Chapter 1 - East Los Angeles College · Intermediate Algebra Chapter 1: Section 1 Examples: 2,3,5,7,11,13,17 … are all prime numbers e) Composite Numbers are

Intermediate Algebra Chapter 1: Section 1

By Viken Kiledjian Section 1 Sets and the Real Number

System:

A set is a grouping of objects. The objects in the set are called its elements. The set that contains no elements is called the null set.

1) SET NOTATION: There are 2 main methods of writing sets.

a) Roster Method: We enclose the elements of the set within braces.

ex. A = {1, 2, 4, 6}

The name of this set is A. To indicate that 2 is an element of this set, we write


2 {1, 2, 4, 6} or 2 A

And to indicate that another number like 5 is not a member of this set, we would write

5 {1, 2, 4, 6} or 5 A

b) Set Builder Notation: We could also express the previous set A like this.

A = {x| x = 1, 2, 4, 6}

This reads the set of all numbers x, such that x = 1, 2, 4, or 6.

Another example would be:

B = {x| x is a positive integer less than 10}

This would represent the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9.

Equality of Sets: Two sets are equal if and only if all of their elements are the same.


2) SUBSETS OF SETS: A set A is a subset of another set B if and only if all of set A’s elements are also elements of set B.

Ex. If A = {1, 2, 3, 4, 5} and B = {1, 4}

Then B is a subset of A and is written

B A

A couple of interesting facts about subsets.

a) every set is a subset of itself!!! Therefore, B B

b) the nullset is a subset of every set!!!

3) UNIONS AND INTERSECTIONS:

a) The Union of two sets is the set of all of their elements combined together. ( The symbol is )

b) The Intersection of two sets is the set of all elements that they share in common. (The symbol is )


Ex: Let A = {1, 2, 4, 6} and B = {x| x is a positive integer less

than 10} and C = {2, 4, 8, 10}, then

A B = {1, 2, 3, 4, 5, 6, 7, 8, 9} and A C = {1,2,4,6,8,10}

A B = {1, 2, 4, 6} and A C = {2, 4}. Therefore,

(A B) C = {1,2,3,4,5,6,7,8,9,10}

Now we have to add the element 10 to the answer that we got

for A union B.

(A B) C = {2, 4, 8} because the set C only has 2, 4, and

8 in common with the union of A and B.

(A B) C = {2, 4} because the set C only has 2 and 4 in

common with the intersection of A and B.

There are many other combinations you can try out for practice!!


4) SETS OF NUMBERS:

a) Natural Numbers include the numbers starting from 1 then

2, 3, 4, 5, all the way to infinity.

b) Whole Numbers include the numbers starting from 0 then 1,

2, 3, 4, all the way to infinity.

c) Integers include numbers such as

-4, -3, -2, -1, 0, 1, 2, 3, 4 going all the way to negative and

positive infinity.

(The Natural Numbers are a subset of Whole Numbers and

Integers. The Whole Numbers are a subset of Integers)

d) Prime Numbers are the natural numbers greater than 1 that

are divisible by themselves and 1.


Examples: 2,3,5,7,11,13,17 … are all prime numbers

e) Composite Numbers are all the numbers that are not prime.

f) Rational Numbers are all the numbers that can be written as the ratio of two integers a/b where b 0.

Ex: - 2, ¾, 0, 11/89 …are all rational numbers

Note: Integers are a subset of Rational Numbers

g) Irrational Numbers are numbers whose decimals never repeat or terminate.

Examples: p, both have non-terminating decimals

Note: Rational Numbers are NOT subsets of Irrationals.

h) Real Numbers are all the non-imaginary numbers which can be written a repeating, terminating or non-terminating decimal.

2


Note: Rational and Irrational Numbers put together comprise the

Real Numbers.

5) ORDERING REAL NUMBERS: To order real numbers imagine putting them on a number line such as

| | | | | | | | | | | | | | |

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

The more to the right a number is the bigger it is. That’s the rule!!

Examples: 4 > 1, 5>-1, 0> -2 because it is to the right of -2

-4 > -7 and -6 < 2 because -6 is to the left of 2

6) INTERVALS: This is yet another way of expressing sets of numbers. There are 3 kinds of intervals.


a) Open Intervals: such as (-2, 5) means all the real numbers

between –2 and 5 but NOT INCLUDING –2 AND 5!!!

b) Half-Open Intervals: such as (-2, 5] means all the real

numbers between –2 and 5 INCLUDING 5 but NOT –2!!!

c) Closed Intervals: such as [-2, 5] means all the real numbers

between –2 and 5 INCLUDING BOTH –2 and 5.

We can now take intersections and unions of such sets. This is Fun!!

Example: Let set A = (-2, 5) and set B = [5, 8) and set C = [1, )

(The symbol means infinity. You can also have - )

Then, A B = the Null Set since the set A does not include 5

A C = [1, 5) since that is what they have in common.


and B C = [5, ) since that is what they share in common.

However, we can now do so unions and they will be different!!

A B = (-2, 8) because this is the sum of what they both have.

A C = (-2, ) because this is the sum of what they both have.

B C = [1, ) because B is a subset of C and its union with

C is equal to the set C

7) ASOLUTE VALUE OF A NUMBER: measures how far the

number is from the origin (which is the number 0 on the number

line). The symbol for absolute value is | |

Ex: |4| = 4, |0|= 0, |-3| = 3, |-(-4)| = 4 because the two

negatives inside the absolute value cancel each other and

absolute value of 4 becomes 4


|-4 – 10| = |-14| = 14

But |-4| - |10| = 4 – 10 = -6 Quite a different answer!!!

Also, |-3| . |4| = (3).(4) = 12

But, -|3| . |-4| = (-3).(4) = -12 because the negative is outside the

absolute value sign. So always watch out for the position of the

negative sign!!!!!!!!!!!

Formal definition of the absolute value:

For any real number x,

x , if x 0 (When the number is positive, just

|x| = leave it alone. When it is negative,

-x , if x < 0 take its negative to make it positive.)


Section 2: This section concerns the Properties of real numbers and

how to add, subtract, divide, and multiply them and evaluate

complex expressions.

1) ADDING, SUBTRACTING, MULTIPLYING, DIVIDING:

Practice the skills you learned in the beginning algebra course!!!

Ex. 5 – 6 = -1, -4 – 7 = -11, -4 + 11 = 7

because the two negatives make a positive and the

the three in the denominator reduces the 6 into a 2

and the two times the previous 2 makes a 4!!!!

7

6

3

2

7

4

6322

15

5

4

15

2

5

4


2) ORDER OF OPERATIONS: In an expression which contains

some addition, subtraction, multiplication and division,

multiplication and division always get the priority unless the

addition and subtraction are put in parentheses!!!!!

Ex. 2 – 3.5 = 2 – 15 = -13 unless the problem is (2-3).5 = -5

4 + 2/6 = 4 + 1/3 = 41/3 unless the problem is (4+2)/6 = 1

notice that I divided the 4 by 4 first since the order of precedence

between a division and multiplication sign is from left to right!!

423213132134432

81

8

23

282

243

2242


3) EVALUATING ALGEBRAIC EXPRESSIONS: This part

essentially reinforces the same skills of the previous part.

Except now you will work with variables.

Ex. Assume a = 2, b = -3, c = -2, then evaluate these expressions

1) a + b x c = 2 + (-3) x (-2) = 2 + 6 = 8 You just plug in the

variables and follow the order of operations rules. Practice the

order of operations before you move on to these!!!!!!

2)

4

74

7

4

310

)1(3

3)2()5(

2)2(3

)3()2()3(2

acb

bcba


1) PROPERTIES OF EXPONENTS: These are the main properties of exponents

a) Product Rule of Exponents: xmxn = xm+n

b) Power Rule of Exponents: (xm)n = xmn and (xy)n = xn yn and

if y is not 0

c) Law of Zero Exponents: if x is not 0, then x0 = 1

d) Law of Negative Exponents: if x is not 0, then

and

n

nn

y

x

y

x

n

n

xx

1 n

nx

x

1


e) Quotient Rule of Exponents: if x is not 0, then

f) Theorem of Reciprocals: if x and y are not 0, then

Here are examples of each one. It is very important to practice

these until you know them by heart!!!!

a) 2324 = 23+4 = 27 = 128, 4(-5)47 = 4(-5 + 7) = 42 = 16

b) (42)3 = 46 = 4096 (notice that you don’t add the 2 and 3!!!)

(4.2)3 = 43 23 = 64.8 = 512 (here you don’t have to use the law;

you can just multiply 4 by 2 and get 8 and 8 cubed is 512)

nm

n

m

xx

x

nn

x

y

y

x


here is another example of

law B: -------------------->

d) and

e)

f)

Here are some more complex examples which combine the Rules!!

27

8

3

2

3

23

33

9

1

3

13

2

2 6444

1 3

3

81333

3 426

2

6

16

49

4

7

4

7

7

42

222

4

6

42

6642232

4222

x

y

x

yyxyx


It takes lot of practice to master this. Good luck and have fun at it!!

34

12312430)1(213

31

023

zx

yzyxzyx

zxy

zyx nnn

n

846

864

)2)(4()2)(2()2)(3(

)2)(4()2)(3()2)(2(2

423

432

2

2

2

2

2

2

yx

yx

yx

yx

yx

yx nnn

1646164610)8(8)4(6)6(4 102422 yxyxyx nnn


2) ORDER OF OPERATIONS AND USING CALCULATORS:

This part builds on the former. There are NO new formulas!!!

Ex. Suppose x = 2.7, y = 3.0 Evaluate the following expressions

1) = x4 y-6 = (2.7)4(3)-6 = 0.0729 use the power

button on your calculator.

(On some calculators the power button is yx button. On

TI calculators, the power button looks like ^ )

2) x-3 + y4 = (2.7)-3 + (3.0)4 = 81.0508

This is the end of Section 3 !!!!!!!!!!!!

232 yx


SCIENTIFIC NOTATION: This section is pretty short!!! It is

intended to illustrate the use of scientific notation. Scientific

Notation is used to express large or small numbers and to make

calculations easier.

Ex .000000456 = 4.56 x 10-7, .0000389 = 3.89 x 10-5

345,023,000,000 = 3.45023 x 1011, 20,000,000,000,000 = 2 x 1013

Notice that you always move the decimal to the right of the 1st digit!

Now, we can also go backwards.

Ex 2.4 x 10-4 = .00024 (you move the decimal point 4 places to

the left)

3.5 x 100 = 3.5 since 10 to the power 0 is equal to 1

4.62 x 105 = 462,000 (move decimal point 5 places to the right!)


Now, we can do some calculation using scientific notation.

Ex. Suppose we want to calculate this expression,

(450,000)(23,000) we can either put this as it is into our

(780,000,000) calculators or we can first convert

them into scientific notation!

(4.5 x 105)(2.3 x 104) Now, combine the powers of ten,

(7.8 x 108) and put the rest into the calculator!

(4.5)(2.3) x 109 = 1.33 x 101 Both methods give the same

(7.8) x 108 answer, so it’s a matter of style!!!

If you put the numbers as they are into the calculator, then the

answer will be 13.3 which you then change to scientific notation.


Here is another example using small numbers this time.

(.0000067)(.0002004) = .000257 if you just put into calculator.

(.00000523) You can then change this into

scientific notation, so the answer will be 2.57 x 10-4. Some

calculators will already give the answer in scientific notation!!

However, you can also do it by first changing all the numbers to

scientific notation and combining the powers of the ten yourself,

and ONLY inputting the other numbers into the calculator.

(6.7 x 10-6)(2.004 x 10-4) = (6.7)(2.004) x 10-10 =

(5.23 x 10-6) (5.23) x 10-6

2.57 x 10(-10 – (-6)) = 2.57 x 10(-10 + 6) = 2.57 x 10-4 Whalla!!


Section 5 is about solving algebraic equations using the properties

of equality and working with repeating decimals, and isolating

a variable in an equation.

1) SOLVING EQUATIONS using the properties of equality.

a) If a = b, then ca = cb (you can multiply both sides of the

equation by the same number)

b) If a = b, then c + a = c + b (you can add the same number to

both sides of the equation)

Given these two properties, you can solve any linear equation by

combining like terms. Like terms are any terms that look

similar and have the same power of the x. There are two types

of equations which have no definite solution.


a) Identity: There are infinite number of solutions

b) Impossible equation or Contradiction There is no solution.

Here is an example of each special case.

a) x(4 – x) = 4x + 2 – (x2 +2) You distribute the x into the 4 – x

4x – x2 = 4x + 2 – x2 – 2 and the negative 1 into the x2 + 2

4x – x2 = 4x – x2 The 2 and the negative 2 cancel and you are

left with an identity because 4x – x2 is always equal to itself

b) 3(x – 6) = 5x - 2(x – 4) Distribute the 3 into the x – 6

3x – 18 = 5x – 2x + 8 and the –2 into the x – 4.

3x – 18 = 3x + 8 Subtract 3x from both sides and you get,

-18 = 8 which is an impossibility!! There is NO SOLUTION


Here are some general examples of linear equations.

1) 4(3 – x) +2 = x - 5 first Distribute the 4 into the 3 – x

12 – 4x + 2 = x – 5 then combine the 12 and 2

10 – 4x = x – 5 then add 4x to both sides (usually it is good practice to collect the x’s to the side which makes their coefficient positive!!)

10 = 5x – 5 now add 5 to both sides!!!

15 = 5x finally divide both side by 5

3 = x this is the answer!!!!

2) The best thing to do here is to

multiply both sides by the LCD (least common denominator) to get rid of all fractions at once!!!!

4

6143

2 xx


Now distribute the 12 into

the paranthesis

8(4x – 1) + 72 = 3x Now distribute the 8!!

32x – 8 + 72 = 3x Now combine the –8 and 72

32x + 64 = 3x Subtract 3x and 64 from both sides!!

29x = -64 Divide by 29

x = - 64/29 Now let’s do one more example!!!

3) Multiply everything by the

LCD which is 12 again!!

)4

(12)6143

2(12

xx

4

3

6

2

4

12

xxx


3(2x – 1) + 2(2 – x) = -3(x + 3) Notice that the negative in front of the x + 3 fraction has to be distributed into the x + 3!!!

6x – 3 + 4 – 2x = -3x – 9

4x + 1 = -3x – 9 You can do the following steps usually in one step. Add 3x to both sides and subtract 1 from both sides. This will collect the x’s to the side which makes their coefficient positive and it will collect all the other things on the other side.

7x = -10 and x = - 10/7

2) REPEATING DECIMALS The general rule of repeating decimals is that any number divided by 9 becomes a repeating decimal. So 4/9 = .444444…. , 7/9 = .77777…, 6/9 = .6666..

)4

3(12)

6

2(12)

4

12(12

xxx


Notice that 6/9 is 2/3 and we already know that 2/3 = .66666…

In this section we shall learn to deal with more difficult repeating decimals such as .566666… where only the 6’s are repeating but the 5 is not. The question is, “What fraction yields the decimal .56666…?” The way to do it is this.

Notice that the repetition of the 6’s begins from the 2nd decimal digit which is the 1/100th place. So

.56666… = 5/10 + 6/90 Try it now on your calculator. Divide 6 by 90 and you will notice that it equals .066666667 and of course 5 divided by 10 equals .5, so now just add 5/10 to 6/90.

45/90 + 6/90 (5/10 is equivalent to 45/90!!!)

51/90 which reduces to 17/30. Now divide 17 by 30 on your calculator and it should equal .566666…..


Let’s do another example.

.4022222… = 40/100 + 2/900 Notice that the 2 begins repeating from the 3rd decimal place which is the 1/1000th place. Divide 2 by 900 on your calculator and it will equal .002222222…

= 360/900 + 2/900 (40/100 is equivalent to 360/900. Just multiply the top and bottom by 9 to achieve the common denominator of 900)

= 362/900 = 181/450 which equals .40222222……

C. ISOLATING A VARIABLE IN AN EQUATION. This section utilizes the same techniques as subsection A, but with variables. Let’s do some examples.

1) A = Lw solve for w. (this is the simplest case.) We just divide both sides by L to isolate w. Therefore, w = A/L


2) y = mx + b This is the slope-intercept form for a straight line.

Suppose, we knew y, x, and b, and wanted to solve for m. Let’s isolate m. First subtract b from both sides,

y – b = mx Now divide both sides by x!!

m = 1/x (y – b) Write m on the left of the equal sign because that’s what you are solving for!!

3) S = n(a + L) This is the equation for the sum in an algebraic

2 series. Suppose we want to isolate a.

2S = n(a + L) First multiply both sides by 2.

2S/n = a + L Divide both sides by n!!!

2S/n – L = a Subtract L from both sides,

a = 2S/n – L Rewrite it so that a is on the left side of the = sign!!


Section 6: This section and the next section are Word Problems. This is everyone’s favorite part of Algebra, right? My technique for word problems is to show you examples rather than talking in the abstract. For each kind of word problem, I’ll show you about 2 examples. It’s up to the student to then practice more homework problems to get it down firmly.

In this section, there are 4 kinds of word problems.

RECREATION:

Ex. 1 A 186-foot cable is to be cut into four pieces. Find the length of each piece if each successive piece is 3 feet longer than the previous one.

Solution: Let x = length of the shortest piece in feet

then x + 3 = length of the next longer piece


then x + 6 = length of the next longer piece

and x + 9 = length of the longest piece.

(Notice that I am adding 3 feet to each one since each successive piece is 3 feet longer.) Now I just add all four pieces and set them equal to 186

x + (x + 3) + (x + 6) + (x + 9) = 186

4x + 18 = 186 ----------> 4x = 168 ----> x = 42

Therefore, the answer is, each successive piece is

42 feet, 45 feet, 48 feet, 51 feet. (You can now check this by adding all these numbers and seeing if they equal 186!!!)

Ex. 2 A 30-foot beam is to be cut into two pieces. The longer pieces is to be 2 feet more than 3 times the shorter piece. Find the length of each piece.


let x = the shorter piece (I can also let x be the longer piece, but

the setup would be different!!!)

then 30 – x = length of longer piece (since the total length is 30 feet, if one of them is x, then the other one is 30 – x, NOT x – 30)

Now use the fact that the longer piece ----> 30 – x is 2 feet more than (this is a + sign) 3 times (this is a Multiplication) the shorter piece. Therefore, this sentence translates into,

30 – x = 2 + 3x Add x to both sides, Subtract 2 from both sides

28 = 4x Divide both sides by 4!!

x = 7 feet Therefore, the final answer is

The shorter piece is 7 feet long, the longer piece is 23 feet long.

(Now you can check your answer. Multiply 7 by 3 and you get 21. Now add 2 to it and you get 23. Whalla, it works out!!!)


BUSINESS PROBLEMS:

Ex. 1 A bedroom set regularly sells for $983. If it is on sale for $737.25, what is the percent of markdown?

Solution: First find the amount of the sale = 983 – 737.25 = 245.75

Now the percent of sale is the Amount of Sale divided by the Original Price times a hundred to change to percentage.

% sale = $245.75 x 100 = 25% sale.

$983.00

Ex. 2 A pension fund owns 12,000 shares in mutual stock funds and mutual bond funds. Currently, the stock funds sell for $12 per share, and the bond funds sell for $15 per share. How many shares of each does the fund own if the total value of the securities is $165,000?


Solution: This is a little similar to the second example under the Recreation Problems.

Let x = number of shares in mutual bonds at $15/share

Then 12,000 – x = number of shares of stocks at $12/share

Then 15x is the total value of the mutual funds and 12(12,000 – x) is the total value of the stocks. They have to add to 165,000.

15x + 12(12,000 – x) = 165,000 Distribute the 12 in the paranthesis

15x + 144,000 – 12x = 165,000 Combine the 15x and the –12x!

3x + 144,000 = 165,000 Subtract 144,000 from both sides.

3x = 21,000 ------> x = 7,000 Therefore the final answer is

There are 7,000 shares of bonds and 5,000 shares of stocks.

You can now check your answer. Multiply 7,000 by $15 and 5,000 by $12 and add them and you should get $165,000.


GEOMETRIC PROBLEMS:

Ex. 1 The width of a rectangular swimming pool is one-third its

length. If its perimeter is 96 meters, find the dimensions of the

pool.

Let w = width of rectangular in meters

Then L = 3w (the opposite of one-third is 3. I did this to avoid

fractions. I’ll show you the setup for the other way also!!)

Now the Perimeter of a rectangle is the equivalent of Circumference

for a circle. It is the distance all the way around the rectangle.

P = L + L + w + w = 2L + 2w therefore

96 = 2(3w) + 2w I substituted 3w in the place of L.

96 = 6w + 2w = 8w Therefore, w = 96/8 = 12 feet


And L = 3w = 3(12) = 36 feet. The final answer is

The width is 12 feet and the length is 36 feet.

((Now check your answer. Twice 36 equals 72 and twice 12 equals 24. If you add 72 to 24, you get 96 feet))

The other setup would be this way.

Let L = length of the rectangle

Then w = L/3 is the width, since it is one-third of the length.

P = 2L + 2w, so 96 = 2L + 2(L/3) = 2L + 2L/3

96 = 6L/3 + 2L/3 Get a common denominator so you can add them

96 = 8L/3 Now multiply both sides by 3/8

3/8(96) = L, and therefore L = 36 feet. Notice how much uglier it was. So, to avoid this, let your variable equal the smaller one!!


Ex. 2 A rectangular picture has a length that is 5 inches longer

than its width. You put a frame on the picture that is 2 inches

wide all around. If the perimeter of the frame is now 70 inches,

what was the original length and width of the unframed picture?

Let w = width of picture

Then L = w + 5 Now, when you put a frame that is 2 inches wide

all around, the width and length of the picture increase by 4

inches. That’s because there are 2 inches on the top and bottom

and 2 inches on the left and right!!!! So the New length and

width of the framed picture are:

New width ----> w + 4 New Length ------> (w + 5) + 4 = w + 9

Therefore, P = 2L + 2w = 2(w + 9) + 2(w + 4) = 70 inches


Distribute the 2’s into the paranthesis and combine like terms:

2w + 18 + 2w + 8 = 70 --------> 4w + 26 = 70 -----> 4w = 44

w = 11 inches and L = 11 + 5 = 16 inches.

Now check your answer: Adding 2 inches on both sides to the length and width make them 15 and 20. Double 15 and you get 30. Double 20 and you get 40. Add 30 to 40 and it equals 70!!!

Ex. 3 If one of two supplementary angles is 40 degrees larger than the other, find the measure of the smaller angle.

Supplementary angles are angles that add to 180 degrees

Complimentary angles are angles that add to 90 degrees

Vertical angles are angles that are congruent (equal in value)

In this problem, we are dealing with 2 supplementary angles.

Let x = the smaller angle -----------> then x + 40 = larger angle


Now add them and set the sum equal to 180 degrees.

x + (x + 40) = 180 ------> 2x = 140 ------> x = 70 degrees

Now check your answer: If the smaller angle is 70 degrees, and the larger is 40 more, therefore the larger is 110 degrees. Now add 70 and 110 degrees and you’ll get 180.

Always get in the habit of checking your answers whenever possible. I will not always show you the steps of checking from now on, but you should do it yourself.

Ex. 4 If the height of a triangle with a base of 10 inches is doubled, the area of the triangle is increased by 90 square inches. Find the height of the original triangle.

Area of a Triangle = ½ base x height

Area of a Rectangle = base x height


Let h = height of original triangle and base = 10 inches

Then A = Original area = ½ (10)h = 5h

Now, the problem says that by doubling the height of the triangle, the original Area will increase by 90 in2. Therefore,

New Area = ½ (base)(new height) = ½ (10)(2h) = 10h And,

New Area = Original Area + 96 = 5h + 90

Now equate the 2 New Areas ----->10h = 5h + 90 ------>5h = 90

Original Height = 18 inches.

LEVER PROBLEMS:

Ex. 1 Two forces --- 110 pounds and 88 pounds --- are applied to opposite ends of an 18 foot lever. How far from the greater force must the fulcrum be placed so that the lever is balanced?

Let x = distance of fulcrum from the greater force of 110 pounds


Then 18 – x = distance of fulcrum from the smaller force of 88 lbs

The principle we have to use here is that the Torque due to a force

is the product of the force x distance to pivot point. The torques

due to these 2 forces must be equal. In this problem, the pivot

point is the fulcrum!!!! Therefore,

110x = 88(18 – x) ------> 1584 – 88x Add 88x to both sides

198x = 1584 --------> x = 8 feet

The fulcrum must be placed 8 feet from the 110 lb force.

Ex. 2 Jim and Bob sit at opposite ends of an 18 foot seesaw, with

the fulcrum at its center. Jim weighs 160 lbs, and Bob weighs

200 lbs. Kim sits 4 feet in front of Jim, and the seesaw balances.

How much does Kim weigh?


Notice that since Jim and Bob both sit at the ends of the seesaw, and Bob weighs more, Bob would naturally win. Therefore, in order to balance the seesaw, Jim needs someone else on his side.

Also, notice that Jim and Bob are both sitting 9 feet away from the fulcrum. The problem does not state this explicitly but it can be ascertained from the wording of the problem!!!!!!

Let W = weight of Kim Now, the question arises: “How many feet is Kim sitting away from the fulcrum???” The answer is

9 – 4 = 5 feet. This is because she is 4 feet IN FRONT OF JIM.

Now, add the torques due to Jim and Kim and set it equal to the torque due to Bob.

160(9) + W(5) = 200(9) ------> 1440 + 5W = 1800 ----> 5W = 360

Kim weighs 72 lbs.


Section 7 has Three more kinds of Word problems. I’ll do several

example of these as well.

INVESTMENT PROBLEMS (Everyone’s favorite !!!!!)

Ex 1 A man invested $14,000, some at 7% and some at 10% annual

interest. The annual income from these investments was $1280.

How much did he invest at each rate?

Let x = amount invested at 7% --> .07

Then (14,000 – x) = amount invested at 10% ---> .10

Remember that the income from an investment is the annual

percentage rate (expressed in decimals) times the amount of the

investment. Therefore,

.07x + .10(14,000 – x) = 1280 Now distribute the .10


.07x + 1,400 - .10x = 1280 Now, combine the .07x and .10x!!

-.03x + 1,400 = 1,280 Subtract 1400 from both sides

-.03x = -120 Divide both sides by -.03

x = 4000 and 14,000 – 4,000 = 10,000

The man invested $4,000 at 7% and $10,000 at 10%.

Ex. 2 A bus driver wants to earn $3500 per year in supplemental

income from an inheritance of $40,000. If the driver invests

$10,000 in a mutual fund paying 8%, what rate must he earn on

the remainder to achieve his goal?

Solution: If he invests 10,000 at 8%, then the remainder will be

30,000, since it has to add up to 40,000.

Let x = rate of yield for the $30,000


Then .08(10,000) + x(30,000) = 3500 which is the total income

800 + 30,000x = 3500 -------> 30,000x = 2700 -----> x = .09

Now convert this to percentage and the answer is

The bus driver must invest the rest of the money at 9%.

UNIFORM MOTION PROBLEMS:

The main formula here is::: Distance = rate x time

Ex 1 A cyclist leaves Las Vegas riding at the rate of 18 mph. One hour later, a car leaves Las Vegas going 45 mph in the same direction. How long will it take the car to overtake the cyclist?

The Question is asking how long the car will take to reach the cyclist, So let t = time the car has been traveling

Then t + 1 = time the cyclist has been traveling, since she set out

1 hour before the car.


The distance traveled by cyclist = 18(t + 1)

The distance traveled by car = 45t Now, set them equal

18(t + 1) = 45t ----> 18t + 18 = 45 ---> 18t = 27 ---> t = 1.5

It will take the car 1.5 hours to catch up to the cyclist.

Ex. 2 Sarah walked north at the rate of 3mph and returned at the rate of 4mph. How many miles did she walk if the round trip took 3.5 hours?

Let t = time for the trip north

Then 3.5 – t = time for the return trip (since the total is 3.5 hours)

The distance traveled north = 3t

The distance for the return trip = 4(3.5 – t) since distance equals rate times time. These two distances must be equal because she ended up where she started, so set them equal to each other!!


3t = 4(3.5 – t) = 14 – 4t ---> 7t = 14 ---> t = 2 hours.

This means the trip north took 2 hours and the return trip took 1.5

hours. The problem wants to know the total round trip distance.

The distance north = 3t = 3(2) = 6 miles

The return distance = 4(3.5 – t) = 4(1.5) = 6 miles (Are you

surprised that it is the same??? You shouldn’t be!!)

Sarah walked a total distance of 12 miles!!!!!

MIXTURE PROBLEMS:

Ex. 1 A mixture of candy is made to sell 89 cents per pound. If 32

pounds of a cheaper candy, selling for 80 cents per pound, are

used along with 12 pounds of a more expensive candy, find the

price per pound of the better candy.


We start with 32 lbs of a candy selling at $.80/per pound, and mix with it

12 lbs of candy selling at x dollars per pound. We end up with 44 lbs

of candy, right??? And the problem tells us that the price of the

mixture = $.89/per pound. The idea here is that:

Price of cheaper candy + Price of better candy = Price of Mixture

.80(32) + x(12) = .89(44) ---> 25.6 + 12x = 39.16 --->12x = 13.56

The better candy costs $1.13/per pound.

Ex 2 How much acid must be added to 60 grams of a solution that is

65% acid to obtain a new solution that is 75% acid?

This is like a chemistry problem. The idea here is that:

Original amount of acid + Amount of added acid = Total amount of acid

Let x = amount of added acid.


The tricky part of this problem is to realize that when you add x

grams of acid to 60 grams of a solution, the total mass of the

mixture become ----> 60 + x!!!!

.65(60) + x = .75(60 + x) ---> 39 + x = 45 + .75x Subtract .75x from

both sides and 39 from both sides ---> .25x = 6 ---> x = 24

24 grams of acid must be added!!

Let me show you how to check this one.

There were 39 grams of acid in the original solution. If you add 24

grams of acid to it, you end up with 63 grams of acid. However,

the total solution will now have a mass of 60 + 24 = 84 grams.

If you have 63 grams of acid in 84 grams of solution, what percentage

is that??? 63/84 = .75 = 75% (Wow, it worked out!!!!)


Ex 3 A student has averaged 77% on 4 tests. What must she earn on the fifth test in order to average 80% in the tests?

This is a good problem to know for your courses!!!

Treat the 77% as points. In other words, the student has averaged 77 points out of 100 on the first 4 tests. That means her total points for the first 4 tests are:::: 4(77) = 308

Now let x = the number of points on the fifth test,

Then 308 + x = total number of points she earned.

The total possible number of points = 500 since there are a total of 5 tests each worth a 100 points.

She needs to get 80% so that is ---> .80(500) = 400 points.

308 + x = 400 ---> x = 92

She needs a 92% on her 5th Test!!!

Intermediate Algebra: Chapter 2 Section 1

By Viken Kiledjian Section 1 covers the topic of the Cartesian

or Rectangular Coordinate System.

Origin . Point P

Quadrant 2 Quadrant 1

x-axis

Quadrant 3 Quadrant 4

y-axis

Every point on this System is given by a

pair of points (x, y). The origin is the

point (0, 0). The point P is roughly

about (3, 5). The 3 is the x-coordinate

of point P and 5 is the y-coordinate.


Graphing Linear Equations: A linear equation is an equation that relates a variable y with another variable x, such that both of their powers are 1. Ex. y = 2x – 5 (notice that the powers of the variables y and x are both 1) To graph this equation one only needs two pairs of (x, y) points since a straight line can be constructed from just 2 points.

We can arbitrarily choose 2 values for x, and solve for the corresponding values of y.

If x = 0 -----> y = 2(0) – 5 = -5 -------> this yields the point (0, -5)

If x = 1 -----> y = 2(1) – 5 = -3 -------> this yields the point (1, -3)

Therefore, the graph will look something like:

(0, -5) means you don’t move to the right,

but you go 5 down.

(1, -3) means you move 1 to the right,

and you go 3 down.


The y-intercept of a line is the point where it crosses the y-axis.

In order to find it, set the value of x equal to zero, and solve for y.

The x-intercept of a line is the point where it crosses the x-axis.

In order to find it, set the value of y equal to zero, and solve for x.

In the previous example, y = 2x – 5,

If we set x = 0, then y = -5 ----> this means the y-intercept = -5

If we set y = 0, and solve for x ----> 0 = 2x – 5 ----> x-intercept = 5/2

This means that the line given by y = 2x – 5 will definitely go through these 2 points --------> (0, -5) and (5/2, 0)

Often we can graph a line by finding its x and y-intercepts!!!!!!

Graphing Lines Parallel to the X- and Y- Axis:

Graphing lines such as ----> y = 5, or x = -3 could sometimes seem strange because they are missing One of the variables.


This is because they are either parallel to the x or y-axis.

Ex. 1 The line y = 5 means that y is always equal to 5 no matter

what the value of x is. This means that you go up 5 and draw a

straight horizontal line parallel to the x-axis such as:

Ex. 2 The line x = -3 means that x is always equal to –3 no matter

what the value of y is. Therefore, go to the left 3 and draw a

straight vertical line parallel to the y-axis such as:


and Section 2 The Midpoint Formula: gives the midpoint between any two points

(x1, y1) and (x2, y2). What you do is basically take the averages of

the x’s and the y’s.

The midpoint is ------>

Ex Find the midpoint between the points (-3, 5) and (2, -7)

Find the average of the x’s -----> (-3 + 2)/2 = -1/2

Find the average of the y’s ------> (5 + -7)/2 = -2/2 = -1

The Midpoint is (-1/2, -1) or (-.5, -1)

Section 2 deals with the idea of Slope. The slope of a line measures

its steepness. You can compare the idea of Slope to the idea of

Grade on a road. Often when you drive on a road, there might be

a sign that says, “7% Downhill Grade: Trucks Use Low Gear”

2,

2

2121 yyxx


It is common to define slope as “Rise over Run” and use “m” as its symbol. The formal equation is:

m = change in y = Dy = y2 – y1

change in x Dx x2 – x1

Ex 1: Find the slope between the points (-2, 6) and (1, 4).

Solution: Let x1 = -2, then y1 = 6. Therefore, x2 = 1, y2 = 4

m = (4 – 6)/(1 - -2) = -2/3 (The slope is negative!!!)

You can also let x1 = 1 and then y1 = 4, x2 = -2, y2 = 6. So the slope,

m = (6 – 4)/(-2 – 1) = 2/(-3) = -2/3 (The answer is the same!!!)

Ex 2: Find the slope of the line in the first example of Section 2:

y = 2x – 5 (You get any two points on this line and apply the slope equation to it. Remember in Section 2, we already found 2 points? They were (0, -5) and (1, -3). Let x1 = 0, so


y1 = -5, x2 = 1, y2 = -3. Therefore, m = (-3 - -5)/(1 – 0) = 2/1

m = 2 (Notice that the slope is the same as the coefficient of the x

in the equation. We’ll come back to this in section 4)

Ex. 3 Find the slope of the equation 4x – 3y = 2 We have to first

find any two arbitrary points. Let x1 = 0 and x2 = 1 and solve for

the corresponding values of y:

4(0) – 3y = 2 ------> -3y = 2 ------> y = -2/3 this is y1!!!

4(1) – 3y = 2 ------> -3y = -2 ------> y = 2/3 this is y2!!! Therefore,

m = (2/3 - -2/3)/(1 – 0) = (4/3)/1 = 4/3

Graphical Interpretation of Slope: Lines which have Positive slope

are “rising” and look like:


Lines which have Negative slope are descending and look like:

Lines which have Zero slope are horizontal such as the line y = 5:

Vertical lines such as x = -3 have Infinite or Undefined slopes:


Besides helping us to draw a line, the slope of a line helps us to

ascertain if it is parallel or perpendicular to another line. Here is

the rule:

The slopes of parallel lines are equal. The slopes of

perpendicular lines are Negative Reciprocals of each other!!

Ex. 1 Tell whether the lines with the following slopes are parallel or

perpendicular to each other or neither.

m1 = 3, m2 = -1/3 Since m2 is the reciprocal of m1, and it is

negative of m1, then these 2 lines are perpendicular

Ex. 2 Tell whether the line PQ is parallel or perpendicular to a line

of slope –2.

P(6, 4) Q(8, 5) -----> m = (5 – 4)/(8 – 6) = ½ which is the

negative reciprocal of –2. Therefore, they are perpendicular!!


Section 3 covers the topic of the different ways of writing the Equations of Lines

The Point-Slope form of the equation of a line is written as:

y – y1 = m(x – x1)

The idea is that if you know the slope of a straight line and you know one point that it goes through, then you can write the general equation of the line.

Ex. 1 If the slope of a line = -2 and it goes through the point (1, -4), what is its general equation??

Solution: x1 = 1, y1 = -4, and m = -2 Therefore,

y – (-4) = -2(x – 1) Now, distribute the –2 into the parenthesis

y + 4 = -2x + 2 ---------> y = -2x – 2 is the answer!!

The Slope Intercept form of the equation of a line is written as:

y = mx + b where m is the slope and b is the y-intercept


The General Form of the equation of a line is written as:

Ax + By = C

(The answer to Example 1 ------> y = -2x – 2 is written in the slope-

intercept form. The slope = -2 and the y-intercept = -2. We could

also write the answer in the general form by adding 2x to both sides.

Then the answer would be 2x + y = -2)

Ex. 2 Find the equation of the line that passes through the two points

(-1, 4) and (3, -6). Express the answer in both Slope-Intercept form

and General Form

Solution: First find the slope m! Let x1 = -1, then y1 = 4, x2 = 3, y2 = -6

Therefore, m = (-6 – 4)/(3 – -1) = -10/4 = -5/2

Then use the point slope form, y – y1 = m(x – x1)

You can let the point (-1, 4) or the point (3, -6) represent the x1 , y1 !


Since we let the point (-1, 4) represent the x1 and y1 while we were computing the slope, I’ll stick to the same practice.

So x1 = -1 and y1 = 4 ---------> y – 4 = -5/2(x - -1) = -5/2(x + 1)

Now, distribute the –5/2 into the parenthesis, y – 4 = (-5/2)x – 5/2

And, add 4 to both sides -----> y = (-5/2)x + 3/2 This is the Slope-Intercept Form where the slope = -5/2 and the y-intercept = 3/2

To write the answer in General Form, we multiply everything by 2, so we can get rid of fractions ------> 2y = -5x + 3 and add 5x to both sides --------> 5x + 2y = 3

The advantage of the Slope-Intercept form is that it reveals the slope of the line, (whether it is inclined upward or downward), and the y-intercept, the point (0, b) that the line goes through.

The advantage of the General form is its aesthetic beauty and its neatness. It avoids the fractions!!


To illustrate the usefulness of the slope-intercept form, let’s see how

it can be used to quickly draw a rough sketch of a line.

Take for Example, the solution to the last example in slope-intercept

form: y = (-5/2)x – 3/2 Since the slope is negative, the graph

is inclined downward, and its y-intercept is –3/2, which means it

goes through the point (0, -3/2). The graph will look like:

Ex. 3 Find the slope and y-intercept of the line determined by the

given equation

5

)3(243

yx


First, multiply everything by 5 to get rid of the 5 on the bottom,

5(3x + 4) = -2(y – 3) Now, distribute the 5 and –2!!

15x + 20 = -2y + 6 --------> 15x + 14 = -2y Now, Divide by –2!!

y = (-15/2)x – 7 Therefore, m = -15/2 and b = -7

Ex. 4 Find the equations of the lines going through the point (2, -5) and parallel and perpendicular to the line given by the equation

3x – 2y = 5. Express the answer in General Form!!

Solution: First, let’s do the line parallel to 3x – 2y = 5

We have to first write this equation in Slope-Intercept form so we can know what its slope is.

3x – 2y = 5 ------>3x = 5 + 2y ------>3x – 5 = 2y ----> y = (3/2)x – 5/2

So, its slope is 3/2. A line which is parallel to it will also have a slope of 3/2. Since the line we are looking for also goes through the point (2, -5), we can use the point-slope form ---------------->


x1 = 2, y1 = -5, and m = 3/2 Therefore, y - - 5 = 3/2(x – 2)

Since, we want the final answer in the General Form, multiply both

sides by 2 to get rid of the fraction!!!

2(y + 5) = 3(x – 2) --------> 2y + 10 = 3x – 6 ------> 2y + 16 = 3x

3x – 2y = 16 Notice that the answer has the same coefficients for

x and y as the original line since it is parallel to the original line!!

Now, let’s do the line perpendicular to the original line!! The slope

of this line will be the Negative Reciprocal (remember section 3?)

of the Slope of the original line which is 3/2. Therefore, the slope

of the perpendicular line to this line will be –2/3.

So, the equation becomes y - - 5 = -2/3(x – 2) Multiply both sides

by 3 since we want the final answer in the General Form.

3(y + 5) = -2(x – 2) ------> 3y + 15 = -2x + 4 --------> 2x + 3y = -11


Section 4 deals with Functions. This is especially useful for Calculus!!

Function Definition: A function is a correspondence between the elements of one set (called the domain) and the elements of another set (called the range), where exactly ONE ELEMENT IN THE RANGE CORRESPONDS TO EACH ELEMENT IN THE DOMAIN.

Relation Definition: A relation is a correspondence between the elements of one set (called the domain), and the elements of another set (called the range), where ONE OR MORE ELEMENTS IN THE RANGE CORRESPONDS TO EACH ELEMENT IN THE DOMAIN.

NOTE: A function is ALWAYS a relation, but a relation is not always a function.

Function are often referred to as mappings of one variable (the domain) unto another variable (the range).


Ex. 1 Give the domain and range of each relation and tell whether

the relation is a function

Golfer Tournament Champion

Hale Irwin United States Open

Jack Nicklaus The Masters

Greg Norman The British Open

Solution: The Domain are the Golfers: Irwin, Nicklaus, Norman

The Range are the Championships they have won: the US Open, The

Masters, the British Open.

This Relation is NOT a function because one member in the Domain

is getting mapped into more than one member in the range.

It would be a function if Jack Nicklaus had ONLY won one of the 3

championships. Suppose, the situation was like this ------>


Golfer Tournament Champion

Hale Irwin United States Open

Jack Nicklaus The Masters

Greg Norman The British Open

Now, the Range would consist of Only 2 members: the US Open and

the British Open. But this relation WOULD BE A FUNCTION.

This is because 2 members in the Domain are getting mapped

into 1 member in the Range. This is acceptable for a function.

Ex. 2 Tell whether the following equations are functions:

a) y = 4x – 1 b) y2 = x + 1 c) y = x2 + 1 d) y = |x| e) x = |y|

Usually, “x” takes the role of the domain and the variable “y” takes

the role of the range. In this example, (a), (c ), and (d) are all

functions but (b) and (e) are not.


Let’s take each one separately,

a) y = 4x – 1 for each value of x, there is Only one value of y

b) y2 = x + 1 for each value of x, there are Two values of y. For Example, suppose x = 3. Then y2 = 4 and y = +2 or –2. One member in the domain (the x-value) cannot get mapped into 2 values in the range (the y-value); therefore, it’s not a function.

c) y = x2 + 1 for each value of x, there is Only 1 value of y. For example, suppose x = 3 again. Then y = 32 + 1 = 10. Therefore, it is a Function. (It is true that two different values of x will yield the same y, but that is Acceptable for a function. Remember the second case in Example 1??

d) y = |x| This is similar to example (c ). For each value of x, there is only 1 value of y. Therefore, it is a Function

e) x =|y| This is similar to example (b). Each value of x is getting mapped into 2 values of y. For example, if x = 3, then y can equal 3 or –3. Therefore, it is NOT A FUNCTION.


Function Notation: We usually write a function as:

y = f(x) which states that “y is a function of x”

Note: It does not mean that “y equals f times x”

In the notation y = f(x), x is the domain of the function f, and y is the range of the function f. Another way of saying it is: x is the independent variable of the function f, and y is the dependent variable because it depends on x via the function f.

Ex. 1 If f(x) = x2 – 3x, find a) f(-1) b) f(w) c) f(w) - f(2w)

a) f(-1) means “f of –1” NOT “f times –1”. Therefore, wherever you see an “x” in the equation, you substitute a “–1” for it!!

f(-1) = (-1)2 – 3(-1) = 1 + 3 = 4

b) f(w) = w2 – 3w You substitute a “w” in place of the “x”

c) f(w) – f(2w) = [w2 – 3w] – [(2w)2 – 3(2w)] Substitute 2w in the place of x and subtract this from f(w).

(w2 – 3w) – (4w2 – 6w) = w2 – 3w – 4w2 + 6w = -3w2 + 3w


Ex. 2 If g(x) = 3x – 1, find a) g(2) – g(-2) b) g(z + 3)

a) g(2) – g(-2) = [3(2) – 1] – [3(-2) – 1] = (6 – 1) – (-6 – 1)

= 5 - -7 = 12

b) g(z + 3) = 3(z + 3) – 1 You just substitute “z + 3” in the = 3z + 9 – 1 = 3z + 8 place of “x”

Ex. 3 Find the domain and range of each function

a) {(0,2) , (1,2) , (3,4)} b) f(x) = 5/(x + 1) c) f(x) = 5x –2

• The domain of the function given by these pair of points is the x-coordinate of the pair ------> 0, 1, 3

The range of this function is the unique values of the y-coordinate -----> 2, 4 (You don’t have to write the 2 twice!!)

b) The domain of this function is the set of allowable values of x. The only Non-Allowable value is the number x = -1 since it would make the denominator equal to 0 which is undefined. The domain is the set of all Real Numbers except –1.


The range of the function f(x) = 5/(x + 1) is harder to decide. It is

best if you know how to graph this function, but we won’t get into

that now. For now, let us suffice by this reasoning: In order for

a fraction to equal zero, the numerator has to equal zero, right?

Therefore, since the numerator of this function is equal to 5, this

function will never be equal to Zero, since its numerator can

never equal zero. Therefore, the range of this function is the set

of all real numbers except 0!!!!!

c) f(x) = 5x – 2 Since x can take on any values, there is no

restriction on the domain. There is nothing in the denominator

that would cause any disasters. Therefore, the domain is the set

of All Real Numbers. For the same reason, the values of the

range can take on any value also. The range is also the set of All

Real Numbers. This is actually a general result: All linear

functions have as their domain and range, the set of all Real #’s


The Vertical Line Test can be used on a graph of a relation to

determine if it is a function. If a vertical line can cross the graph

of a relation more than once, then it is NOT A FUNCTION.

Conversely, if a vertical line can ONLY CROSS IT ONCE, then

the graph represents a function.

Ex. 1 Determine if the following graph is a function. (I am omitting

the x- and y – axis from the graph because there will be too many

lines and it might get confusing)


Notice that ALL of the vertical lines cross the graph of this relation

ONLY ONCE. Therefore, this graph represents a function.

Ex. 2 Determine if the following graph is a function also.

Notice that all of

The Vertical Lines

Cross the graph at

Two places. Therefore,

This graph does not represent a function.

Ex. 3 Determine if the following graph is a function.

Since the vertical lines

Sometimes cross it once, and

Other times cross it 3 times,

it is NOT a function!!

Intermediate Algebra Chapter 3 Section 1

By Viken Kiledjian This chapter deals with solving systems of

equations. Later on in the chapter, we shall introduce the technique of Matrices and Determinants for solving equations.

Section 1 In this section, we will solve a system of 2 equations with 2 unknowns. I will illustrate the Substitution Method and the Addition Method and I’ll solve some Word Problems.

There are 2 special situations to be aware of:

An Inconsistent System results in an impossible situation and therefore there are NO SOLUTIONS FOR IT.

A System with Infinitely Many Solutions results when the 2 equations are a redundant repetition of each other.

Intermediate Algebra Chapter 3 Section 1 Ex. 1 Solve this system of 2 equation by the Substitution Method:

3x – 2y = -10 In this method, we have to Solve for one of

6x + 5y = 25 the variables and substitute it in the other

equation. It is usually easiest to solve for the variable that DOES

NOT HAVE A COEFFICIENT IN FRONT OF IT. In this

example, since all of the variables have a coefficient in front of it,

it does not matter which one we solve for or isolate. Let’s solve

for the x in the upper equation:

3x = 2y – 10 --------> x = (2y – 10)/3 Now, plug this in for the

6(2y – 10) + 5y = 25 x in the bottom equation!!!!

3 The 3 divides into the 6!!!

2(2y – 10) + 5y = 25 Distribute the 2 into the parenthesis!!

4y – 20 + 5y = 25 --------> 9y = 45 --------> y = 5

Now, plug this in the equation for the x: x = [2(5) – 10]/3 = 0

Intermediate Algebra Chapter 3 Section 1 Ex. 2 Solve the following 2 equations by the Addition Method

a) 2y – 3x = -13 1) First, rearrange the 17 and the 4y in the

b) 3x – 17 = 4y bottom equation to group the variables.

b) 3x – 4y = 17 2) Now, rewrite the top equation so that the

a) –3x + 2y = -13 x comes first and then the y.

3) Now, just add the equations and the x’s

-4y + 2y = 17 + (-13) will cancel, because 3x + (-3x) = 0

-2y = 4 ---------> y = -2 4) Now, plug this value of y into either

equation “a” or “b” to solve for x.

3x – 17 = 4(-2) = -8 I’ll choose equation “b” because the

3x = 17 – 8 = 9 coefficient of the x is positive in that

x = 3 equation!!


Ex. 3 Solve the following 2 equations by BOTH methods:

a)(2/3)x – (1/4)y = -8 First, let’s multiply the Top equation by 12

b)(1/2)x – (3/8)y = -9 and the Bottom by 8 to get rid of All Fractions

a) 12(2/3)x – 12(1/4)y = 12(-8) -------> 8x – 3y = -96 That’s better!!

b) 8(1/2)x – 8(3/8)y = 8(-9) --------> 4x – 3y = -72 Now we’re talking!!

Substitution Method: Solve for x from the top equation:

8x = 3y – 96 ------> x = (3y – 96)/8 = (3/8)y – 12 Since 96/8 = 12

Now, plug this value of x into the bottom equation:

4[(3/8)y – 12] – 3y = -72 --------> 4(3/8)y – 4(12) – 3y = -72 -------->

(3/2)y – 48 – 3y = -72 --------> (3/2)y – 3y = 48 – 72 --------->

(-3/2)y = -24 Multiply both sides by (-2/3) and y = (-2/3)(-24) = 16

Plug this value of y into the bold equation for x above:

x = (3/8)y – 12 = (3/8)(16) – 12 = 6 – 12 = -6

Intermediate Algebra Chapter 3 Section 1 I think the Addition method for this problem will be Faster!!!

a) 8x – 3y = -96 It looks like the “y” variable is the easiest one

b) 4x – 3y = -72 to get rid of because its coefficient is the same

in both equations. However, we have to multiply one of the

equations by (-1) so that the coefficients of the “y” will be

opposite of each other and the “y” will cancel when we add the 2

equations. So, let’s multiply the Top equation by (-1)!!

-8x + 3y = 96 I simply changed the sign of each member!!

4x – 3y = -72 Now, add the 2 equations, and the y’s cancel

-8x + 4x = 96 –72 --------> -4x = 24 --------> x = -6 Whalla!!

Now, plug this in to either equation and solve for y. Let’s plug it into

the changed equation above because the coefficient of the y is

positive: -8x + 3y = 96 -------> -8(-6) + 3y = 96 ------->

48 + 3y = 96 ----------> 3y = 48 ---------> y = 16 Great!!

Intermediate Algebra Chapter 3 Section 1 Ex. 4 Solve this 2 equations by the fastest and easiest method:

(3/x) - (2/y) = -30 First, let z = 1/x and w = 1/y so that you don’t

(2/x) – (3/y) = -30 have to deal with variables in the denominator

3z – 2w = -30

2z – 3w = -30 I think the best way to solve this is by the addition

method, but since none of the coefficients are the same, we have

to do some manipulations. Let’s decide to eliminate the w. If we

multiply the top equation by 3 and the bottom equation by (-2),

then the sign of the w’s will be exactly opposite and they’ll

cancel when we add the equations.

3(3z) – 3(2w) = 3(-30) ---------------> 9z – 6w = -90 Now, we can

(-2)(2z) – (-2)(3w) = -2(-30) --------> -4z + 6w = 60 add them!!

5z = -30 --------> z = -6 Therefore, x = -1/6 !!!!!

Intermediate Algebra Chapter 3 Section 1 Now, plug this value of z into either equation and solve for w:

Let’s do the top equation: 9(-6) – 6w = -90 -------> -54 – 6w = -90

-6w = -36 -------> w = 6 Therefore, y = 1/6 !!!!!!!

Now, let’s do some Word Problems utilizing these techniques:

Ex 1: A sporting goods salesperson sells 2 fishing reels and 5 rods

for $270. The next day, the salesperson sells 4 reels and 2 rods

for $220. How much does each cost?

Solution: Let “x” be the cost of each fishing reel and “y” the cost

of each rod. Then the 2 equations become:

2x + 5y = 270 It is best to use the addition method here and

4x + 2y = 220 eliminate the x variable. Multiply the top

equation by (-2) to make the coefficient of the x’s equal.

(-2)(2x) + (-2)(5y) = -2(270) -------------> -4x – 10y = -540

Now, add this to the bottom equation: -10y + 2y = -540 + 220

Intermediate Algebra Chapter 3 Section 1 -8y = -320 -------> y = 40 Now, plug this into either equation to

solve for “x”. Let’s plug it into the top equation:

2x + 5(40) = 270 -------> 2x + 200 = 270 ---------> x = 35

Therefore, the reels cost $35 each and the rods cost $40 each.

Ex. 2 In a right triangle, one acute angle is 15 degrees greater than

two times the other acute angle. Find the difference between the

angles.

Solution: Let x be the value of the greater acute angle and y the

value of the lesser acute angle. Since this is a right triangle, we

know that x and y have to add to 90 degrees since the triangle

already has another 90 degree angle. Here are our 2 equations:

x + y = 90 It is easiest to use the Substitution Method here.

x = 15 + 2y Plug the 2nd equation for x into the 1st equation for x

(15 + 2y) + y = 90 -------> 15 + 3y = 90 ---------> y = 25

Intermediate Algebra Chapter 3 Section 1 Now, plug this value of y into the bottom equation for x:

x = 15 + 2(25) = 75

The Small acute angle is 25 degrees and the Large is 75 degrees.

Ex. 3 A Hard Candy costs $2/lb and a Soft Candy costs $4/lb. How

many pounds of each must be mixed to obtain 60 pounds of candy

that is worth $3 per pound.

Solution: Let x be the pounds of Hard Candy and y the pounds of Soft

Candy. Here are the 2 equations:

x + y = 60 (Since there has to be a total of 60 pounds of candy)

2x + 4y = 3(60) (Since the cost of the total candy is $3 times 60 lbs)

Here we can either use the Substitution or Addition Method with the

same ease. Let’s use the Substitution method since the coefficient

of the x or y on the top equation is 1. Let’s solve for x in the top

equation: x = 60 – y Now, plug this into the bottom equation x:

Intermediate Algebra Chapter 3 Section 1 2(60 – y) + 4y = 180 -------> 120 – 2y + 4y = 180 --------> 2y = 60

y = 30 Now, plug this into the equation for x ----> x = 60 – 30 = 30

Therefore, we should mix 30 lbs of Soft and 30 lbs of Hard Candy.

This makes sense since the total mixture is going to cost the exact average. The average of $2/lb and $4/lb is $3/lb.

Ex. 4 Mixture A has 12% Protein and 9% Carbohydrates and Mixture B has 15% Protein and 5% Carbohydrates. A Farmer keeps some animals on a strict diet. Each animal is to receive 15 grams of protein and 7.5 grams of Carbohydrates. How many grams of each mixture should be used to provide the correct nutrients for each animal?

Solution: Let “x” be the number of grams of Mixture A and “y” the number of grams of Mixture B. Here are the 2 equations:

.12x + .15y = 15 The Protein amount of should add to 15 grams.

.09x + .05y = 7.5 The Carbohydrates should add up to 7.5 grams.


and Section 2 Now, multiply both equations by 100 to get rid of decimals:

12x + 15y = 1500 Reduce this equation by dividing it by 3!!

4x + 5y = 500 Multiply the top equation by (-1) and add this to

9x + 5y = 750 the bottom equation to eliminate the “y” variable

-4x – 5y = -500 Now, add them and you get:

5x = 250 --------> x = 50 Plug this into the top equation:

4(50) + 5y = 500 --------> 200 + 5y = 500 ---------> y = 60

The Farmer needs 50 grams of Mixture A and 60 grams of Mix B

SECTION 2 deals with 3 equations and 3 unknowns. There are again 2 kinds of Special situations to be aware of:

An Inconsistent System has No Solution. Usually it is caused by an impossible situation of 2 equations and 2 unknowns at the end.

For Example: 2x – 3y = 1 and 2x – 3y = 0 This is impossible.

Intermediate Algebra Chapter 3 Section 2 A System with Dependent Equations has Infinite solutions. It is

usually caused by a redundancy in the 3 equations when you add them. For Example, you might end up with something like:

3x + y = 4 and 3x + y = 4 This is always true !!!!!!!!

Ex. 1 Solve this system of 3 equations and 3 unknowns:

2x + 3y + z = 2 I’ll solve this problem using the Substitution

4x + 6y + 2z = 5 Method, and the next problem using the Addition

x – 2y + z = 3 Method. Solve for x from the bottom equation:

x = 3 + 2y – z Now, plug this into the x’s of the 1st and 2nd Eq’s:

2(3 + 2y – z) + 3y + z = 2 ----------> 6 + 4y – 2z + 3y + z = 2 ----->

6 + 7y – z = 2 ---------> 7y – z = -4 Now, do the same to the 2nd Eq:

4(3 + 2y – z) + 6y + 2z = 5 -------> 12 + 8y – 4z + 6y + 2z = 5 --->

12 + 14y – 2z = 5 --------> 14y – 2z = -7 We have transformed the 3 equations with 3 unknowns into a system of 2 equations and 2 unknowns. Let’s solve this by the Addition Method:

Intermediate Algebra Chapter 3 Section 2 Multiply the first equation by (-2) to make the coefficient of the “z”

equal to 2, so that it can cancel the “z” of the second equation:

(-2)7y – (-2)z = (-2)(-4) --------> -14y + 2z = 8 Now add this to the 2nd equation ---------> 0 = 8 –7 = 1 This is an impossibility

This is an inconsistent system and has No Solution!!

Ex 2: Solve this system of 3 equations and 3 unknowns:

x – (1/5)y – z = 9 Multiply everything by 5 to get rid of fractions

(1/4)x + (1/5)y – (1/2)z = 5 Multiply everything by 20 (the LCD)

2x + y + (1/6)z = 12 Multiply everything by 6

5x – y – 5z = 45

20(1/4)x + 20(1/5)y – 20(1/2)z = 20(5) -------> 5x + 4y – 10z = 100

12x + 6y + z = 72 I am going to solve this system using the Addition Method just like I promised in the last example. Let’s get rid of “z”. First, let’s multiply the 3rd equation by 5:

5(12x) + 5(6y) + 5(z) = 5(72) --------> 60x + 30y + 5z = 360

Intermediate Algebra Chapter 3 Section 2 Now, add this to the 1st equation to get rid of “z”:

5x + 60x – y + 30y = 45 + 360 ---------> 65x + 29y = 405

Now, let’s get rid of “z” from the 2nd and 3rd equations because the

sign of the “z” is ALREADY OPPOSITE in them. We just have

to multiply the 3rd equation by 10 to make the coefficient of the

“z” equal to 10:

10(12x) + 10(6y) + 10(z) = 10(72) -------> 120x + 60y + 10z = 720

Now, add them: 5x + 120x + 4y + 60y = 100 + 720 --------->

125x + 64y = 820 Now, let’s use the Substitution method to

solve for x from the top equation and plug it into the bottom x:

65x = 405 – 29y ---> x = (405 – 29y)/65 Plug this into the bottom x

125[(405 – 29y)/65] + 64y = 820 Multiply everything by 65!!

125(405 – 29y) + 4160y = 53,300 Distribute the 125!!

Intermediate Algebra Chapter 3 Section 2 50,625 – 3625y + 4160y = 53,300 ------> 535y = 2675 -----> y = 5!

Wow!! That was long and included lots of big numbers. Now,

plug this value of y into the top equation of the previous page:

65x + 29(5) = 405 ----------> 65x + 145 = 405 -----> x = 4 !!!!

Now, plug this into the 3rd original equation because the coefficient

of the “z” is 1 in there:

12(4) + 6(5) + z = 72 -------> 48 + 30 + z = 72 ------> z = -6!!!

Now, let’s do some Word Problems Involving 3 equations and 3

unknowns:

Ex 1: The sum of three integers is 48. If the first integer is doubled,

the sum is 60. If the second integer is doubled, the sum is 63.

Find the integers.

Solution: Let x be the first integer, y the second, and z the third.

Intermediate Algebra Chapter 3 Section 2 x + y + z = 48 It is easiest to use the Addition Method here,

2x + y + z = 60 and to eliminate the “z” variable. Just

x + 2y + z = 63 multiply the 1st and 2nd equations by (-1)!!

-x – y – z = -48 and -2x – y – z = -60 Add them to the 3rd eq:

2y – y = 63 – 48 -----> y = 15 and x – 2x + 2y – y = 63 – 60 ---->

-x + y = 3 -------> x = y – 3 Therefore, x = 12 Therefore,

15 + 12 + z = 48 -------> z = 21

Ex. 2: A factory manufactures three types of footballs at a monthly cost of $2425 for 1125 footballs. The manufacturing costs for the three types of footballs are $4, $3, and $2. These footballs sell for $16, $12, and $10, respectively. How many of each type are manufactured if the monthly profit is $9275?

Solution: Let x, y, and z, represent the number of 3 kinds of footballs

x + y + z = 1125 Since the total number of balls is 1125

Intermediate Algebra Chapter 3 Section 2 4x + 3y + 2z = 2425 Since the cost of each is $4, $3, and $2

(16 – 4)x + (12 – 3)y + (10 – 2)z = 9275 Since the profit is the

12x + 9y + 8z = 9275 difference between the selling price and the cost

For this problem, the easiest method would probably be the Substitution method. Let’s solve for “z” in the first equation and substitute that expression in the 2nd and 3rd equations:

z = 1125 – x – y Therefore, 4x + 3y + 2(1125 – x – y) = 2425 --->

4x + 3y + 2250 – 2x – 2y = 2425 ------> 2x + y = 175 and

12x + 9y + 8(1125 – x – y) = 927 --->12x + 9y + 9000 – 8x – 8y = 9275

4x + y = 275 Now, multiply the Top bold equation by (-1) and add to the Bottom bold equation and you get: -2x + 4x = -175 + 275 ------>

2x = 100 ------> x = 50 footballs Plug this into the Top Bold equation:

2(50) + y = 175 -------> y = 75 footballs Plug these into the expression for z and you get: z = 1125 – 50 – 75 = 1000 footballs.

Intermediate Algebra Chapter 3 Section 3 Section 3 is about Matrices. Matrices can be used to solve a system

of 2 equations or 3 equations. In Section 5, we shall be

introduced to yet another tool, called Determinants, to help solve

equations. If you go on further in Math, you might take a class

called Linear Algebra which covers the topic of Matrices and

Determinants to much greater detail.

A Matrix is any rectangular array of numbers. The numbers in a

matrix are called its elements. Here are some examples:

A is a 2 x 2 Matrix (read “2 by 2” Matrix) since

it has 2 rows and 2 columns. Hence, it is also called a Square

Matrix. B is a 2 x 4 Matrix since it has 2 rows and 4 columns

and B is a 3 x 1 Matrix since it has 3 rows and only 1 column.

64

32A

2431

3012B

1

0

3

C

Intermediate Algebra Chapter 3 Section 3 We can use the following Row Operations to solve equations:

1) Any two rows of a matrix can be interchanged.

2) Any row of a matrix can be multiplied by a nonzero constant.

3) Any row of a matrix can be changed by adding a nonzero constant multiple of another row to it.

(Basically, these Row Operations are akin to how we were solving equations using the Addition Method. Sometimes, we would multiply an equation by some constant in order to add it to another equation to cancel a certain variable.)

Let’s solve two of the examples from Section 1 and another example from Section 2 using the technique of Matrices:

Here is Example 2 from Section 1 after we regroup the variables:

-3x + 2y = -13 In Section 2, we used the Addition method for this

3x – 4y = 17 problem and our answer was: x = 3, y = -2

Intermediate Algebra Chapter 3 Section 3 Here is how we would do it with Matrices: We first write the 2 x 2

Matrix containing the coefficients of the variables x and y.

This is called the Coefficient Matrix. Now we

add the numbers to the right side of the equal sign

on to this Matrix and it becomes an Augmented Matrix.

This Augmented Matrix looks like the 2 equations

but without the variables. Now, we do a series of

Row Operations in order to express this matrix in a Triangular

Form. A Triangular form is a Matrix that has Zeros below its

main Diagonal. Here is an example of a Triangular Matrix and

its main diagonal. In the top Matrix, if we

add Row 1 to Row 2, the (-3) will cancel the

(3) and we will achieve the Triangular Form:

43

23

1743

1323

12400

3120

2132

Intermediate Algebra Chapter 3 Section 3 Here is what we would end up with, if we add ALL ELEMENTS

OF ROW 1 TO ROW 2 while still keeping ROW 1 THE SAME.

because (2) + (-4) = -2 and (-13) + (17) = 4.

Now, we divide ALL ELEMENTS of Row 2 by (-2) and achieve the final Triangular Form:

This is now equivalent to the following 2

equations:

-3x + 2y = -13

y = -2 Now, just plug the value of y into the top equation and solve for x ------> -3x + 2(-2) = -13 ------> -3x = -9

Therefore, y = -2 and x = 3 which was our solution in Section 2

Now let’s do Example 3 of Section 1: Here are the 2 equations:

8x – 3y = -96 In Section 1, we got x = -6 and y = 16 for the answer

4x – 3y = -72

420

1323

210

1323

Intermediate Algebra Chapter 3 Section 3 Here is the Augmented Matrix for this problem:

Now, multiply Row 1 by (-1/2) and add the

result to Row 2. What will happen is this:

8(-1/2) + 4 = 0 (Which is what we want!!!)

-3(-1/2) + (-3) = -3/2 and -96(-1/2) + (-72) = -24 Therefore,

the new Triangular Form Matrix is:

Notice that we have kept Row 1 intact. Now,

multiply Row 2 by (-2/3) and you’ll get:

This is equivalent to the two equations:

8x – 3y = -96 and y = 16

Now, plug this value of y into the equation and solve for x:

8x – 3(16) = -96 ---------> 8x – 48 = -96 --------> 8x = -48

Therefore, x = -6 and y = 16 which was our solution in Section 1

7234

9638

242/30

9638

1610

9638

Intermediate Algebra Chapter 3 Section 3 Now, let’s do one hard example from Section 2. Here are the 3

equations from Example 2 of Section 2 in their final form:

5x – y – 5z = 45 We obtained a solution of x = 4, y = 5, z = -6.

5x + 4y – 10z = 100 First write the Augmented Matrix consisting

12x + 6y + z = 72 of the coefficients and the right sides:

1) First, let’s eliminate the 5 in the 2nd row,

and 1st Column. Multiply the 1st row by (-1)

and Add the result to the 2nd row. You get:

5(-1) + 5 = 0 Which is what we want!!

-1(-1) + 4 = 5 and -5(-1) + (-10) = -5 and 45(-1) + 100 = 55

2) Second, let’s eliminate the 12 in the 3rd row

and 1st Column. Multiply the 1st row by (-12/5)

and Add the result to the 3rd row. You get:

721612

1001045

45515

721612

55550

45515

Intermediate Algebra Chapter 3 Section 3 5(-12/5) + 12 = 0 Which is what we want!!!

-1(-12/5) + 6 = 12/5 + 6 = 42/5 and -5(-12/5) + 1 = 13

and 45(-12/5) + 72 = -108 + 72 = - 36

3) Now, multiply every element in the

3rd row by 5 to get rid of fractions.

The new Matrix becomes:

4) Now, divide every element in the

2nd row by 5 to make the “First

Number” of the row a 1.

5) Now, let’s eliminate the 42 in the

3rd row and 2nd Column. Multiply

every element of Row 2 by (-42)

and add the result to Row 3. You get:

36135/420

55550

45515

18065420

55550

45515

18065420

11110

45515

Intermediate Algebra Chapter 3 Section 3 1(-42) + 42 = 0 Which is what you want!!

-1(-42) + 65 = 107 and 11(-42) + (-180) = -462 - 180 = -642

6) Lastly, Divide Row 3 by 107.

The Final Matrix becomes:

This is equivalent to the result:

z = -6. Now, plug this into the

equation represented by the 2nd row:

y – z = 11 ------> y – (-6) = 11 --------> y + 6 = 11 -----> y = 5

Finally, plug these results for y and z into the equation represented

by the 1st row: 5x – y – 5z = 45 --------> 5x – 5 – 5(-6) = 45 --

---> 5x – 5 + 30 = 45 -----------> 5x + 25 = 45 -------> x = 4

Therefore, the answer is the same as what we got before:

64210700

11110

45515

6100

11110

45515

Intermediate Algebra Chapter 3 Section 4 Section 4 introduces yet another approach to solving equations.

This is my personal favorite. I think it is quicker and easier than

any of the other ones. It is called Determinants.

A Determinant is a number that is associated with any square

matrix. If A is the following general square matrix:

Then, its Determinant, written as |A|, is given by:

|A| = ad – bc (Don’t confuse the way a Determinant

is written with the Absolute Value sign. You can’t take the

absolute value of a Matrix, but you can take its Determinant)

Notice that you basically Cross Multiply the numbers and subtract

them. For example, if A is the following specific matrix:

Then, |A| = 2(4) – 3(-1) = 8 + 3 = 11. Now, let’s

introduce a theorem which helps us solve equations.

dc

ba

43

12

Intermediate Algebra Chapter 3 Section 4 Cramer’s Rule for Two Equations in Two Variables:

The solution of the system of 2 equations: ax + by = e

is x = Dx/D and y = Dy/D cx + dy = f

where D is the Determinant of the matrix formed by the

coefficients of the variables. ----------> D =

Dx is the Determinant of the matrix formed by

replacing the coefficient of the “x” variable by the numbers to the

right side of the equal sign. ---------> Dx =

Dy is the Determinant of the matrix formed by

replacing the coefficient of the “y” variable by the numbers to

the right side of the equal sign. ---------> Dy =

Now, let’s solve the same two examples from

Section 1 with this method and see if we get the same answer:

dc

ba

df

be

fc

ea

Intermediate Algebra Chapter 3 Section 4 Here is Example 2 from Section 1:

-3x + 2y = -13 Our answer was x = 3 and y = -2.

3x – 4y = 17 Let’s calculate D, Dx , Dy :

D = = -3(-4) – 3(2) = 12 – 6 = 6!!

Dx = = -13(-4) – 17(2) = 52 – 34 = 18!!

Dy = = -3(17) – 3(-13) = -51 + 39 = -12!!!

Now, use Cramer’s rule: x = Dx/D = 18/6 = 3 Amazing, right?

y = Dy/D = -12/6 = -2 See how easy that was?

Here is Example 3 from Section 1 in its final rearranged form:

8x – 3y = -96 and 4x – 3y = -72 We got x = -6, y = 16

43

23

417

213

173

133

Intermediate Algebra Chapter 3 Section 4 First Calculate D, Dx , and Dy again.

D = = 8(-3) – 4(-3) = -24 + 12 = -12

Dx = = -96(-3) – (-72)(-3) = 288 – 216 = 72

Dy = = 8(-72) – 4(-96) = -576 + 384 = -192

Now, use Cramer’s rule again. x = Dx/D = 72/(-12) = -6 and

y = Dy/D = -192/(-12) = 16 That’s quick and easy, wow!!!!!

Now, let’s go on to 3 Equations and 3 Unknowns. Of course, these

are tougher but still it is faster than the other methods:

First, let’s learn how to take determinants of 3 by 3 matrices:

34

38

372

396

724

968

Intermediate Algebra Chapter 3 Section 4 The way to take the Determinant of a 3 x 3 Matrix is to expand it into

Three “2 x 2 Determinants” along any row or any column.

Usually, it is customary to expand along the 1st row. This technique is called Expanding by Minors. Here is how it works:

Notice a couple of things:

1) The signs are alternating. First, it’s “a” times the determinant of its minor then it is MINUS “b” times the determinant of its minor, and then finally it is “c” times the determinant of its minor. So it is ---------> + - +

2) Next, notice that the minor of “a” is formed by crossing out the column and row that “a” is located in and writing the remaining 4 numbers. The same goes with the minor of “b” and the minor of “c”. This is illustrated in the next slide----------------->

hg

edc

ig

fdb

ih

fea

ihg

fed

cba

Intermediate Algebra Chapter 3 Section 4 Here is how to form the Minor of “a”:

Notice that the remaining 4 numbers are

e , f, h, i so you take the Determinant of

this 2 x 2 Matrix just like the formula in the previous slide shows.

Here is how to form the Minor of “b”:

Notice that the remaining 4 numbers are

d , f, g, i so you take the Determinant of

this 2 x 2 Matrix just like the formula in the previous slide.

Finally, here is how you form the Minor of “c”:

Now, you take the Determinant of the

2 x 2 Matrix formed by “d”,”e”, “g”, “h”.

Let’s do an Example of a 3 x 3 Determinant with numbers.

ihg

fed

cba

ihg

fed

cba

ihg

fed

cba

Intermediate Algebra Chapter 3 Section 4 Solve this 3 x 3 Determinant:

I am going to expand this along

the 1st row first, then I’ll show

you a faster way.

= 2[(3)(2) – (4)(1)] – 3[(0)(2) – (-2)(1)] – 1[(0)(4) – (-2)(3)]

= 2(6 – 4) – 3(0 + 2) – 1(0 + 6) = 2(2) – 3(2) – 1(6) = 4 – 6 – 6 = -8

The Faster Way is to expand along the 1st column because there

is a “0” in that column.

242

130

132

42

30)1(

22

103

24

132

242

130

132

13

13)2(

24

130

24

132

242

130

132

Intermediate Algebra Chapter 3 Section 4 Notice that I alternated the signs again: + going down the column.

Here is how you form the minors for -

the numbers “0” and “-2”. +

and I don’t have to do the 2nd minor

because 0 times anything is 0:

= 2[(3)(2) – (4)(1)] – 2[(3)(1) – (3)(-1)] = 2(6 – 4) – 2(3 + 3) ------>

= 2(2) – 2(6) = 4 – 12 = -8 Wow, we got the same answer!!!!!!

The only time you should expand along another row or column

besides the 1st row is if there is a “0” in that row or column.

Now, let’s move into solving 3 Equations with 3 unknowns.

Cramer’s rule for 3 equations is similar to his rule for 2 equations.

x = Dx/D y = Dy/D z = Dz/D

242

130

132

242

130

132

Intermediate Algebra Chapter 3 Section 4 Let’s go back to our famous example from Section 3 and solve it using

Determinants this time: The problem was:

5x – y – 5z = 45 We obtained a solution of x = 4, y = 5, z = -6

5x + 4y – 10z = 100 using the Addition Method and the Matrix

12x + 6y + z = 72 Method, but it took a lot of work.

Here are the Determinants D, Dx , Dy , Dz

D =

= 5[(4)(1) – (6)(-10)] + 1[(5)(1) – (12)(-10)] – 5[(5)(6) – (12)(4)]

= 5(4 + 60) + 1(5 + 120) – 5(30 – 48) = 5(64) + 1(125) – 5(-18) = 535

Dx =

612

45)5(

112

105)1(

16

1045

1612

1045

515

672

4100)5(

172

10100)1(

16

10445

1672

104100

5145

Intermediate Algebra Chapter 3 Section 4 Notice that Dx is obtained by replacing the 1st column (which

contains the coefficients of the x variable) by the numbers on the

right side of the equal sign. The same goes with Dy and Dz

Dx = 45[(4)(1) – (6)(-10)] + 1[(100)(1) – (72)(-10)] – 5[(100)(6) –

(72)(4)] = 45(64) +1(820) – 5(312) = 2140

At this point, we can check to see if we are doing it correctly:

x = Dx/D = 2140/535 = 4 This is Correct, so let’s proceed!!!!!

Dy =

= 5[(100)(1) – (72)(-10)] – 45[(5)(1) – (12)(-10)] – 5[(5)(72) –

(12)(100)] = 5(820) – 45(125) – 5(-840) = 2675

y = Dy/D = 2675/535 = 5 This is also Correct, so let’s proceed!!

7212

1005)5(

112

10545

172

101005

17212

101005

5455


Dz =

= 5[(4)(72) – (6)(100)] + 1[(5)(72) – (12)(100)] + 45[(5)(6) – (12)(4)] = 5(-312) + 1(-840) + 45(-18) = -3210

Therefore, z = Dz/D = -3210/535 = -6 Which is also correct!!

This example of course is long and laborious but it serves to show you how we can solve even a hard problem with all 3 methods and they yield the same solution. When you encounter a 2 equation or 3 equation problem, it is totally up to you how you want to solve it. Use the method you are most comfortable with.

That’s the End!!

612

45)45(

7212

1005)1(

726

10045

72612

10045

4515


By Viken Kiledjian

Section 1: covers the topic of solving

linear inequalities. It is pretty much the

same as solving equations with ONE

exception. When you divide/multiply

by a negative number, you must

change the sign of the inequality.

Here are the different kinds of Inequality:

< means “less than”

> means “greater than”

means “less than or equal to”

means “greater than or equal to”

Ex 1: Solve for x and give the result in

interval notation:

-2x + 6 16


First, subtract 6 from both sides:

-2x 10 and now divide both sides by –2 and CHANGE THE

x -5 INEQUALITY SIGN!!

The answer in interval notation is

Ex 2: -9(h – 3) + 2h < 8(4 – h) First, Distribute the –9 and 8!!

-9h + 27 + 2h < 32 – 8h ---------> -7h + 27 < 32 – 8h

Now, add 8h to both sides and subtract 27 from both sides!!!

h < 5 or you can write the answer as

Remember, the bracket sign means you include the number and the

parenthesis means you don’t include the number.

COMPOUNT INEQUALITIES are a shorthand way of writing 2

inequalities at once. For example, if x < 3 AND x > 0, we could

write this combined inequality as 0 < x < 3. In interval notation,

this is equivalent to writing (0 , 3)

]5,(

)5,(


However, you can’t use Compound Inequalities to express OR

statements. For example, if x < 0 OR x > 3, these are mutually

exclusive intervals. You can’t say x < 0 AND x > 3. Therefore,

it would be wrong to write this inequality as 0 > x > 3. The best

way to write it is: x < 0 OR x > 3. In interval notation, you can

express it as U It is the union of the 2 sets of

numbers from minus infinity to Zero and 3 to positive infinity.

Ex 3: Solve this compound inequality and give the result in interval

notation. 2 < -t – 2 < 9

Solution: The above compound inequality is equivalent to:

-t – 2 > 2 AND -t –2 < 9 Solve each one separately

-t > 4 AND -t < 11 Multiply both sides by (-1) and change signs

t < -4 AND t > -11 The solution can be written in compound

inequality form or interval form -----> -11 < t < -4 or (-11 , -4)

)0,( ),3(


Ex 4: Solve this compound inequality: x – 1 2x + 4 < 3x – 1

Again, express this as 2 inequalities and solve them separately

2x + 4 x –1 AND 2x + 4 < 3x –1

x -5 AND -x < -5 -----> x > 5 Notice that I changed the sign

Now, this one is kind of tricky. What is the intersection of these

two sets?? (Review chapter 1 if you don’t remember how to do

this). The intersection is --------> x > 5 or

Ex 5: Solve this inequality: 3x + 4 < -2 OR 3x + 4 > 10

Solution: Solve each one separately as before:

3x < -6 OR 3x > 6 Therefore, the answer is:

x < -2 OR x > 2 ------------>

If this problem had been: 3x + 4 < -2 AND 3x + 4 > 10, The

answer would have been x < -2 AND x > 2 which is

IMPOSSIBLE. The answer would have been the NULL SET.

),5(

),2()2,(


Let’s do some Word Problems Involving Inequalities:

Ex 1: The cost of renting a truck is $29.95 for the first hour and

$8.95 for each additional hour. How long can a person have the

truck if the cost is to be less than $110? (Part of an hour is

considered a Full hour)

Solution: Let x = the number of additional hours beyond the 1st hour

The total cost of renting must be less than $110.Here is the Equation

29.95 + 8.95x < 110 -----------> 8.95x < 80.05 ------> x < 8.94 hrs

Since, part of an Hour is considered a Full hour, if the person drives

for 8.94 hours, beyond the original 1st hour, he/she will be

charged for 9 Additional hours. Therefore, they should only drive

it for 8 Additional hours!!!!! Adding 1 hour to this yields 9 hours.

The Total Number of Hours must be less than 9 hours


Ex 2: A student who can afford to spend up to $2000 sees the following advertisement: Big Sale on Computer for $1695.95 and All CD-ROMs for $19.95. If she buys the computer, find the greatest number of CD-ROMs that she can buy.

Solution: let x = the number of CD-ROMs, then the total cost is:

1695.95 + 19.95x 2000 (Notice that the wording “up to” is NOT the same as the wording “less than”. “Up to” includes the number whereas “less than” does NOT include the number.)

19.95x 304.05 ----------> x 15.24 (However, you can’t buy fractional CDs, so you have to round this number down)

The student can buy a maximum of 15 CD-ROMs.

Ex 3: An excavating company charges $300 an hour for the use of a backhoe and $500 an hour for the use of a bulldozer. The company employs one operator for 40 hours per week. If the company wants to take in at least $18,500 each week, how many hours per week can it schedule the operator to use the backhoe?


In this problem, Part of an hour counts as a full hour.

Solution: Let x = the number of hours for the use of the backhoe,

then 40 –x = the number of hours for the use of the bulldozer

(Since the total # of hours must be 40 hrs) Therefore, the total cost is

300x + 500(40 – x) 18,500 (Note that the phrase “at least”

translates into a “greater or equal” sign.) Now, Distribute the 500!

300x + 20,000 – 500x 18,500 ------> -200x + 20,000 18,500

--------> -200x -1500 (Now divide by –200 and change the sign)

x 7.5 (Now, round this down to the nearest integer, since part

of an hour is considered an hour)

The backhoe can only be used for 7 hours Maximum. The

bulldozer should get 33 hours Minimum.


Section 2 is about Absolute Values.

Definition of Absolute Value: The absolute value of a number

measures its distance to the origin of the number line which is the

number Zero. Its symbol is the | | two vertical lines.

Ex. |3| = 3 and |-2| = 2 (Since the number 3 is 3 units away from

the origin and the number –2 is 2 units away from the origin)

Here is the Formal Algebraic Definition of the Absolute Value:

If x 0, then |x| = x (You simply keep the same number)

If x < 0, then |x| = -x (You take the negative of the number)

Graphing Absolute Value Functions The absolute value function

looks like this --------> y = f(x) = c|x – a| + b

Here is the Graph of y = |x|

It looks like a “V” whose

Vertex is the origin.


Now, let’s learn how to graph a general absolute value function:

y = c|x – a| + b The “a” and the “b” shift the graph in the

following manner

If a > 0, the graph is shifted to the right by an amount equal to “a”

If a < 0, the graph is shifted to the left by an amount equal to “a”

If b > 0, the graph is shifted up by an amount equal to “b”

If b < 0, the graph is shifted down by an amount equal to “b”

If c > 0, the graph opens upward

If c < 0, the graph opens downward

Ex 1: Graph the following function: y = |x – 2| + 3

Shift the Vertex 2 to

the right and 3 up


Ex. 2 Graph the following function: y = -|x + 2|

Shift the Vertex 2 to the

Left and invert the graph

So that it opens Downward.

Now, let’s move into

Absolute Value Equations:

If k > 0, then |x| = k is equivalent to x = k or x = -k

Ex 1: Solve this equation for x: |x| = 3

Solution: x = 3 OR x = -3 (Because the absolute value of 3 and the absolute value of –3 are both equal to 3!!!)

Ex 2: Solve this equation for x: |x + 4| = 8

Solution: x + 4 = 8 OR x + 4 = -8 (Solve these separately)

x = 4 OR x = -12 (You can check these results by putting them

into the original equation) |4 + 4| = |8| = 8 AND |-12 + 4| = |-8| = 8


Ex 3: Solve the following equation for x: |2 – x| + 3 = 5

Solution: First, Subtract 3 from both sides in order to isolate the

Absolute Value --------> |2 – x| = 2 Now, split this up!!

2 – x = 2 OR 2 – x = -2 (Subtract 2 from both sides!!)

-x = 0 OR -x = -4 (Multiply both sides by (-1)!!!)

x = 0 OR x = 4 (Check these to make sure it is correct)

Equations with Two Absolute Values: If a and b represent some

algebraic expressions, then the equation |a| = |b| is equivalent to

the pair of equations --------> a = b OR a = -b

So, in other words, You solve it EXACTLY the same way as a

Single Absolute Value Equation!!!!!!!!!!!!!! Here is one example

Ex: |7x + 12| = |x – 6| Split these up and solve them separately

7x + 12 = x – 6 OR 7x + 12 = -(x – 6) = -x + 6

6x = -18 -------> x = -3 OR 8x = -6 ---------> x = -6/8 = -3/4


Absolute Inequalities of the Form |x| < k:

If k > 0, then

|x| < k is equivalent to -k < x < k (x < k AND x > -k)

|x| k is equivalent to –k x k (Same here but with equal sign)

(Note: If k < 0, then there is NO SOLUTION to the above equation,

since Absolute Value of a number can never be negative and

hence can NEVER be less than a negative number!!!!!)

Ex: Solve for x and express the answer in interval notation also.

|(1/2)x – 3| - 4 < 2 (Add 4 to both sides to isolate the absolute value)

|(1/2)x – 3| < 6 (Now split this into two inequalities according to the

above theorem and solve separately)

(1/2)x – 3 < 6 AND (1/2)x – 3 > -6 (Add 3 to both sides!!)

(1/2)x < 9 ---------> x < 18 AND (1/2)x > -3 --------> x > -6

Therefore, the solution is ------> -6 < x < 18 or (-6 , 18)


Absolute Inequalities of the Form |x| > k:

If k 0, then

|x| > k is equivalent to x < -k OR x > k

|x| k is equivalent to x -k OR x k

(Note that if k < 0, then you CHANGE it to zero and solve the

equation like that. For example: If |x – 2| > -4, then change the

-4 into 0 and now solve the equation as |x – 2| 0. This is

because Absolute Values are at least greater than or equal to Zero)

Ex: Solve for x in the following equation.

|-1 –2x| > 5 (Split this up into two inequalities and solve separately)

-1 – 2x < -5 OR -1 –2x > 5 (Add 1 to both sides)

-2x < -4 OR -2x > 6 (Divide by –2 and change the signs)

x > 2 OR x < -3 ---------->

),2()3,(


Section 3 is about Graphing Linear Inequalities. This section is

pretty short and easy!!

Graphing Linear Inequalities: Let’s do a few examples.

Ex 1: Graph the inequality y < 2x –1

Solution: First, you graph this line on the x-y axis just like in

chapter 2. Since there is NO equal sign in this problem, you

make the graph a DOTTED LINE.

Now, the next thing you do is to see

which region you have to shade ----->

the one above or the one below this line??

The easiest way to do this is to test the point

(0,0) in the original equation ------> 0 < 2(0) – 1 ??? This leads to

0 < -1 which is Not True. Therefore, you have the shade the

region BELOW THE LINE since(0,0) did not satisfy the inequality.


Ex 2: Graph the Inequality 2x -3y - 12

Solution: Add 3y to both sides and you get -----> 2x + 3y -12

Now, graph this equation as in chapter 2. However, this time make

the line SOLID because the inequality has AN EQUAL SIGN

included in it.

Again, check the point (0,0)

in the inequality ------------>

2(0) + 3(0) -12 ????

0 < -12 ?? Since this is Not True,

you must shade the region below this line.

Graphing Compound Inequalities: Let’s do a few examples of

these. Remember that compound inequalities can be of two types.

The AND type or the OR type. So, we’ll do 1 example of each.


Ex 1: Graph the compound inequality –3 < y -1

Solution: This is the type AND. It is equivalent to the two

inequalities y > -3 AND y -1 . So graph each line

separately as in chapter 2.

Since the first inequality y > -3

does NOT include an equal sign,

make this line DASHED. Now,

shade the region BETWEEN these.

You don’t have to do ANY TESTING

for this, since it is an AND Problem.

You know that the solution MUST BE between these two graphs!!!!

Ex 2: Graph this compound inequality x < -3 OR x > 2

Solution: This is obviously of the type OR. Graph both lines first.

Intermediate Algebra Chapter 4 Section 3 & 4

In this problem, both of the

inequalities don’t have an

equal sign. Therefore, both

lines must be Dashed.

Since x < -3, shade to the

Left of the left line and

since x > 2, shade to the Right of the right line!!!!!!

SECTION 4: covers the topic of Systems of Inequalities. This is

very much related to the previous section’s compound inequalities

topic. When there are more than 1 linear inequality, you shade

the region of the INTERSECTION OF THEIR SOLUTIONS.

Ex 1: y x - 2 AND y 2x + 1

Solution: Both of these lines are going to be SOLID.


Draw each line separately and shade its solution. The Final

Solution is the Region where the two shaded regions overlap.

For the 1st equation, we

shade the region

below the line,

and for the 2nd

equation, we

shade the region

above. It is also

good to calculate

their intersection point to know exactly how to draw the graph.

Set their y’s equal to each other --------> x – 2 = 2x + 1 -------->

-x = 3 ---------> x = -3 Therefore, y = x – 2 = -3 –2 = -5


Ex 2: x + 2y < 3 AND 2x + 4y < 8

Solution: First, divide the 2nd equation by (2) to reduce it. We end

up with -------> x + 2y < 4 Notice that, the left part of this

equation is the same as the 1st equation. These are two parallel

equations because they have the same slope (remember chapter 2?)

Therefore, they are NOT going to intersect. Draw each dashed line

and shade their respective regions. The final solution will be the

cross-hatched region where the 2 regions overlap.

Shade the lower

region for

each graph

because the

point (0,0)

satisfies both graphs.


Ex 3: 2x + y 2 AND y x AND x 0

Solution: Now, we have 3 equations. We are going to end up with 3

solid lines and 3 regions to shade. The final answer will the

intersection of ALL 3 regions. The first 2 lines will intersect at

the point ------> 2x + x = 2 ------> x = 2/3 and therefore, y = 2/3

The first and the 3rd line will intersect at x = 0 and therefore, y = 2

The 2nd and the 3rd line will intersect at x = 0 and therefore, y = 0

For the 1st equation, we

must shade the region

below the line. For the

2nd equation, shade the

region above the line,

and for the 3rd, shade the

region to the right of the vertical line.


By Viken Kiledjian Section 1: In this section, we define what a

Polynomial is and how to find its power.

A Monomial is an expression such as

Axbyczd such that “b”, “c”, and “d” are All positive integers or zero. They can NOT be fractions or zero. The Degree of the Monomial is the sum of these integers ---------> b + c + d

Ex 1: 5x2y3z is a Monomial of Degree =

2 + 3 + 1 = 6

Ex 2: -2x1/2y3z2 is Not a Monomial because the power of the x is a fraction

A Polynomial is the sum of 2 or more Monomials. A Polynomial with 2 ----->


Monomials is called a Binomial and with 3 Monomials is called a

Trinomial. The Degree of a Polynomial is equal to the Degree of

the Monomial with the Largest Degree.

NOTE: You don’t add the Degrees of each Monomial to come up

with the Degree of a Polynomial. You just take the largest one!!!!

Ex 1: 2x3y2z + 5xy8 is a Polynomial of 2 terms. Therefore, it can

also be called a Binomial.

Degree of 1st Monomial = 3 + 2 + 1 = 6

Degree of 2nd Monomial = 1 + 8 = 9 ---------> Therefore, the

Degree of this Binomial = 9 (You don’t add 9 & 6)

Ex 2: -x2y3 + z8 + w2t4x2 is a Polynomial of 3 terms. Therefore, it

can also be called a Trinomial. Degree of 1st Monomial = 5

Degree of 2nd Monomial = 8 and Degree of 3rd Monomial = 8

Therefore, the Degree of this Trinomial = 8


Section 2: In this section, we learn how to add and subtract multiples

of Polynomials.

Polynomials can be added by adding the Monomials that they contain

only if these Monomials are exactly the same. These are called

Like or Similar Monomials.

Ex 1: (2x2y3 + 3z8) – (4x2y2 + 2z9) These 2 Binomials CANNOT be

added/subtracted because none of their respective Monomials are

Alike or Similar

Ex 2:(2x2y3 + 3z8) – (4x2y3 + 2z9) Distribute the (-1) into Parenthesis

2x2y3 + 3z8 – 4x2y3 – 2z9 Combine the 1st and 3rd terms (2 – 4 = -2!!)

-2x2y3 + 3z8 – 2z9 You can’t combine the 2nd and 4th terms

because they are not exactly the same.

Ex 3: 2(5xy4z2 – 2xy2 + 8z5) – 3(2xy4z2 – 4xy2 + z5) First, Distribute

the 2 and the (-3) into both Polynomials.


and Section 3 10xy4z2 – 4xy2 + 16z5 – 6xy4z2 + 12xy2 - 3z5 Now, you can combine

the 1st and 4th terms, 2nd and 5th terms, and 3rd and 6th terms

since they are all Similar Monomials.

The Final answer is -----------> 4xy4z2 + 8xy2 + 13z5

Section 3: This Section covers the topic of Multiplying Polynomials.

This section is harder and longer than section 2.

1) Multiplying Monomials by Monomials: When you multiply two

monomials, you add the powers of the variables that they share

in common. You also multiply their coefficients.

Ex: (2x2y4)(-4x3yz2) = (2)(-4)x2 + 3 y4 + 1 z2 = -8x5y5z2

2) Multiplying Monomials by Polynomials: Multiply the

Monomial by each term of the Polynomial. In other words,

Distribute the Monomial into the Polynomial.

Ex 1: -2x2yz3(x4 – 3yz2 + 2x2y3) = -2x2+4yz3 – (2)(-3)x2y1+1z3+2 – (2)(2)x2+2y1+3z3 = -2x6yz3 + 6x2y2z5 – 4x4y4z3

Ex 2: ab-2c-3(a-4bc3 + a-3b4c3) This example is NOT with Polynomials since there is a negative power, but we can still utilize the same technique of multiplying these.

= a1-4b-2+1c-3+3 + a1-3b-2+4c-3+3 = a-3b-1c0 + a-2b2c0 (c0 = 1!!!)

= a-3b-1 + a-2b2 Or 1/a3b + b2/a2

Ex 3: a2n(an + a2n) = a2n+n + a2n+2n = a3n + a4n

3) Multiplying Polynomials by Polynomials: Multiply each term of one Polynomial by each term of the other Polynomial. If both Polynomials are Binomials, there will be 4 multiplications. If one of them is a Binomial and the other is a Trinomial, there will be 6 multiplications overall. If both of them are Trinomials, there are 9 multiplications. We’ll do a few examples to illustrate.

Ex 1: (2x – 3y)(3x + 6y) = (2x)(3x) + (2x)(6y) – (3y)(3x) – (3y)(6y) = 6x2 + 12xy – 9yx – 18y2 = 6x2 + 3xy – 18y2


Intermediate Algebra Chapter 5 Section 3 Ex 2: (5x-4 – 4y2)(5x2 – 4y-4) = (5)(5)x-4+2 + (5)(-4)x-4y-4 – (4)(5)y2x2

– (4)(-4)y2-4 = 25x-2 -20x-4y-4 – 20x2y2 + 16y-2 OR

25/x2 – 20/x4y4 – 20x2y2 + 16/y2 (This is the answer w/out negative exponents)

Ex 3: (xn + yn)(xn + y-n) = xn+n + xny-n + ynxn + yn-n =

x2n + xny-n + xnyn + y0 = x2n + xn/yn + xnyn + 1

Ex 4: (x – 3y)(x2 + 3xy + 9y2) This is a binomial times a trinomial

= (x)(x2 + 3xy + 9y2) – (3y)(x2 + 3xy + 9y2) Distribute each one now

= x3 + 3x2y + 9xy2 – 3x2y – 9xy2 – 27y3 There are 6 terms. Now,

= x3 – 27y3 combine like terms, and the middle ones cancel out!!

Ex 5: (x – 2y – 3z)(3x + 2y + z) This is a trinomial times a trinomial

= x(3x + 2y + z) – (2y)(3x + 2y + z) – (3z)(3x + 2y + z)

3x2 + 2xy + xz – 6xy – 4y2 – 2yz – 9xz – 6yz – 3z2 Combine like terms

= 3x2 – 4xy – 8xz – 4y2 – 8yz – 3z2 This is the simplest form!!


4) Special Products: are short ways of doing a few specific kind of

multiplications. Although you can still do these multiplications

the regular way as in step3, it is recommended that you learn

these short cuts to speed up your work.

There are 3 of them. Here are the formulas:

(a + b)2 = a2 + 2ab + b2 “a” and “b” can stand for any 2 expressions

(a – b)2 = a2 – 2ab + b2 For example: a = 4x and b = 3y

(a + b)(a – b) = a2 – b2 Let’s do a few examples to illustrate this.

Ex 1: (4x – 3y)2 Here a = 4x and b = 4y Use formula 2 above

= (4x)2 – 2(4x)(3y) + (3y)2 = 16x2 – 24xy + 9y2

Here is the regular way of doing this:

(4x – 3y)2 = (4x – 3y)(4x – 3y) = (4x)2 – (4x)(3y) – (3y)(4x) + (3y)2

= 16x2 – 12xy – 12xy + 9y2 = Gives us the same answer!!!!!!!!!!!

Intermediate Algebra Chapter 5 Section 3 Ex 2: (5x – 2y)(5x + 2y) Here a = 5x and b = 2y Use Formula 3

= (5x)2 – (2y)2 = 25x2 – 4y2 That’s pretty quick, wouldn’t you say!!

Here is the regular way of doing it:

(5x – 2y)(5x + 2y) = (5x)2 + (5x)(2y) – (2y)(5x) – (2y)2 = 25x2 +10xy

– 10xy – 4y2---> the middle terms cancel and we get the same answer

Ex 3: [x + (2a – b)]2 Here a = x and b = 2a – b Use Formula 1

(x)2 + 2(x)(2a – b) + (2a – b)2 Now apply Formula 2 on the 3rd term

x2 + 4xa – 2xb + (2a)2 – 2(2a)(b) + b2 where a = 2a and b = b

x2 + 4xa – 2xb + 4a2 – 4ab + b2 There are NO Like terms to combine

Ex 4: (3y + 1)2 + (2y – 4)2 Use Formula 1 on the 1st term and

= (3y)2 + 2(3y)(1) + 12 + (2y)2 – 2(2y)(4) + 42 F. 2 on the 2nd term.

= 9y2 + 6y + 1 + 4y2 – 16y + 16 = 13y2 – 10y + 17

Hopefully we have done enough examples to give you a flavor for this!! Remember, the ultimate way of learning is to Practice a lot.


Section 4 is about Factoring. Basically it is the reverse process of

what we did in Section 3. Instead of multiplying out monomials

and polynomials, we are going to look for ways of factoring

things out or combining things so that we can work backwards.

1) Factoring out the Greatest Common Factor (GCF) The first

step is to always look for the GCF. This includes the greatest

number and greatest powers of the variables that ALL the terms

of the polynomial share in common.

Ex 1: 13ab2c3 – 26a3b2c The GCF = 13ab2c because both terms

have at least a 13 in them and “a” to the power of 1 and “b” to

the power of 2 and “c” to the power of 1. Factor out this GCF!

= 13ab2c(c2 – 2a2) You leave in the parenthesis whatever is left over

Ex 2: 18y2z2 + 12y2z3 – 24y4z3 The GCF = 6y2z2 Factor this out!!

6y2z2(3 + 2z – 4y2z) Leave in the parenthesis whatever is left over


Sometimes the book gives you specific directions on what to factor

out in some of their problems. Here are some examples of these:

Ex 1: Factor out the negative of the GCF

-56x4y3z2 – 72x3y4z5 + 80xy2z3 Here the GCF is 8xy2z2 so factor out

the negative of this and switch all the signs. This is useful when the

= -8xy2z2(7x3y + 9x2y2z3 – 10z) original expression has too many

negative signs. Notice that the polynomial in the parenthesis of

our answer has ONLY 1 negative sign.

Ex 2: Factor out y-4 from the expression 7y4 + y

= y-4(7y8 + y5) Notice that if you Redistribute the y-4 back into the

parenthesis, you’ll get ----> 7y-4+8 + y-4+5 = 7y4 + y1 The best

way to do these problems is to ask yourself what you need to put in

the parenthesis so that if you Redistribute the factored out

expression, you’ll end up with the Original expression!!!!!!!


Ex 3: Distribute t-4ns from the expression t8ns3 – s2 – t-3ns

= t-4ns(t12ns2 – t4ns – tn) Notice that if you Redistribute the t-4ns, you’ll

get -----> t-4n+12ns3 – t-4n+4ns2 – t-4n+ns = t8ns3 – t0s2 – t-3ns (t0 = 1)

2) Factoring by Grouping: is the second major technique used in

simplifying and factoring expressions. You group together like

terms and factor the GCF from those and then you might be able

to factor out one more time from the entire expression.

Ex 1: x2 + 4y – xy – 4x Group the 1st and 3rd terms and the 2nd and

x2 – xy + 4y – 4x 4th terms. Now factor out the GCF of each pair.

x(x – y) + 4(y – x) Now, change the sign of (y – x) = -(x – y)!!!

x(x – y) – 4(x – y) Finally, factor out the expression (x – y)

(x – y)(x – 4) You can check your answer this way: Multiply out

these 2 binomials and you should get the original expression!!!


Ex 2: ab – b2 – bc + ac – bc – c2 Here, group the 1st and 2nd terms,

3rd and 4th terms, and 5th and 6th terms.

= b(a – b) – c(b – a) – c(b + c) Notice that I had to change the

signs in the 2nd and 3rd parenthesis because I factored out a negative.

Now, switch the 2nd parenthesis using the theorem (b – a) = -(a – b)

= b(a – b) + c(a – b) – c(b + c) Factor out the (a – b)!!!

= (a – b)(b + c) – c(b + c) Now, factor out the (b + c)

= (b + c)(a – b – c)

Ex 3: a2c3 + ac2 + a3c2 – 2a2bc2 – 2bc2 + c3 Here, we have to

factor out all common Monomials first and then use the Grouping

method. The GCF = c2 So, factor this out of all the expressions!!

c2(a2c + a + a3 – 2a2b – 2b + c) = c2(a3 + a2c – 2a2b + a + c – 2b)

I just rearranged the terms so that I can now factor out an “a2” from

the first three terms. Notice what will happen!!!!!!!!!!!!!!


and Section 5 = c2[a2(a + c – 2b) + (a + c – 2b)] Notice that the expression (a + c

– 2b) appears twice inside the brackets and we can factor it out!!

= c2(a + c – 2b)(a2 + 1)

Section 5: is about factoring 3 Special Forms. Here they are:

a2 – b2 = (a – b)(a + b) This is called the Difference of Two Squares

a3 – b3 = (a – b)(a2 + ab + b2) This is the Difference of Two Cubes

a3 + b3 = (a + b)(a2 – ab + b2) This is the Sum of Two Cubes

a2 + b2 = NOT Factorable. Let’s do some examples.

Ex 1: 16x4 – 81y2 We can put this in the a2 – b2 form if we write it

like this -------------> (4x2) – (9y)2 where a = 4x2 and b = 9y

Therefore the solution is (4x2 – 9y)(4x2 + 9y)

Ex 2: 64r6 – 121s2 = (8r3)2 – (11s)2 = (8r3 – 11s)(8r3 + 11s)


Ex 3: 16a4b3c4 – 64a2bc6 First, let’s factor out the GCF = 16a2bc4

= 16a2bc4(a2b2 – 4c2) = 16a2bc4[(ab)2 – (2c)2]

= 16a2bc4(ab – 2c)(ab + 2c)

Ex 4: 8x6 + 125y3 = (2x2)3 + (5y)3 Now apply the formula for

a3 + b3 ----------> (2x2 + 5y)[(2x2)2 – (2x2)(5y) + (5y)2]

= (2x2 + 5y)(4x4 – 10x2y + 25y2)

Ex 5: x9 + y9 = (x3)3 + (y3)3 Here a = x3 and b = y3

= (x3 + y3)[(x3)2 – (x3)(y3) + (y3)2] = (x3 + y3)(x6 – x3y3 + y6)

Now, reapply the formula to the expression in the first parenthesis

= (x + y)(x2 – xy + y2 )(x6 – x3y3 + y6)

Ex 6: x6m – y3n = (x2m)3 – (yn)3 Here a = x2m and b = yn

= (x2m – yn)[(x2m)2 + (x2m)(yn) + (yn)2]

= (x2m – yn)(x4m + x2myn + y2n)


and Section 6 Ex 7: m – 2n + m2 – 4n2 Here, apply the a2 – b2 formula to the 3rd

and 4th terms first. Then factor out anything that is common.

= (m – 2n) + (m – 2n)(m + 2n) Factor out the (m – 2n)!!!!!!!!

= (m – 2n)(1 + m + 2n)

Section 6: This section is about factoring Trinomials of the form

ax2 + bx + c where a, b, c may be any numbers.

“a” is often called the “Lead Coefficient”. When a = 1, the factoring

becomes quicker and easier. Otherwise, it is trickier and harder.

Let’s first concentrate on cases where a = 1 -----> x2 + bx + c

Ex 1: x2 + 4x + 3 Here b = 4 and c = 3. This is the easiest case

because both “b” and “c” are positives. Therefore, the 2

parenthesis will be BOTH positives --------> = (x + 3)(x + 1)

Notice that the factors of 3 are 3 and 1 and they add up to 4!!!


Ex 2: x2 – 7x + 12 This is a little harder but not much. Here “b” is

a negative and “c” is a positive. In these cases, BOTH

parenthesis will have a Negative sign. You have to ask yourself

the question, “What are the factors of 12 which add up to 7?”

It is 4 and 3, right? Therefore the answer is (x – 4)(x – 3)

Ex 3: x2 + 3x – 28 This is even a little harder. Here “b” is positive

and “c” is negative. In these cases, One of the parenthesis will be

positive and the other one a negative. The question now should be

“What are the factors of 28 whose difference is 3?”

It is 7 and 4, right? Therefore the answer is (x + 7)(x – 4)

Notice that the 7 has the + sign because the “b” = +3.

If the question was ------> x2 – 3x – 28, the answer is (x – 7)(x + 4)

If the question was ------> x2 + 4x – 28, the answer would be Prime

or NOT Factorable because no factors of 28 have a difference of 4.


Now, let’s move on to cases where the “a” is not equal to 1.

Ex 1: 6x2 + 17x + 12 Here both “b” and “c” are positive, so both

Parenthesis will have a positive sign. However, we can NOT use

the question, “What factors of 12 add up to 17?” anymore,

because the “a” = 6 will come into play.

The answer to this question could either be (6x + ?)(x + ?) or

(3x + ?)(2x + ?) because the factors of 6 are 6 and 1 or 3 and 2.

Then, you have to ask yourself “What factors of 12 can I put in the

question marks so that 6 times one of the factors + 1 times the

other factor = 17 OR 3 times one of the factors + 2 times the other

factor = 17?” Here is a Hint that will make your work quicker.

The number that you put in the question mark should not make

that parenthesis factorable. In other words, you should NOT

end up with something like ----> (6x + 2)(x + 6) because you

can factor a 2 out of the 1st parenthesis.


Therefore, the only likely possibilities of factoring this trinomial are:

(6x + 1)(x + 12), (3x + 4)(2x + 3) These are the ONLY

combinations that lead to Non-Factorable parenthesis. However

the 1st of these leads to (6)(12) + 1 = 73 for the Middle Term.

The 2nd one leads to (3)(3) + (4)(2) = 17 for the Middle Term and

that’s what we want. The answer is (3x + 4)(2x + 3)

Ex 2: 7x2 – 23x + 6 Here “b” is negative and “c” is positive.

Therefore, Both parenthesis will have a Negative Sign like before.

The possibilities are (7x – 1)(x – 6) or (7x – 6)(x – 1) or

(7x – 2)(x – 3) or (7x – 3)(x – 2) All of these lead to Non-

factorable parenthesis but they yield a Different answer for the

Middle Term “b”. We want the “b” to come out to -23.

The first one yield -43, the second one -13, the 3rd one -23, the 4th one

-17. Therefore, the answer is (7x – 2)(x – 3)


Ex 3: -90x2 + 2 – 8x Here a = -90, b = -8, and c = +2. It’s all out

of order and the “a” is negative. So, let’s factor out a “-2” so that

“a” becomes positive and let’s rearrange it so that it is in the

right order. Now, it becomes -2(45x2 + 4x – 1) Now, it’s good!!

Since, “b” is positive and “c” is negative now, One of the

parenthesis will be positive and the other one negative like before.

The choices are (45x + 1)(x – 1) or (45x – 1)(x + 1) or (15x + 1)

(3x – 1) or (15x – 1)(3x + 1) or (9x + 1)(5x – 1) or (9x – 1)(5x + 1)

The 1st one yields a Middle term of -44, the 2nd one 44, the 3rd one

-12, the 4th one 12, the 5th one -4, and the 6th one 4.

Therefore, the final answer is -2(9x – 1)(5x + 1)

As you see, it’s a matter of practice and getting used to it. The more

you do, the faster you get at it and you don’t have to try out every

possibility. Now, let’s do some that have an extra level of trickiness.


Ex 1: c4 – 8c2 – 9 Here, you treat it as if it was c2 – 8c – 9 except

that the answer will have c2 in the two parenthesis. “b” is

negative and “c” is negative, Therefore the 2 parenthesis will

have Opposite signs, as before. Try out the different possibilities

and you will eliminate all of them but one. The answer is

(c2 – 9)(c2 + 1) because the Middle Term needs to be Negative 8.

Now, we can go one more step and factor the 1st parenthesis.

The final answer is (c – 3)(c + 3)(c2 + 1)

Ex 2: 12y4n + 10y2n + 2 First, factor out a 2 from everything.

= 2(6y4n + 5y2n + 1) Treat this as if it was 6y2 + 5y + 1 except that

you will put “y2n” in the parenthesis instead of “y”.

The final answer is 2(3y2n + 1)(2y2n + 1)

Ex 3: (x – y)2 + 3(x – y) – 10 Here, treat this as if it was

x2 + 3x – 10 which would be factored as (x + 5)(x – 2) except --->


Replace the “x” in the answer by “x – y”.

The final answer is (x – y + 5)(x – y – 2) That wasn’t too bad!!!

Ex 4: x2 – 6x + 9 – 4y2 Whenever you see 4 or more terms, this is a

clue that you should first use Grouping Technique.

Notice that the 1st three terms comprise a Perfect Square Trinomial.

x2 – 6x + 9 = (x – 3)2 Therefore, x2 – 6x + 9 – 4y2 = (x – 3)2 – 4y2

= (x – 3)2 – (2y)2 This is the Difference of Two squares.

= (x – 3 – 2y)(x – 3 + 2y)

Ex 5: a2 + 14a – 25b2 + 49 Here, we need to use Grouping again

but first we need to rewrite it like this ------------->

a2 + 14a + 49 – 25b2 Now, the 1st three terms are a perfect square

= (a + 7)2 – 25b2 This is the Difference of Two squares again.

= (a + 7)2 – (5b)2 = (a + 7 – 5b)(a + 7 + 5b)


Section 7: is a catch-all section which summarizes the techniques of

the previous few sections. Here are the steps you should follow:

1. Factor out all common monomials first, the GCF.

2. If there is 4 or more terms, group 2 or 3 of them together.

3. See if there are any Perfect Square or Cube forms like:

(a – b)2 = a2 – 2ab + b2 or (a + b)2 = a2 + 2ab + b2 or

a2 – b2 = (a – b)(a + b) or a3 – b3 = (a – b)(a2 + ab + b2) or

a3 + b3 = (a + b)(a2 – ab + b2)

4. If non of the above forms exist, try to factor the trinomials.

Ex 1: 3x2y + 6xy2 – 12xy First factor out the GCF = 3xy

= 3xy(x + 2y – 4) You can’t do anything further with this!!!

Ex 2: 12x2 + 14x – 6 First factor out the GCF = 2

= 2(6x2 + 7x – 3) = 2(3x – 1)(2x + 3)

Intermediate Algebra Chapter 5 Section 7 Ex 3: 100z2 – 81t2 Here, there are no Common Monomials.

However, this falls into the category of Difference of Two Squares.

= (10z)2 – (9t)2 = (10z – 9t)(10z + 9t)

Ex 4: a2x2 + b2y2 + b2x2 + a2y2 Here, there are is no GCF again!

Since there are 4 terms, I will group the 1st and the 3rd and the 2nd and the 4th terms. So rewrite it as = a2x2 + b2x2 + b2y2 + a2y2

= x2(a2 + b2) + y2(b2 + a2) Now, factor out the a2 + b2 term!!!!

= (a2 + b2)(x2 + y2)

Ex 5: 6x2 – 63 – 13x Here, there are no GCF’s again. Just rewrite it as = 6x2 – 13x – 63 and Factor it out using the techniques we already learned (3x + 7)(2x – 9)

Ex 6: x2 + 10x + 25 – y8 = (x + 5)2 – (y4)2 Here, I grouped the 1st three terms into a Perfect Square and rewrote the 4th term as a square also. Now, the expression is the Difference of Two Squares -------------> (x + 5 – y4)(x + 5 + y4)


Ex 7: x5 – x2 – 4x3 + 4 Here, there is no GCF again. Since there are

4 terms, I have to group them. If I group the 1st three terms, I can

factor out an “x2” from them. Notice what happens ---------->

x2(x3 – 1 – 4x) + 4 Now, I can group the 1st and the 3rd terms in the

parenthesis and factor out an “x”.

x2[x(x2 – 4) – 1] + 4 Now, factor out the x2 – 4 and you get.

x2[x(x – 2)(x + 2) – 1] + 4 It led to a DEAD END!!!!!!!!!!!!!!

Therefore, in the original expression, I should group the 1st two

terms and the 2nd two terms. Now, it will work out much better.

x2(x3 – 1) – 4(x3 – 1) Now factor out the x3 – 1 expression!!

= (x3 – 1)(x2 – 4) Now, use the special forms to factor these, walla!

(x – 1)(x2 + x + 1)(x – 2)(x + 2)

Ex 8: 4a-2 – 12a-1 + 9 Here, treat “a-1” as if it was the variable x.

Rewrite this expression with x -----------> 4x2 – 12x + 9


This is a perfect square which equals (2x – 3)2 Now, replace the x

with the original variable: = (2a-1 - 3)2 = (2/a – 3)2

Ex 9: x-4 – y-4 Treat this as if it was x4 – y4 = (x2 – y2)(x2 + y2) =

(x – y)(x + y)(x2 + y2). However, replace this answer with negative

powers of x and y. ( x becomes x-1 , x2 becomes x-2 , same for y)

= (x-1 – y-1)(x-1 + y-1)(x-2 + y-2) = (1/x – 1/y)(1/x + 1/y)(1/x2 + 1/y2)

Ex 10: x4 + x3 – 2x2 + x – 1 Here there is no GCF. Since there are

5 terms, we need to do some Grouping. Group the 1st three terms

and factor out an “x2” from them. You end up with --------------->

= x2(x2 + x – 2) + (x – 1) Factor the trinomial in the 1st Parenthesis

= x2(x – 1)(x + 2) + (x – 1) Factor out the expression (x – 1)

= (x – 1)[x2(x + 2) + 1] Distribute the x2 into the inner parenthesis

= (x – 1)(x3 + 2x2 + 1) That is it!!!! You can always multiply out

your factored expression and check your answer to see if it’s right.


Section 8: In this section, we use the techniques of factoring to solve

equations. To do this, we use the Zero-Factor Theorem which

states that ------> If ab = 0, then a = 0 or b = 0

Ex 1: x2 – 9 = 0 First factor the expression x2 – 9

(x – 3)(x + 3) = 0 Now, treat (x – 3) as “a” and (x + 3) as “b”

Therefore the answer is: x – 3 = 0 OR x + 3 = 0 Now, solve each

individual equation for x and you get ---------> x = 3 or x = -3

Ex 2: x2 – 3x = 0 Factor this expression first.

x(x – 3) = 0 Treat x as “a” and (x – 3) as “b” Use the Zero-

Factor theorem and you get x = 0 OR x – 3 = 0 Now, solve

for x in each equation and you get. x = 0 OR x = 3

Ex 3: 5x2 = 6 – 13x First gather everything to the left side and

factor it out. 5x2 + 13x – 6 = 0 -------------> (5x – 2)(x + 3) = 0

5x – 2 = 0 OR x + 3 = 0 Therefore, the answer is x = 2/5 OR -3


Ex 4: x(x/11 – 1/7) = 6/77 Multiply everything by 77 to get rid of

fractions. x(7x – 11) = 6 Now, distribute the x into the

parenthesis. 7x2 – 11x = 6 Take the 6 to the left and factor the

Trinomial. 7x2 – 11x – 6 = 0 --------> (7x + 3)(x – 2) = 0

Therefore, 7x + 3 = 0 OR x – 2 = 0 x = -3/7 OR x = 2

Ex 5: x4 – 10x2 + 9 = 0 Factor this trinomial by treating the x4 as if

it was “x2” and the x2 as if it was “x”. You end up with --------->

(x2 – 9)(x2 – 1) = 0 Now, each parenthesis is further factorable.

(x – 3)(x + 3)(x – 1)(x + 1) = 0 The answer is x = 3, -3, 1, -1

Ex 6: x2 = -4x3(3x + 5)/3 Cross-multiply the 3 to the other side.

3x2 = -4x3(3x + 5) = -12x4 – 20x3 Take everything to the left side.

12x4 + 20x3 + 3x2 = 0 Factor out the x2

x2(12x2 + 20x + 3) = 0 Now, factor the trinomial in the parenthesis

x2(6x + 1)(2x + 3) = 0 So, the answer is x = 0, -1/6, -3/2


Now, let’s solve some Word Problems with these techniques.

Ex 1: The sum of the squares of two consecutive positive integers is

85. Find the integers.

Let x = the smaller integer, then x + 1 = the larger integer.

Therefore, x2 + (x + 1)2 = 85 Now, we have to solve for x.

Expand the (x + 1)2 = x2 + 2x + 1 and add this to the x2. You get ---

-----> 2x2 + 2x + 1 = 85 Subtract 85 from both sides.

2x2 + 2x – 84 = 0 Divide everything by 2.

x2 + x – 42 = 0 Factor this trinomial (x + 7)(x – 6) = 0

Therefore, x = -7 or x = 6. Since, x has to be a positive integer,

discard the 1st answer. Therefore, x = 6 and x + 1 = 7

The smaller integer is 6 and the larger integer is 7.

Ex 2: One side of a rectangle is three times longer than another. If

its area is 147 square centimeters, find its dimensions.


Let w = the width, then l = the length = 3w (3 times longer)

The Area = lw = (3w)w = 3w2 = 147 Take the 147 to the left side.

3w2 – 147 = 0 Divide everything by 3 --------------> w2 – 49 = 0

(w – 7)(w + 7) = 0 Therefore, the width = 7 or -7. Discard the

negative answer. Width = 7 cm , Length = 3w = 21 cm

Ex 3: A woman plans to use one-fourth of her 48 foot by 100 foot

backyard to plant a garden. Find the perimeter of the garden if

the length is to be 40 feet longer than the width.

Let w = width of her garden, then l = 40 + w.

The Area of her garden will be ¼ of the Area of her backyard =

(48)(100) = 4,800 square feet. ¼ of this equals 1,200 square feet.

Therefore, A = lw = (40 + w)w = 1200 ---------> w2 + 40w = 1200

w2 + 40w – 1200 = 0 ---------> (w + 60)(w – 20) = 0 ------------>

w = 20 or w = -60 Discard the negative answer again.


Therefore, the w = 20 and l = 40 + w = 60

The Perimeter = 2l + 2w = 2(60) + 2(20) = 120 + 40 = 160 feet

Ex 4: After how many seconds will an object hit the ground if it was

thrown straight up with an initial velocity of 208 feet per second?

Solution: The equation that you have to use here is h = vt – 16t2

h = the height above the ground that a projectile reaches

v = the initial velocity of the projectile, t = time of flight

In our problem, h = 0 (when it hits the ground, its height above the

ground will be zero) and v = 208 ft/sec Therefore, the equation

becomes ----------> 0 = 208t – 16t2 Factor 16t out of this.

16t(13 – t) = 0 Therefore, t = 0 or t = 13 Discard the 1st answer

Because it represents the original time when the projectile was

launched. The projectile will hit the ground in 13 seconds.


By Viken Kiledjian Section 1: is about simplifying Rational

Expressions. A Rational Expression is any

expression which is the Quotient of two

Polynomials.

This section, even this chapter, are based on the

techniques that we learned from the last

chapter. Let’s so some example to illustrate.

Ex 1: Simplify Each Rational Expression

Ex 2:

2

5

2

345

23

45

7

5

7

5

21

15

c

ba

c

ba

cb

ba

2)2()2(

)2(233

2

x

x

x

x

x

xx


Ex 3: In this example, we are going to utilize a technique of

Factoring out a (-1) -----------> (a – b) = -(b – a)

Ex 4:

Ex 5:

Ex 6:

1)(

)(

))((

))((

pr

pr

qppr

rpqp

3

1

)93)(3(

93

27

932

2

3

2

xxxx

xx

x

xx

)2)(32(

)12)(32(

)62(

)12)(32(

26

38422

2

xx

xx

xx

xx

xx

xx

x

x

x

x

2

12

)2(

12

)]3(4)3([

)6(2

1243

12222

2

23

2

xxx

xx

xxx

xx

Section 2: covers the topic of Multiplying and Dividing Rational

Expressions which is not much different than Simplifying them. I

will do 1 example from this section which involves Dividing and

Multiplying expressions.

Ex:

Intermediate Algebra Chapter 6 Section 1 &

Section 2

2

2

)2)(2(

)2(2

)4)(3(

)2)(3(22

xxx

x

xx

xx

123

32

1

9124

1

3762

22

2

2

aa

aa

a

aa

a

aa

123

32

9124

1

1

3762

2

22

2

aa

aa

aa

a

a

aa

1

1

)1)(13(

)1)(32(

)32)(32(

1

)1)(1(

)32)(13(

aaa

aa

aa

a

aa

aa


Section 3: is not as short as Section 2. It involves Adding and

Subtracting Rational Expressions which is a totally Different

process than simplifying, multiplying, or dividing them.

When you add 2 or more Fractions, they must ALL have the same

Denominator which is called the LCD (least common

denominator). If they don’t, then you need to multiply each

denominator by an appropriate number or expression so that it

becomes equal to the LCD. When you do this, you have to also

multiply the numerator by the same expression or number so you

won’t change the Fraction.

Ex 1: Here, the LCD = 120. You need to multiply

8 by (15) in order to achieve this LCD. You

need to multiply 6 by (20) and 10 by (12). However, you have to

also multiply their numerators by these numbers so that the

values of the Fractions are not altered.

10

3

6

1

8

5

59/120 is NOT simplifiable any more so it is the Final Answer.

Ex 2: Here, they already have the Same

Denominator, so just combine the

Numerators, Factor the 9 and cancel.

Ex 3: Here, the LCD = 10ab. Multiply the 5a by

(2b) and the 2b by (5a) to achieve this LCD

Ex 4: Here, the LCD = (x +3)(x + 6).


120

59

120

362075

120

36

120

20

120

75

10)12(

3)12(

6)20(

1)20(

8)15(

5)15(

yx

y

yx

x

99

9)(999

yx

yx

yx

yx

ba 2

3

5

2

ab

ab

ab

a

ab

b

ba

a

ab

b

10

154

10

15

10

4

2)5(

3)5(

5)2(

2)2(

6

4

3

7

x

x

x


Since the Numerator is not Factorable

any further, this is the Final Answer.

Ex 5:

Here, we had to do some work on the original problem in order to

determine what the LCD is. Now, it is evident that the LCD = (x

– 2)(x + 2). The 2nd fraction already has this LCD so nothing

needs to be done with it. You just need to multiply the top and

bottom of the 1st fraction by (x + 2).

)6)(3(

124427

)6)(3(

4)3(

)3)(6(

7)6( 2

xx

xxx

xx

xx

xx

x

)6)(3(

42194 2

xx

xx

)2)(2(

5

2

4

4

5

2

4

4

5

2

422

xxxxxxx

)2)(2(

34

)2)(2(

584

)2)(2(

5

)2)(2(

4)2(

xx

x

xx

x

xxxx

x


Ex 6: Whenever an expression does not have

a Denominator, you can put a (1) for its

Denominator and the LCD = x – 1

Ex 7:

Here, the LCD

is (x + 3)(x – 2)(x – 5). Multiply the top and bottom of each

fraction by whatever it needs to achieve this LCD!!!!!!!

2541

23xx

x

1

23495

1

545423 23232

x

xxx

x

xxxx

1)1(

)54)(1(

1

23

1

54

1

23 22

x

xxx

x

xx

x

107

32

152

53

6

12222

xx

x

xx

x

xx

x

)2)(5(

32

)3)(5(

53

)2)(3(

12

xx

x

xx

x

xx

x

Section 4: is about the topic of Complex Fractions which are

fractions within fractions. Examples of complex fractions include.

Intermediate Algebra Chapter 6 Section 3 &

Section 4

)2)(5)(3(

)32)(3(

)3)(5)(2(

)53)(2(

)2)(3)(5(

)12)(5(

xxx

xx

xxx

xx

xxx

xx

)2)(3)(5(

)932()10113()5112( 222

xxx

xxxxxx

)5)(2)(3(

143

)2)(3)(5(

910531111232 2222

xxx

xx

xxx

xxxxxx

9

23

5

xy

x

49

2

24

xx

xy

yx

2

52


The basic idea to Simplify or Reduce them is to multiply the TOP AND BOTTOM fractions by their COMMON LCD.

Ex 1: In this complex fraction, the Common LCD of

both fractions is 9, since 3 goes into 9. So,

multiply both the top and bottom fractions by 9!!

Notice how much we have simplified the original fraction!!!

Ex 2: In the 2nd example of the previous slide, the Common LCD of both fractions is 9x. So, multiply both the top and bottom fractions by 9x!!!

9

23

5

xy

x

yxy

x

xy

x

xy

x

2

15

2

15

2)1(

5)3(

9

2)9(

3

5)9(

xx

x

xx

x

xx

x

xx

xx

xx

362

1836

)4)(9()9

2)(9(

)2)(9()4

)(9(

)49

2)(9(

)24

)(9(

49

2

24

2


Ex 3: In the 3rd example, the Common LCD = xy.

Let’s do a couple more examples that look a little different.

Ex 4: Here, we rewrote the original Complex

fraction so the Common LCD will be

more apparent. It is equal to xy(x + y)

yxx

xy

xxyy

xy

yxy

xxy

xy

xy

yxxy

xy

yx22

52

))(()2

)((

)5

)(()2

)((

)2

)((

)52

)((

2

52

yx

yx

yx

yx

1

11

)( 1

11

xy

yx

xy

yxxyxy

yxyxxy

yxyxxy

2)()()(

1)]([

)11

)](([


Ex 5: Here, we have a Fraction within a fraction

within a fraction. We need to first simplify the

top fraction and then multiply both the top and

bottom fractions by their common LCD.

Now, the Common LCD = b(b + a)

b

ab

a

1

1

21

b

aab

b

b

a

bb

a

b

1

21

1

))(1(

)(21

22

2

22

222 32

)()(

2)(

)1)((

)2

1)((

ab

bab

aabbab

bbab

abaabb

babb

b

aabb

ab

babb

))((

)3(

abab

abb

Intermediate Algebra Chapter 6 Section 4 and

Section 5 Ex 6:

Section 5: In this section, we solve equations which include Rational

Expressions. The technique to solve them is to multiply both sides

of the equal sign by the Common LCD.

I will do one example with variables and the rest, word problems.

)3(

1

)1)(3(

2

)1)(3(

2

)1)(1(

1

3

1

22

234

2

1

1

2

22

xxx

xxxx

xxx

xxx

)]1)(1()1(2[

)]1(2)3[(

])3(

1

)1)(3(

2)[1)(3)(1(

])1)(3(

2

)1)(1(

1)[1)(3)(1(

xxx

xx

xxxxxx

xxxxxxx

2222 )1(

5

)1(

5

12

5

122

223

x

xOR

x

x

xx

x

xx

xx


Ex: Here, we have to first factor

the trinomials as before.

Then we can determine what the LCD is and multiply both sides by

this LCD. First, rewrite the trinomials then factor!!

The LCD = (a + 2)

(a + 3)(a – 2)

aa

a

a

a

aa

a

6

32

4

52

56

23222

6

32

4

52

65

23222

aa

a

a

a

aa

a

)2)(3(

32

)2)(2(

52

)3)(2(

23

aa

a

aa

a

aa

a

]

)2)(2(

52

)3)(2(

23)[2)(3)(2(

aa

a

aa

aaaa

])2)(3(

32)[2)(3)(2(

aa

aaaa

)2)(32()3)(52()2)(32( aaaaaa


Now, combine like terms and solve for “a”

Let’s solve some word problems that involve this technique.

Ex 1: A proofreader can read 50 pages in 3 hours, and a second

proofreader can read 50 pages in 1 hour. If they both work on a

250 page book, can they meet a six-hour deadline?

Solution: When people work together, their rates of accomplishing

a task add up.

The rate for person A = 50/3 = 16.6666667 pages/hr

The rate for person B = 50/1 = 50 pages/hr

The combined rate = 250/t (We want to see if t is less than 6)

aaaaaaaaa 6342155626342 222

443143 22 aaaa 410 a 5

2a


Now, let’s add up their rates and solve for “t”

The LCD = 3t so multiply everything by this.

50t + 150t = 750 t = 3.75

Since t < 6 hours, they can meet this deadline.

Ex 2: Sally can clean the house in 6 hours, and her father can clean the house in 4 hours. Sally’s younger brother, Dennis, can completely mess up the house in 8 hours. If Sally and her father clean and Dennis plays, how long will it take to clean the house?

Sally’s rate for cleaning = 1/6 house/hour

The Father’s rate for cleaning = ¼ house/hour

Dennis’s rate of messing up the house = 1/8 house/hour

Combined rate for cleaning house = 1/t house/hr

t

250

1

50

3

50

)250

(3)1

50

3

50(3

ttt


When we add up their rates, we have to put a Minus sign before Dennis’ rate because he is messing up the house.

It will take 24/7 = 3 and 3/7 hours to clean the house.

Ex 3: Two trains made the same 315 mile run. Since one train traveled 10 mph faster than the other, it arrived 2 hours earlier. Find the speed of each train.

Solution: The basic equation is Distance = rate x time

Let x = rate of slower train. Then x + 10 = rate of faster train

Time for slower train = 315/x

Time for faster train = 315/(x + 10)

We also know that the faster train’s time is 2 hrs less than the slower trains time. Therefore, the equation becomes

t

1

8

1

4

1

6

1 )

1(24)

8

1

4

1

6

1(24

ttt 24364 ttt


Therefore, x = -45 or 35. We discard the negative answer.

The slower train has a speed of 35 mph and the faster train has a

speed of 45 mph.

Ex 4: If three times a number is subtracted from four times its

reciprocal, the result is 11. Find the number.

Let x = the number. Then, 1/x is its reciprocal. The equation is

2315

10

315

xx]2

315)[10()

10

315)(10(

xxx

xxx

)10(2)10(315315 xxxx xxxx 2023150315315 2

03150202 2 xx 01575102 xx 0)35)(45( xx

1131

4 xx

)11(]34

[ xxx

x xx 1134 2

04113 2 xx 0)4)(13( xx 43

1 orx


Section 6: is about Proportion and Variation. Here are some

definitions we will need to understand this section.

A Ratio is the quotient of two numbers.

A Proportion is an expression which indicates that 2 ratios are

equal to each other.

“a” and “d” are the Extremes

“b” and “c” are the Means of the proportion

Here is a very important theorem about proportions which helps us

solve them.

The product of the Extremes equals the product of the Means

Another way to state this is that “We can cross multiply the

Denominator of one ratio with the Numerator of the other ratio

and vice versa”.

d

c

b

a

d

c

b

a


Here are some examples that utilize this technique:

Ex 1: After we Cross-Multiply, we end up with

4(x + 1) = 6(x – 1) 4x + 4 = 6x – 6

10 = 2x and x = 5 (That’s pretty easy, isn’t it???)

Ex 2: After we Cross-Multiply, we get

10 = -2x(x + 6) 10 = -2x2 – 12x

2x2 + 12x + 10 = 0 x2 + 6x + 5 = 0 (x + 1)(x + 5) = 0

Therefore, x = -5 or -1

Let’s get to the topic of Variation and give some definitions.

y = kx means that “y varies directly with x” or “y is directly

proportional to x”. “k” is the constant of proportionality.

Lots of physical quantities in the universe vary directly with each

other. Ex: Circumference with Diameter C = pD

4

6

1

1

x

x

5

2

6

2 x

x


y = k/x means that “y varies inversely with x” or “y is inversely

proportional to x”. “k” is the constant of proportionality.

Ex: Supply and Demand in Economics are quantities that vary

inversely. When the Demand of a product rises, the Supply goes

down and vice versa.

y = kxz is called a Joint Variation because “y varies jointly with x

and z” or “y is directly proportional to both x and z”. “k” is the

constant of proportionality as usual.

Ex: The Area of a rectangle varies jointly with the Width and

Length A = LW. Here, the constant of proportionality = 1.

y = kx/z is called a Combined Variation because “y varies directly

with x and inversely with z”.

Ex: The Pressure of a gas varies directly with its Temperature and

inversely with its Volume. P = kT/V


Now, let’s do some Word Problems that utilize Variation.

(I am not doing the Word Problems involving Proportion because

they are relatively easy and straight-forward)

Ex 1: An object in free fall travels a distance s that is directly

proportional to the square of the time t. If an object falls 1024

feet in 8 seconds, how far will it fall in 10 seconds?

Solution: The sentence above can be translated into the following

equation: s = kt2 Now, put in s = 1024 feet and t = 8 seconds

and solve for “k”.

1024 = k(8)2 = 64k k = 16 s = 16t2

Now, put in t = 10 and solve for s: s = 16(10)2 = 1600 feet

Ex 2: The value of a car usually varies inversely with its age. If a

car is worth $7000 when it is 3 years old, how much will it be

worth when it is 7 years old.


Let v = value of the car and t = the age of the car.

Then the equation becomes: v = k/t Now, put in v = $7000 and t =

3 years and solve for “k”. 7000 = k/3 k = 21000

Therefore, the equation is now v = 21,000/t. Now, put in t = 7

years and solve for v: v = 21,000/7 = $3,000.

Ex 3: The volume of a rectangular solid varies jointly with its

length, width, and height. If the length is doubled, the width is

tripled, and the height is doubled, by what factor is the Volume

multiplied?

Solution: V = kLWH If you replace L with 2L, W with 3W, and H

with 2H,what happens to V?

Vnew = k(2L)(3W)(2H) = 12(kLWH) = 12Vold

The Volume gets multiplied by a factor of 12.


Section 7: is about dividing Polynomials. First, we start out dividing

a Monomial by a Monomial. Second, we divide a Polynomial by

a Monomial. Lastly, we divide a Polynomial by a Polynomial. In

this last case, we employ a similar technique as in the case of

Long Division.

Ex 1:

As you notice, the basic idea is to subtract the powers of the

variables and leave the final answer with positive powers.

Ex 2:

6

76733)3(4

33

34

9

7

9

7

9

7

81

63

b

ababa

ba

ba

11

4

11

22

11

23

11

42223

12

12

12

8

12

4

12

1284

yx

y

yx

yx

yx

yx

yx

yyxyx

)1(41)1(2)1(2)1(2)1(3

3

2

3

yxyxyx


Remember that 1/x-1 = x1. Now, simplify each division and leave

the answer with Positive exponents.

Now, we will do some examples of the third kind. Before doing that

however, let’s review how long division works.

Suppose we wanted to divide 4056 by 72. The 4056 is called the

Dividend and the 72 the Divisor. The answer of the division is

called the Quotient. If the 72 does not divide perfectly into the

4056, then there will be a Remainder.

The general equation is: Dividend = Quotient + Remainder

Divisor Divisor

In the next slide, I illustrate how to do this division.

534

513114

3

2

33

2

3xy

x

y

y

xyx

yxyx


You put the 72 to the left of the 4056 and a bar between them.

Then you ask yourself “How many times does the 72 go into the

405?” The answer is 5 times. Then you put the 5 over the 4056

and multiply it by 72 which gives 360. Now, write 360 below the

405 of the 4056 number. Subtract 405-360 and you get 45. Now,

bring down the 6 and attach it to the 45 and you get 456. Now,

repeat the process and you get a remainder of 24. This is

illustrated in the following picture

In light of the general equations, what this

means is: 4056 = 56 + 24

72 72

24/72 = .333333 and indeed, when you

do this division in the calculator, you get

56.33333333

56

0024

04320456

36004056

72


Let’s use this technique to divide a Polynomial by Polynomial.

Ex 1:

Step 1: “x” goes into (4x3) how many times? Answer: 4x2

Step 2: Put the (4x2) on top and multiply it by (x + 1).

Step 3: Write this answer below the 4x3 + x2 and Subtract.

Step 4: Bring down the (-3x) from the original Polynomial.

Step 5: “x” goes into (-3x2) how many times? Answer: -3x

Step 6: Put the (-3x) on top and multiply it by (x + 1).

Step 7: Write this answer below the -3x2 – 3x and Subtract.

1

734 23

x

xxx7341 23 xxxx

xx

xx

xxxx

x

04

30

444

1

2

23

23

23


Here is the picture of what happens so far:

Everything cancelled out when we

did the subtracting. Therefore, we

are done. Otherwise, we would

have had to do one more cycle.

The remainder is the 7 at the end of the original Polynomial.

Therefore, the Answer is

Ex 2:

Step 1: “2x” goes into (6x3) how many times? Answer: 3x2

Step 2: Put the (3x2) on top and multiply it by (2x + 3).

Step 3: Write this answer below the 6x3 + 23x2 and Subtract.

034

000

330330

04434

1

2

23

23

23

23

23

xx

xxx

xxxxxx

xxxxx

x

1

734

1

734 223

xxx

x

xxx

0272363232

62327 2332

xxxx

x

xxx


Step 4: Bring down the (27x) from the original Polynomial.

Step 5: “2x” goes into (14x2) how many times? Answer: 7x

Step 6: Put the (7x) on top and multiply it by (2x + 3).

Step 7: Write this answer below the 14x2 + 27x and Subtract.

Step 8: Bring down the Missing (0) from the original Poly.

003

27140

09627236

32

2

23

23

23

xx

xxx

xxxxxx

x

073

0600

021140027140

0096027236

32

2

23

23

23

23

23

xx

xxx

xxxxxx

xxxxxx

x


Step 9: “2x” goes into (6x) how many times? Answer: 3

Step 10: Put the (3) on top and multiply it by (2x + 3).

Step 11: Write this answer below the 6x + 0 and Subtract.

Step 12: Whatever the Subtraction yields is the Remainder.

373

9000

96000600

021140027140

0096027236

32

2

23

23

23

23

23

23

23

xx

xxx

xxxxxx

xxxxxx

xxxxxx

x

32

9373

32

62327 232

xxx

x

xxx


By Viken Kiledjian Section 1: is about Radical Expressions. We

first begin with the Square root expression

then we generalize to any nth order

expression.

Square Roots: if x2 = y, then x is the Square

Root of y denoted as

Actually, x can also equal the Negative of this

because when you square a negative, it

becomes a positive. The positive square root

of a number is called the Principle Square

Root.

What happens when you take the square root of

a square of a number?

If x > 0, then

If x < 0, then

yx

2x

xx 2

xx 2


In other words, If you square a number and then take its square root,

the result is always the Positive of the original number. This is

equivalent to the Absolute Value Function.

Therefore,

Ex 1:

Ex 2:

You don’t need the Absolute Value signs around the x2 because it is

always positive.

Ex 3:

Ex 4:

We can’t take the square root of -49 yet. We’ll learn how to deal

with these in Chapter 8.

The Nth Root of a Number is Written:

(n is a positive integer greater than 1).

xx 2

xxx 41616 22 2244 552525 xxxx

5)5(2510 22 xxxx

beraginaryNumxx Im4949 22

n x


If n is an Even integer, then we have to put Absolute Value signs

around our answer if the answer is x, x3, or any Odd power of a

variable as was state in the previous slide.

If n is an Odd integer, then we don’t ever have to put Absolute Value

signs because there is Only 1 odd root of a number.

The “n” is called the Index of the Radical Sign and the “x” is the

Radicand. When the n = 2, we call the radical a Square Root and

we don’t write the n. When the n = 3, we call the radical a Cube

Root but we begin to write the n.

What happens when you take the nth root of the nth power of a

variable?

As state above, when “n” is even You end up with

the original variable but you have to include the Absolute Value

signs.

When “n” is odd, No Absolute Value Sign needed!!!!

n nxxxn n

xxn n


Section 2 Ex 1:

Notice that you can take the Odd root of a Negative Number. It will

not cause any imaginary numbers to appear!!

Ex 2: You can use your Calculator

to take these general roots.

Ex 3: We don’t need absolute

values around the x2 !!!!

Section 2: is a continuation of the same concepts of Section 1

A Rational Exponent is any fractional exponent of a variable.

Examples:

bababa 23 33 633 36 7343343

2

3

32

243

32

2435

5

5

38181

1 2

4

4 8

48 xx

x

21

xx 31

3 xx 43

4 3 xx 32

64

6 4 xxx


Rational Exponents are another way of writing Radicals. They are

easier to understand and manipulate than radicals.

Whenever we want to Simplify a Radical Expression, it is often

easier to convert it to its equivalent Rational Expression and

simplify it and then convert it back to a Radical Expression.

In this section, we will also utilize all of the previous laws of

exponents that we learned in Chapter 1 and used in Ch 5.

Let’s do some examples of Simplifying Radicals and Rationals.

Ex 1:

Ex 2:

xxxx 33)()27()27( 3

1)3(3

133

13

13

3

2)3(3

2)3(2

31

32

332

332

32

33 27)()(27)27( bababa

22222 93 baba


Ex 3:

Ex 4:

43

43

443

43

443

4

16

)81(

16

81

81

16 yy

y

8

27

2

3

16

81 3

3

33

3

41

43)4(

3

41

yyy

34

32

2

31

32)2(

32

2

31

31

31)3(

31

32

2

31

3

2

3

8

)27(

)8(

)27(

yx

xy

yx

yx

xy

yx

y

xyxyx

4

3

4

3

2

3 31

131

2

34

31

321


Section 3 Now, let’s do some examples involving Distributing and multiplying

a Rational Expression by another Expression.

Ex 1:

Ex 2:

Section 3: is about Simplifying and Adding and Subtracting Radical

Expressions. We will again use the same techniques and

properties as we used in Sections 1 and 2.

34

35

34

32

34

35

32

34

43)43( xxxxxxxx

34

3234

35

34

32

34

4343 xxxxxx

2

12

12

12

12

21

21

xxxxxx

100121

21

21

21

21

21

21

21

xxxxxxxxxxxx

xx

xx 21

11 11


However, this time we will not always have a perfect radicand where

the Radical Sign cancels after simplification.

Ex 1:

Notice that the final answer still has a radical sign that is not

simplifiable any further. We will use 2 very important properties

of Radicals:

Multiplication property of radicals:

Quotient property of radicals:

When we add radicals together, we first simplify each one as much as

possible and then we add Like Radicals.

Ex 2:

6264)6)(4(24

nnn yxxy

n

n

n

y

x

y

x

21624)2)(16()2)(4(328

262422


Let’s do some more complex examples of Simplifying and Adding

and Subtracting Radicals.

Ex 3:

Ex 4:

Ex 5:

Ex 6:

bbab

ab

ab

ab416

7

112

7

112 233

)2)(64()2)(64(128 424253 abbabababa

abbaabbaabba 28282)64( 2242

2

3 2

31

63

3 2

3 6

3 2

36

2

5

11

125

11

125

11

125

11

b

a

b

a

b

a

b

a

225229)2)(25(2)2)(9(50218

213210232)5(223


Ex 7:

Ex 8:

Notice that you can’t combine the with the

Treat it as if it was: 4x – 8y + 12y = 4x + 4y

Ex 9:

4444 )3)(16()3)(81(4482434

444444444 310323123)2(3)3(43163814

)2)(144()2)(64()5)(16(28812880

245421228542144264516

5 2

3 433 433 43 43 43 4 27852785 xyxyxyxyxyxy

3 333 33 43 43 43 4 444325 yxyxyyxyxyxyxy

33 )4()(4 xyyyxy


Section 4: is about Multiplying and Dividing Radicals and

Rationalizing Radicals.

We will use the Multiplication and Quotient property of radicals.

Let’s do some examples of multiplying radicals first, then we’ll learn

the concept of Rationalizing radicals.

Ex 1:

Another approach to this problem would be the following:

I am usually going to use the 1st approach where I Multiply the

radicals first, then I simplify the combined radical.

Ex 2:

57549)5)(49()5)(7)(7()35)(7(357

575495)7)(7(577)5)(7(7357

3 23 23 333 23 232)27()2)(9)(3( yxxyxxyxx

3 533 4 )18)(3(183 yxxyx


Ex 3:

Ex 4: We can use the Special Form formula

(a – b)2 = a2 – 2ab + b2

Rationalizing Fractions:

When the denominator of a fraction contains a non-reducible Radical

sign, we rationalize the fraction by multiplying BOTH the

Numerator and the Denominator by the Conjugate of the Radical

expression. After the process is done, all of the radical signs will

only be found in the Numerator.

103)2()5(7375327 yyyyyy

1029211063521 yyyyy

2353 x

315253335253

22

xxxx

915615 xx


The conjugate of a Radical Expression is another Radical Expression

that when multiplied by the original will yield an integer and will

eliminate the Radical sign.

Ex: The conjugate of is itself, because

The conjugate of is also itself, because

The conjugate of is because

The conjugate of is because

We multiplied 24 by the number needed to achieve the smallest

possible perfect cube root number. The perfect cubes are 8, 27,

64, 216, 343, etc. This is trickier than the square root case. The

best way is to divide the smallest perfect cube root by the original

number (24) until you get an integer. 24 does not go into 27 and

64 but does into 216.

x xxx

x3 xxx 333

3 x3 2x xxxx 3 33 23

3 224x3 9x

xxxx 6216924 3 333 2


The conjugate of a Binomial Radical expression such as

is because (a + b)(a – b) = a2 – b2

Let’s do some example of Rationalizing.

Ex 1:

Or we can do it another way even easier:

23

12323232322

23

abc

cba

abcabc

abccab

abc

cab

10

50

1010

105

10

5 23222

2

2

10

25

10

)2)(25( 222b

abc

babc

abc

bcba

2

2

22

2

2210

5

10

5 22 bbbb

abc

cab

abc

cab


Ex 2:

Ex 3:

Ex 4:

Ex 5:

2

16

2

16

22

2)1(

2

1 5

5 5

5

5 45

5 4

5

)2)(2(33

3233

2323

233

23

3

32343

323

yxyx

yxyx

yx

yx

yx

yxyx

yyxx

yyyxyxxx

2

13

13

1313

1313

1313

13

13

t

t

tt

tt

tt

t

t


Section 5: is about Solving Equations which contain 1, 2, or 3

Radical Expressions. The basic idea is to square both sides of the

equal sign until you get rid of all Radicals.

We use the following property of square roots:

In cases where you have Radicals with indexes other than 2, we use

the appropriate property such as:

Let’s do some examples involving ONLY 1 radical.

Ex 1: Add 2 to both sides and you’ll get.

Now, square both sides of the equal sign.

Ex 2:

xx 2

xx 3

3 xx 4

4 xx 5

5

52136 x

7136 x

6366491367136 22

xxxx

444

444 2221022210 yy

3)2)(16(2102)2(2104

44 yyy


Ex 3: Square both sides of the equal sign

You can factor the (-1) out of the –s – 3 and get –(s + 3). When you

square this, it is just like squaring (s + 3) since -1 squared is

+1!!!! Notice that 2 squared is 4;don’t forget that.

Sometimes, when you get two answers, one of them is NOT a true

solution. You have to check them both to know which one is the

solution or if they both are solutions.

CHECK: s = -11

s = 1

So, the answer is ONLY s = -11.

Now, let’s do examples with Two Radicals.

ss 523

)5(496523 222

sssss

0111042096 22 sssss

0)1)(11( ss 11s 1s

1628)11(523)11(

42415231


Ex 1:

Now, that was easy because there was nothing else besides the two

radicals. When there are other stuff, we usually separate the

Radicals on either side of the equal sign and then square both

sides. This eliminates one of the Radicals but not the other. Then,

we expand the other side and Isolate the remaining Radical and

Square both sides Again. It is a pretty arduous task but well worth

the effort!!

Ex 2:

(s = 0 does NOT work)

22

326326 xxxx

133326 xxxx

16141614 ssss

162614161422

sssss

222

666262 ssssssss

60)6(062 sssss


Ex 3:

There is NO Solution to this problem because the Square Root of a

Number can not be negative.

Let’s do some examples of equations with 3 Radicals or Radicals

within Radicals.

Ex 1:

248248 xxxx

444)4(824822

xxxxx

24448 xx

22

48384838 xxxxxx

4)83(8382)8( xxxxx

228382483824 xxxxxx


Ex 1: (Continuation)

When you check the answers, 68/13 Does NOT check out.

You get a Negative on the left side of the equal sign in the original

equation. Therefore, the answer is Only x = 4

Ex 2:

)83)(8(41682 xxxx

02721201325612812168 222 xxxxxx

4,13

680)4)(6813( xxxx

2

2

112112 yyyy

22

112112112 yyyyyy

22 1212112)1(2 yyyyyyy

20)2(044)1(4 222 yyyyyy

Section 1: covers the topics of Completing

the Square and the Quadratic Formula.

The Square Root Property:

If c > 0, then the equation x2 = c has two

solutions given by and

We can use this property to solve problems

like the following:

Ex 1: 5x2 – 49 = 0

5x2 = 49 Therefore, x2 = 49/5 and

Ex 2: (x + 3)2 – 7 = 0 (x + 3)2 = 7


By Viken Kiledjian

cx cx

5

57

5

57

55

57

5

7

5

49 Orx

3773 xx


In situations where there are more than one power of x, we can either try to Factor the expression like we did in Chapter 5 or we can Complete the Square. Of course, whenever the expression is factorable, it is easier and faster to Factor it. Refer to Chapter 5 for examples of factorable equations. Now we will learn to complete the square. These are the steps:

1) Make sure that the coefficient of x2 is 1. If it is not, then divide both sides of the equal sign by its coefficient to make it 1.

2) Take the Constant (last) term of the quadratic equation to the opposite side of the equal sign if it is not there already.

3) Take ½ of the Coefficient of x and square it.

4) Add this squared value to both sides.

5) Factor the Perfect Square trinomial expression.

6) Using the Square Root property as in the previous slide, square root both sides of the equation and solve for x.


Ex 1: x2 + 6x + 5 = 0 (Even though this is a Factorable expression,

we will Complete the Square for purposes of learning. If we

factor it, we’ll get (x + 5)(x + 1) = 0. Therefore, x = -5 or -1.)

1) The coefficient of x2 is already 1.

2) Take the Constant term to the other side: x2 + 6x = -5

3) Take ½ of the coefficient of x and square it: 6/2 = 3 and 32 = 9

4) Add this to both sides: x2 + 6x + 9 = -5+9 = 4

5) Factor the Perfect Square trinomial. (x + 3)2 = 4

6) Use the square root property:

So we got the same answer as we would have gotten with factoring!!

Ex 2: 11w – 10 = 3w2 3w2 – 11w + 10 = 0

1) Divide everything by 3: w2 – 11/3w + 10/3 = 0

2) Take the Constant term to the other side: w2 – 11/3w = -10/3

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3) Take ½ of the coefficient and square it: (11/3)/2 = 11/6 and

(11/6)2 = 121/36

4) Add this to both sides: w2 – 11/3w + 121/36 = -10/3 + 121/36

w2 – 11/3w + 121/36 = 1/36

5) Factor the perfect square trinomial: (w – 11/6)2 = 1/36

[When you factor the perfect square trinomial, the number in the

Parenthesis will ALWAYS be the number you got when you divided

the Coefficient of x by 2. In this case, it is 11/6]

6) Use the Square Root property to solve for x:

As you noticed, this method is relatively slow and impracticable.

However, we can come up with a general formula for solving any

Quadratic equation by using the Completing the Square Method.

3/526/116/16/136

16/11 orww


This general formula is known as the Quadratic Formula. For any

quadratic equation of the form, ax2 + bx + c = 0, the solution is:

The term inside the Radical sign, b2 – 4ac, is

known as the Discriminant. If the

Discriminant is Less than zero, then the solution for x is two

Imaginary Numbers. If the Discriminant = 0, then there is Only 1

solution to the quadratic equation. This solution is called a Double

Root, and it indicates that the original quadratic equation is a Perfect

Square root trinomial.

If the Discriminant is Positive and a perfect square, then there are 2

Rational solutions for x, and it indicates that the original quadratic

equation is Factorable. Lastly, if the Discriminant is Positive but

NOT a perfect square, then there are 2 Irrational solutions for x, and

it indicates that the original equation is Not Factorable.

a

acbbx

2

42


Ex 1: 4w2 + 6w + 1 = 0 Here, a = 4, b = 6, c = 1. Therefore,

Ex 2: 3x = x2/2 + 2 6x = x2 + 4 x2 - 6x + 4 = 0

Here, a = 1, b = -6, c = 4

8

526

8

206

8

16366

)4(2

)1)(4(466 2

x

4

53

8

532

x

2

526

2

206

2

16366

)1(2

)4)(1(4)6()6( 2

x

53

2

532

x


Section 2: In this section, we learn how to graph quadratic

equations, which are parabolic in shape.

We first learn how to graph the basic equation, y = x2. Then, we

learn the techniques of Translating (moving) this graph up/down and

left/right and Inverting the graph upside down and Making it more

Open or Closed.

The graph of y = x2 is a parabola whose Vertex is the origin and it

opens upward and goes through the points (1,1) and (-1,1).

It looks like the following:

1) The graph of y = -x2 is the upside

down graph opening toward the

bottom. Therefore, a Negative sign

in front of the x2 Inverts the graph.


2) The graph of y = (x – 3)2 is Translated to the Right 3 units. Its

vertex is the point (3, 0). This is because when x = 3, y = 0.

Its graph looks like the following.

The graph of y = (x + 2)2 is

Translated 2 units to the Left.

Its vertex is the point (-2, 0).

This is because when x = -2,

y = 0. Its graph looks like this.

Therefore, equations of the form

y = (x – h)2 entail Horizontal

Translations and yield parabolas

Whose Vertex = (h, 0)


3) The graph of y = x2 + 4 is translated Upward by 4 units. Its

vertex = (0, 4) because when x = 0, y = 4.

Its graph looks like the following:

The graph of y = x2 – 2 is

Translated Downward by 2 units.

Its Vertex = (0, -2) because when

x = 0, y = -2. Its graph looks like this:

Therefore, equations of the form

y = x2 + k entail Vertical

Translations and yield parabolas

Whose Vertex = (0, k)


4) The graph of y = 3x2 has a vertex at (0, 0) because when

x = 0, y = 0. It opens upward because the coefficient of x2 is

positive. However, it is more Closed and Steeper than the graph of

y = x2 because it goes through the points (1, 3) and (-1, 3). Its graph

looks like the following:

The graph of

y = x2/4 has a vertex at (0,0)

because when x = 0, y = 0.

It also opens upward, but it

is more Open and Wider than the

Graph of y = x2 because it goes through the

Points (2, 1) and (-2, 1). Its graph looks like this:


Therefore, the graph of the equation y = ax2 is more Closed and

Steeper if a > 1 and more Open and Wider if a < 1.

Now, let’s do a General One which combines some of these effects.

If the original equation is NOT given in factored form, we have to

use the method of Completing the Square from Section 1 to factor it.

Ex 1: Graph the equation: y = 3x2 – 3

Solution: This will be Translated downward 3 units and more

Closed than y = x2. It will go through the points (1, 0) and (-1, 0)


Ex 2: Graph the equation: y = 3x2 + 6x

Solution: We have to Complete the Square to Factor it!!

1) Factor the 3 out: y = 3(x2 + 2x)

2) Take ½ of the Coefficient of x and Square it: 2/2 = 1 and 12 = 1

3) Add this to the Expression inside the Parenthesis to Make it a

Perfect Square: y = 3(x2 + 2x + 1)

4) Since 3(1) = 3, Subtract 3 from this equation so that the

Original Quadratic remains unchanged!!!!!

y = 3(x2 + 2x + 1) – 3 = 3(x + 1)2 – 3

The Graph of this Equation is Translated

to the left 1 unit, down 3 units

and opens upward and is

More Closed and Steeper than y = x2.


Ex 3: Graph the equation: y – 2 = 3x2 + 4x

1) Factor the 3 out: y - 2 = 3(x2 + 4/3x)

2) Take ½ of the Coefficient of x and Square it: (4/3)/2 = 2/3 and

(2/3)2 = 4/9

3) Add this to the Expression inside the Parenthesis to Make it a

Perfect Square: y – 2 = 3(x2 + 4/3x + 4/9)

4) Since 3(4/9) = 4/3, Subtract 4/3 from this equation so that the

Original Quadratic remains unchanged and add 2 to both sides!!

y = 3(x + 2/3)2 – 4/3 + 2 = 3(x + 2/3)2 + 2/3

The Graph of this equation is

Translated to the Left 2/3 units,

Up 2/3 units and is Steep like

The graph on the previous page.

Documents

Intermediate Algebra Chapter 1 - East Los Angeles College · Intermediate Algebra Chapter 1: Section 1 Examples: 2,3,5,7,11,13,17 … are all prime numbers e) Composite Numbers are