Alg Complete Solutions

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COLLEGE ALGEBRASolutions to Practice Problems

Paul Dawkins


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College Algebra

Table of Contents

Preface ............................................................................................................................................ 2Preliminaries .................................................................................................................................. 3

Integer Exponents ...................................................................................................................................... 3Rational Exponents .................................................................................................................................... 6Real Exponents .........................................................................................................................................11Radicals .....................................................................................................................................................12Polynomials ...............................................................................................................................................20Factoring Polynomials ..............................................................................................................................24Rational Expressions .................................................................................................................................33Complex Numbers ....................................................................................................................................40

Solving Equations and Inequalities ............................................................................................ 44Solutions and Solution Sets .......................................................................................................................44Linear Equations .......................................................................................................................................47Application of Linear Equations ...............................................................................................................54Equations With More Than One Variable .................................................................................................61Quadratic Equations Part I .....................................................................................................................66Quadratic Equations Part II ....................................................................................................................75Solving Quadratic Equations : A Summary ..............................................................................................84Application of Quadratic Equations ..........................................................................................................86Equations Reducible to Quadratic Form ...................................................................................................91Equations with Radicals ............................................................................................................................96Linear Inequalities...................................................................................................................................100Polynomial Inequalities ...........................................................................................................................106Rational Inequalities ...............................................................................................................................113Absolute Value Equations .......................................................................................................................121Absolute Value Inequalities ....................................................................................................................126

Graphing and Functions ........................................................................................................... 130Graphing .................................................................................................................................................130Lines ........................................................................................................................................................137Circles .....................................................................................................................................................146

The Definition of a Function ...................................................................................................................152Graphing Functions .................................................................................................................................164Combining Functions ..............................................................................................................................170Inverse Functions ....................................................................................................................................173

Common Graphs ....................................................................................................................... 180Lines, Circles and Piecewise Functions ..................................................................................................180Parabolas .................................................................................................................................................180Ellipses ....................................................................................................................................................193Hyperbolas ..............................................................................................................................................200Miscellaneous Functions .........................................................................................................................206Transformations ......................................................................................................................................207Symmetry ................................................................................................................................................213Rational Functions ..................................................................................................................................217

Polynomial Functions ................................................................................................................ 223Dividing Polynomials .............................................................................................................................223Zeroes/Roots of Polynomials ..................................................................................................................228Graphing Polynomials .............................................................................................................................231Finding Zeroes of Polynomials ...............................................................................................................238Partial Fractions ......................................................................................................................................245

Exponential and Logarithm Functions .................................................................................... 256Exponential Functions .............................................................................................................................256Logarithm Functions ...............................................................................................................................262

2007 Paul Dawkins i http://tutorial.math.lamar.edu/terms.aspx


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Solving Exponential Equations ...............................................................................................................274Solving Logarithm Equations .................................................................................................................281Applications ............................................................................................................................................294

Systems of Equations ................................................................................................................. 300Linear Systems with Two Variables .......................................................................................................300Linear Systems with Three Variables .....................................................................................................306Augmented Matrices ...............................................................................................................................308

More on the Augmented Matrix ..............................................................................................................311Non-Linear Systems ................................................................................................................................323

Preface

Here are the solutions to the practice problems for my Calculus I notes. Some solutions will have more orless detail than other solutions. The level of detail in each solution will depend up on several issues. Ifthe section is a review section, this mostly applies to problems in the first chapter, there will probably not

be as much detail to the solutions given that the problems really should be review. As the difficulty levelof the problems increases less detail will go into the basics of the solution under the assumption that ifyouve reached the level of working the harder problems then you will probably already understand thebasics fairly well and wont need all the explanation.

This document was written with presentation on the web in mind. On the web most solutions are brokendown into steps and many of the steps have hints. Each hint on the web is given as a popup however inthis document they are listed prior to each step. Also, on the web each step can be viewed individually byclicking on links while in this document they are all showing. Also, there are liable to be some formattingparts in this document intended for help in generating the web pages that havent been removed here.These issues may make the solutions a little difficult to follow at times, but they should still be readable.

2007 Paul Dawkins ii http://tutorial.math.lamar.edu/terms.aspx


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Preliminaries

Integer Exponents

1. Evaluate the following expression and write the answer as a single number without exponents.2 26 4 3 +

SolutionThere is not really a whole lot to this problem. All we need to do is the evaluations recalling the properorder of operations.

( )2 26 4 3 36 4 9 36 36 0 + = + = + =

Be careful with the first term and recall that,

( ) ( )2 26 6 36 36 = = =

If wed wanted the minus sign to also get squared wed have written,

( )2

6 36 =

Always remember to be careful with exponents. The only thing that gets the exponent is the number/term

immediately to the left of the exponent. If we want to include minus signs on numbers with exponentsthen we need to add in parenthesis.

2. Evaluate the following expression and write the answer as a single number without exponents.

( )

( )

4

22 2

2

3 2

+

SolutionThere is not really a whole lot to this problem. All we need to do is the evaluations recalling the proper

order of operations.

( )

( ) ( ) ( )

4

2 2 22 2

2 16 16 16

1699 4 133 2

= = =

++

Remember that we need to do the evaluations inside the parenthesis in the denominator before we dealwith the overall exponent that is on the parenthesis.

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3. Evaluate the following expression and write the answer as a single number without exponents.0 2

1 2

4 2

3 4


( ) ( ) ( )0 2 0 1 21 2 2

1 3 164 2 4 3 412

3 4 2 4

= = =

It is almost always going to be best to first get rid of negative exponents prior to doing any of the rest ofthe evaluation work. Also, dont forget to reduce any resultant fractions down as much as possible.

4. Evaluate the following expression and write the answer as a single number without exponents.1 12 4 +


1 1 1 1 32 4

2 4 4

+ = + =

It is almost always going to be best to first get rid of negative exponents prior to doing any of the rest ofthe evaluation work. Also, make sure you can add/subtract fractions! Were going to be running into alot of fractions here and you need to be able to work with those.

5. Simplify the following expression and write the answer with only positive exponents.

( ) 2

4 52w v

SolutionThere is not really a whole lot to this problem. All we need to do is use the properties from this section todo the simplification.

( )10 10

24 5 2 8 10

2 8 82 2

2 4

v vw v w v

w w

= = =



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Note that there are several paths (i.e.the order in which you chose to use the properties) you can take todo the simplification. Each will end up with the same answer however and so you dont need to getexcited if you chose a different order in which to use the properties than we did here.


4 1

6 3

2x y

x y


4 1 4 6 10

6 3 3 1 4

2 2 2x y x x x

x y y y y

= =

Note that there are several paths (i.e.the order in which you chose to use the properties) you can take todo the simplification. Each will end up with the same answer however and so you dont need to getexcited if you chose a different order in which to use the properties than we did here.


2 10

7 3

m n

m n


2 10 7 3 5

7 3 2 10 7

m n m n m

m n m n n

= =


( )( )

32 4

1 7

2

6

p q

q p

Solution



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There is not really a whole lot to this problem. All we need to do is use the properties from this section todo the simplification.

( )( )

32 4 3 6 4 1 7 4 1 1 5 5

1 1 1 7 3 67

2 2 6 6 3

6 2 8 46

p q p q p q q p q pq

q p pq p

= = = =

Dont try to do use too many properties all at once. Sometimes it is very easy to use too many propertiesall in one step and make a mistake. Theres nothing wrong with using only a single property or two witheach step.


42 1 3

8 6 4

z y x

x z y


4 4 4 42 1 3 2 8 5 4 5 16 20

8 6 4 3 6 1 4 4 5 5 20

z y x z x x z y z y

x z y x z y y z y x x

= = = =

In this case since there was a fair amount of simplification that could be done on the fraction inside theparenthesis so we decided to do that simplification prior to dealing with the exponent on the parenthesis.

Rational Exponents

1. Evaluate the following expression and write the answer as a single number without exponents.1

236

Hint : Recall that

1

nb is really asking what number did we raise to the nto get b. Or in other words,1

? is equivalent to ?nnb b= =

Solution

For this problem we know that2

6 36= and so we also know that,



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1

236 6=

Note that if you arent sure of the answer to these kinds of problems all you really need to do is set up2? 36=

and start trying integers until you get the one you need.

2. Evaluate the following expression and write the answer as a single number without exponents.

( )1

3125

Hint : Recall that

1

nb is really asking what number did we raise to the nto get b. Or in other words,1

? is equivalent to ?nnb b= =

Solution

For this problem we know that 35 125= . Therefore we also know that ( )3

5 125 = and so we further

know that,

( )1

3125 5 =

Note that if you arent sure of the answer to these kinds of problems all you really need to do is set up3? 125=

and start trying integers until you get the one you need. We also know that because the result is anegative number we had to have a negative number to start off with since we cant turn a positive numberinto a negative number simply by raising it to an integer.


216

Hint : Dont forget your basic exponent rules and how the first two practice problems worked. Also, becareful with minus signs in this problem.

Step 1First, lets write the problem as,

3

216

so we arent tempted to bring the minus sign into the exponent.



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Now, lets recall our basic exponent rules and note that we can easily write this as,

33 1

2 216 16 =

Step 2Now, recalling how the first two practice problems worked we can see that,

1

216 4=

because 24 16= .

Therefore,

( )( ) ( )3

3 3 13

2 2 216 16 16 4 64 64 = = = = =

Sometimes the easiest way to do these kinds of problems when you first run into them is to break them upinto manageable steps as we did here.


327

Hint : Dont forget your basic exponent rules and how the first two practice problems worked.

Step 1Lets first recall our basic exponent rules and note that we can easily write this as,

5

35 5

13

3

1 127

27 27

= =


1

327 3=

because 33 27= .

Therefore,



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( )

5

35 5 5

13

3

1 1 1 127

243327 27

= = = =

Sometimes the easiest way to do these kinds of problems when you first run into them is to break them upinto manageable steps as we did here.


29

4



1122

1

2

9 9

44

=


1 1

2 2

9 3 4 2= =

Therefore,1122

1

2

9 9 3

4 24

= =


3

8343





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21

322 233 3

2 21

33

3438 343 343

343 88 8

= = =


1 1

3 3343 7 8 2= =

Therefore,2

1

322 2233 3

2 2 213 3

3438 343 343 7 49

343 8 2 48 8

= = = = =

7. Simplify the following expression and write the answer with only positive exponents.2

1 33 4a b

Solution

There isnt really a lot to do here other than to use the exponent properties from the previous section to dothe simplification.

211 23

3 2 641

6

aa b a b

b

= =


54x x

SolutionThere isnt really a lot to do here other than to use the exponent properties from the previous section to dothe simplification.

1 1 1 11

5 4 5 204x x x x

= =



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1 73 2

1

3

q p

q p


3 331 10 101 7 77

3 3 3 3 72

1 1 3 9

3 2 2 14

q p q q q q

pp p pq p

= = =


11 632

2 7

3 4

m n

n m


1 1 1 11 11 1 7 96 6 6 63 62 2 4 4

2 2 1 9 37 1

3 3 3 4 84

m n m m m n n

nmn m n n m

= = = =

Real Exponents


( ) 2.4

0.1 0.3x y

Solution



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There is not really a whole lot to this problem. Do not get excited about the fact that the exponents arentintegers or rational numbers. The properties from the integer exponent section still work! So, all we needto do is use them to do the simplification.

( ) ( )( ) ( )( )0.72

2.4 0.1 2.4 0.3 2.40.1 0.3 0.24 0.72

0.24

yx y x y x y

x

= = =


( ) ( )3 1.8

0.15 4x y

SolutionThere is not really a whole lot to this problem. Do not get excited about the fact that some of theexponents arent integers or rational numbers. The properties from the integer exponent section still

work! So, all we need to do is use them to do the simplification.

( ) ( ) ( )( ) ( )( )3 1.8 0.15 3 4 1.80.15 4 0.45 7.2

0.45 7.2

1x y x y x y

x y

= = =


1.53.2 0.7

6.4 1.9

p q

q p

SolutionThere is not really a whole lot to this problem. Do not get excited about the fact that the exponents arentintegers or rational numbers. The properties from the integer exponent section still work! So, all we needto do is use them to do the simplification.

( )1.5 1.5

3.2 0.7 3.2 6.4 1.91.5

5.1 5.7 7.65 8.55

6.4 1.9 0.7 7.65 8.55

1p q p q pp q p q

q p q p q

= = = =

Radicals

1. Write the following expression in exponential form.



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7 y

SolutionAll this problem is asking us to do is basically use the definition of the radical notation and write this in

exponential form instead of radical form.

1

7y


3 2x

SolutionAll this problem is asking us to do is basically use the definition of the radical notation and write this inexponential form instead of radical form.

( )1

2 3x


6 ab


( )1

6ab

Be careful with parenthesis here! Recall that the only thing that gets the exponent is the term immediatelyto the left of the exponent. So, if wed dropped parenthesis wed get,

1 1

66 6ab a b a b

= =

which is most definitely not what we started with. The only way to make sure that we understand thatboth the aand the bwere under the radical is to use parenthesis as we did above.



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2 3w v


( )1

2 3 2w v

Recall that when no index is written on the radical it is assumed to be 2.

Also, be careful with parenthesis here! Recall that the only thing that gets the exponent is the termimmediately to the left of the exponent and so we need parenthesis on the whole thing to make sure thatwe understand that both terms were under the root.

5. Evaluate : 4 81

Hint : Recall that the easiest way to evaluate radicals is to convert to exponential form and then also recallthat we evaluated exponential forms in the Rational Exponent section.

SolutionAll we need to do here is to convert this to exponential form and then recall that we learned how toevaluate the exponential form in the Rational Exponent section.

1

44 481 81 3 because 3 81= = =

6. Evaluate : 3 512



( ) ( )1

33 3512 512 8 because 8 512 = = =



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7. Evaluate : 3 1000



1

33 31000 1000 10 because 10 1000= = =

8. Simplify the following expression. Assume thatxis positive.

3 8x

Step 1Recall that by simplify we mean we want to put the expression in simplified radical form (which wedefined in the notes for this section).

To do this for this expression well need to write the radicand as,

( )3

8 6 2 2 2x x x x x= =

Step 2Now that weve gotten the radicand rewritten its easy to deal with the radical and get the expression in

simplified radical form.

( ) ( )3 3

3 3 38 2 2 2 2 2 23 3x x x x x x x= = =

9. Simplify the following expression. Assume thatyis positive.

38y



( )( )3 28 4 2y y y=



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Step 2Now that weve gotten the radicand rewritten its easy to deal with the radical and get the expression insimplified radical form.

( )( )3 2 28 4 2 4 2 2 2y y y y y y y= = =

10. Simplify the following expression. Assume thatx,yandzare positive.

7 20 114 x y z



( ) ( )4 4

7 20 11 4 20 8 3 3 4 5 2 3 3x y z x y z x z x y z x z= =


( ) ( ) ( ) ( )4 4 4 4

7 20 11 4 5 2 3 3 4 5 2 3 3 5 2 3 34 44 4 4x y z x y z x z x y z x z x y z x z= = =


6 7 23 54x y z



( )( ) ( ) ( ) ( )3 3

6 7 2 6 6 1 2 3 2 2 254 27 2 3 2x y z x y y z x y yz= =




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( ) ( ) ( ) ( ) ( )3 3 3 3

6 7 2 3 2 2 2 3 2 2 2 2 2 23 3 33 354 3 2 3 2 3 2x y z x y yz x y yz x y yz= = =


3 2 3 54 44 8x y x y z

Step 1Remember that when we have a product of two radicals with the same index in an expression we firstneed to combine them into one root before we start the simplification process.

( )( )3 2 3 5 3 2 3 5 5 4 54 4 444 8 4 8 32x y x y z x y x y z x y z= =

Step 2Now that the expression has been written as a single radical we can proceed as we did in the earlierproblems.

The radicand can be written as,

( )( )5 4 5 4 4 4 432 2 2x y z x y z xz=


3 2 3 5 5 4 5 4 4 4 4 4 44 4 4 44 8 32 2 2 2 2x y x y z x y z x y z xz xyz xz= = =

13. Multiply the following expression. Assume thatxis positive.

( )4 3x x

SolutionAll we need to do here is do the multiplication so here is that.

( ) ( ) 24 3 4 3 4 3 4 3x x x x x x x x x = = =

Dont forget to simplify any resulting roots that can be. That is an often missed part of these problems.



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( )( )2 1 3 4x x+

Solution

All we need to do here is do the multiplication so here is that.

( )( ) ( ) 22 1 3 4 6 8 3 4 3 2 8 3 2 8x x x x x x x x x x+ = + = + = +

Dont forget to simplify any resulting roots that can be. That is an often missed part of these problems.


( )( )3 32 23

2 4x x x+

SolutionAll we need to do here is do the multiplication so here is that.

( )( )3 3 3 3 3 32 2 2 2 2 23 3 33 3 33 2 43

3 3 33 2 33 3

3 23 3

2 4 4 8 2

4 8 2

4 8 2

4 8 2

x x x x x x x x x

x x x x

x x x x x

x x x x x

+ = +

= +

= +

= +

Dont forget to simplify any resulting roots that can be. That is an often missed part of these problemsand when dealing with roots other than square roots there can be quite a bit of work in the simplificationprocess as we saw with this problem.

16. Rationalize the denominator. Assume thatxis positive.

6

x

Solution

For this problem we need to multiply the numerator and denominator by x in order to rationalize the

denominator.



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2

6 6 6 6x x x

xx x x x= = =


3

9

2x

Solution

For this problem we need to multiply the numerator and denominator by ( )2

3 2x in order to rationalize

the denominator.

( )( )

( )( )

( )

2 2 23 3 3 3 2

3 3 2 33 3

2 9 2 9 29 9 9 42 22 2 2 2

x x x xx xx x x x

= = = =

18. Rationalize the denominator. Assume thatxandyare positive.

4

2x y+

SolutionFor this problem we need to multiply the numerator and denominator by 2x y in order to

rationalize the denominator.

( )( )( )

4 22 4 84 4

42 2 2 2 2

x yx y x y

x yx y x y x y x y x y

= = =

+ + +


10

3 5 x

Solution

For this problem we need to multiply the numerator and denominator by 3 5 x+ in order to rationalizethe denominator.



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( )( )( )

10 3 510 10 3 5 30 50

9 253 5 3 5 3 5 3 5 3 5

xx x

xx x x x x

++ += = =

+ +

Polynomials

1. Perform the indicated operation and identify the degree of the result.

Add 3 24 2 1x x + to 27 12x x+

Step 1Here is the operation were being asked to perform.

( ) ( )3 2 24 2 1 7 12x x x x + + +

Note that the parenthesis are only there to illustrate how each polynomial is being used in the indicatedoperation and are not needed (or used) in general.

Heres the result of the operation.

( ) ( )3 2 2 3 24 2 1 7 12 4 5 12 1x x x x x x x + + + = + + +

Step 2

Remember the degree of a polynomial is just the largest exponent in the polynomial and so thedegree of the result of this operation is three.


Subtract 6 24 3 2z z z + from 6 210 7 8z z +


( )6 2 6 210 7 8 4 3 2z z z z z + +

Be careful with the order here! We are subtracting the first polynomial from the second and that impliesthe order weve got here. Also be careful with the parenthesis on the second polynomial. We aresubtracting the whole polynomial and so we need to have the parenthesis to do that.



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( )6 2 6 2 6 2 6 2

6 2

10 7 8 4 3 2 10 7 8 4 3 2

14 10 2 8

z z z z z z z z z z

z z z

+ + = + +

= +

Step 2

Remember the degree of a polynomial is just the largest exponent in the polynomial and so thedegree of the result of this operation is six.


Subtract23 7 8x x + + from 4 37 12 1x x x+


( )4 3 27 12 1 3 7 8x x x x x+ + +

Be careful with the order here! We are subtracting the first polynomial from the second and that impliesthe order weve got here. Also be careful with the parenthesis on the second polynomial. We aresubtracting the whole polynomial and so we need to have the parenthesis to do that.


( )4 3 2 4 3 24 3 2

7 12 1 3 7 8 7 12 1 3 7 8

7 3 19 9

x x x x x x x x x x

x x x x

+ + + = + +

= + +

Step 2

Remember the degree of a polynomial is just the largest exponent in the polynomial and so thedegree of the result of this operation is four.


( )4 212 3 7 1y y y +

Step 1All we need to do is multiply the 12ythrough the second polynomial. Here is the result of doing that.

( )4 2 5 312 3 7 1 36 84 12y y y y y y + = +



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Step 2

Remember the degree of a polynomial is just the largest exponent in the polynomial and so thedegree of the result of this operation is five.


( )( )23 1 2 9x x+

Step 1All we need to do is foil out the two polynomials. Here is the result of doing that.

( )( )2 2 33 1 2 9 2 6 9 27x x x x x+ = +

Step 2Remember the degree of a polynomial is just the largest exponent in the polynomial and so thedegree of the result of this operation is three.


( )( )2 22 3w w w+ +

Step 1

All we need to do is foil out the two polynomials. Here is the result of doing that.

( )( )2 2 4 3 22 3 3 6 2w w w w w w w+ + = + + +

Step 2



( ) ( )6 64 3 4 3x x x x +

Step 1All we need to do is foil out the two polynomials. Here is the result of doing that.

( )( )6 6 12 24 3 4 3 16 9x x x x x x + =



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Step 2

Remember the degree of a polynomial is just the largest exponent in the polynomial and so thedegree of the result of this operation is twelve.


( )2

33 10 4y

Step 1Remember that this is just another way of writing,

( ) ( )( )2

3 3 33 10 4 3 10 4 10 4y y y =

Now all we need to do is foil out the two polynomials. Here is the result of doing that.

( ) ( )( ) ( )2

3 3 3 3 6 3 63 10 4 3 10 4 10 4 3 100 80 16 300 240 48y y y y y y y = = + = +

Be careful with dealing with the three! Make sure you take care of the exponent first (i.e.make sure youmultiply out the product first) before you multiply the three through the result.

Step 2

Remember the degree of a polynomial is just the largest exponent in the polynomial and so thedegree of the result of this operation is six.


( )( )2 22 3 8 7x x x x+

Step 1Remember that the foil method only works for binomials and these are both trinomials (i.e.they eachhave three terms).

So, all we need to do is multiply each term in the second polynomial by each term in the first polynomial.

Here is the result of doing that.

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

4 3 2

2 3 8 7 3 8 7 3 8 7 6 16 14

3 5 21 9 14

x x x x x x x x x x x x

x x x x

+ = + + +

= + +

Step 2



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Subtract ( )2

23 1x + from 3 26 9 13 4x x x


( )2

3 2 26 9 13 4 3 1x x x x +

Now, before we actually do the subtraction we need to actually multiply out the second term before we dothe subtraction. Here are the results of all these operations.

( ) ( )2

3 2 2 3 2 4 2

3 2 4 2

4 3 2

6 9 13 4 3 1 6 9 13 4 3 2 1

6 9 13 4 3 6 3

3 6 15 13 7

x x x x x x x x x

x x x x x

x x x x

+ = + +

=

= +

Step 2


Factoring Polynomials

1. Factor out the greatest common factor from the following polynomial.

7 4 36 3 9x x x+

Step 1The first step is to identify the greatest common factor. In this case it looks like we can factor a 3 and an

3x out of each term and so the greatest common factor is

33x .

Step 2Okay, now lets do the factoring.

( )7 4 3 3 46 3 9 3 2 3x x x x x x+ = +



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Dont forget to also identify any numbers in the greatest common factor as well. That can often greatlysimplify the problem for later work (when we have later work for the problem anyway.).


3 8 10 4 5 27 2a b a b a b +

Step 1

The first step is to identify the greatest common factor. In this case it looks like we can factor an 3a and

a2

b out of each term and so the greatest common factor is3 2

a b .


( )3 8 10 4 5 2 3 2 6 7 2 27 2 7 2a b a b a b a b b a b a + = +


( ) ( )3 5

2 22 1 16 1x x x+ +

Step 1The first step is to identify the greatest common factor. In this case it looks like we can factor a 2 and an

( )32 1x + out of each term and so the greatest common factor is ( )322 1x + .


( ) ( ) ( ) ( )( )3 5 3 2

2 2 2 22 1 16 1 2 1 8 1x x x x x x+ + = + +

Dont get excited if the greatest common factor has more complicated terms in it as this one did. Thegreatest common factor wont always be just variables to powers.


( ) ( )2 2 6 4 4 12x x x x +

Step 1



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The first step is to identify the greatest common factor and in this case well need to be a little careful. Ifwe just do a quick glance we might be tempted to just say the greatest common factor is justxsince thereis clearly anxin both terms.

However, notice that we can factor a 2 out of the 4 12x in the second term to get,

( ) ( ) ( ) ( )2 22 6 4 4 12 2 6 8 2 6x x x x x x x x + = +

Upon doing this we see that not only do we have anxin both terms we also have a 2 6x in both terms

and so the greatest common factor in this case is ( )2 6x x .


( ) ( ) ( ) ( ) ( ) ( )2 22 6 4 4 12 2 6 8 2 6 2 6 8x x x x x x x x x x x + = + = +

Sometimes we need to do a little pre factoring work on a polynomial in order to determine just what thegreatest common factor is. It wont happen often, but it does need to be done often enough that we cantforget about it.

5. Factor the following polynomial by grouping.

3 4 67 7x x x x+ + +

Step 1

The first step here is to group the first two term and the last two terms as follows.

( ) ( )3 4 67 7x x x x+ + +

Step 2

We can now see that we can factor a 7xout of the first grouping and an 4x out of the second grouping.Doing this gives,

( ) ( )3 4 6 2 4 27 7 7 1 1x x x x x x x x+ + + = + + +

Step 3

Finally we see that we can factor an ( )21x x+ out of both of the new terms to get,

( )( )3 4 6 2 37 7 1 7x x x x x x x+ + + = + +



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6. Factor the following polynomial by grouping.

4 318 33 6 11x x x+

Step 1The first step here is to group the first two term and the last two terms as follows.

( ) ( )4 318 33 6 11x x x+ +

Be careful with the last grouping. Because both of the terms were negative we needed to factor out an -as we did the grouping.

Step 2

We can now see that we can factor a 3 out of the first grouping and an 3x out of the second grouping.Doing this gives,

( ) ( )

4 3 3

18 33 6 11 3 6 11 6 11x x x x x x+ = + +

Step 3

Finally we see that we can factor a 6 11x + out of both of the new terms to get,

( ) ( )4 3 318 33 6 11 6 11 3x x x x x+ = +

7. Factor the following polynomial.2 2 8x x

Step 1The initial form for the factoring will be,

( )( )x x+ +

and the factors of -8 are,

( )( ) ( ) ( ) ( ) ( ) ( )( )1 8 1 8 2 4 2 4

Step 2Now, recalling that we need the pair of factors from the above list that will add to get -2. So, we can seethat the correct factoring will then be,

( )( )2 2 8 4 2x x x x = +



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8. Factor the following polynomial.2 10 21z z +


( )( )z z+ +

and the factors of 21 are,

( )( ) ( )( ) ( ) ( ) ( ) ( )1 21 1 21 3 7 3 7

Step 2Now, recalling that we need the pair of factors from the above list that will add to get -10. So, we can seethat the correct factoring will then be,

( )( )2 10 21 3 7z z z z + =

9. Factor the following polynomial.2 16 60y y+ +


( )( )y y+ + and the factors of 60 are,

( )( ) ( )( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) ( )

1 60 2 30 3 20 4 15 5 12 6 10

1 60 2 30 3 20 4 15 5 12 6 10

Sometimes there are a lot of factors that we need to deal with. As you get more practice you will start tobe able to do most of this in your head and wont need to actually write all of the factors down.

Step 2Now, recalling that we need the pair of factors from the above list that will add to get 16. So, we can seethat the correct factoring will then be,

( ) ( )2 16 60 6 10y y y y+ + = + +



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10. Factor the following polynomial.25 14 3x x+

Step 1There are only two positive factors of 5 so the initial form for the factoring will be,

( )( )5x x+ +


( )( ) ( ) ( )1 3 1 3

Step 2After some trial and error we see that the correct factoring will then be,

( )( )25 14 3 5 1 3x x x x+ = +

11. Factor the following polynomial.26 19 7t t

Step 1There are two sets of positive factors of 6 and so we will have one of the two following possible initialforms for the factoring.

( )( ) ( )( )2 3 6t t t t + + + +


( )( ) ( )( )1 7 1 7


( )( )26 19 7 2 7 3 1t t t t = +

12. Factor the following polynomial.24 19 12z z+ +

Step 1There are two sets of positive factors of 4 and so we will have one of the two following possible initialforms for the factoring.



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( )( ) ( )( )2 2 4z z z z+ + + +

and the factors of 12 are,

( )( ) ( )( ) ( ) ( ) ( )( ) ( )( ) ( )( )1 12 2 6 3 4 1 12 2 6 3 4


( )( )24 19 12 4 3 4z z z z+ + = + +

13. Factor the following polynomial.2

14 49x x+ +

SolutionWe can do this in the manner of the previous problems if we wanted to. On the other hand we can notice

that the constant is a perfect square and that ( )2 7 14= and so we can see that this is one of the specialforms.

Therefore the factoring of this polynomial is,

( )22 14 49 7x x x+ + = +

Note that while you dont need necessarily need to know the special forms if you do and can easilyrecognize them it will make the factoring easier.

14. Factor the following polynomial.24 25w

SolutionWe can do this in the manner of the previous problems if we wanted to. On the other hand we can noticethat the we have a difference of perfect squares and so this is one of the special forms.


( )( )24 25 2 5 2 5w w w = +




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15. Factor the following polynomial.281 36 4x x +

SolutionWe can do this in the manner of the previous problems if we wanted to. On the other hand we can notice

that the constant is a perfect square and the coefficient of the 2x is also a perfect square. We can also

notice that that ( ) ( )2 9 2 36= and so we can see that this is one of the special forms.


( )2281 36 4 9 2x x x + =


16. Factor the following polynomial.6 33 4x x+

Step 1Dont let the fact that this polynomial is not a quadratic. That doesnt mean that we cant factor thepolynomial.

For this polynomial we can see that ( )2

3 6x x= and so it looks like we can factor this into the form,

( )( )3 3x x+ +

At this point all we need to do is proceed as we did with the quadratics we were factoring above.

Step 2After writing down the factors of -4 we can see that we need to have the following factoring.

( )( )6 3 3 33 4 4 1x x x x+ = +

Step 3

Now, we need to be careful here. Sometimes these will have further factoring we can do. In this case wecan see that the second factor is a difference of perfect cubes and we have a formula for factoring adifference of perfect cubes.


( ) ( ) ( )( ) ( )6 3 3 3 3 23 4 4 1 4 1 1x x x x x x x x+ = + = + + +



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17. Factor the following polynomial.5 4 33 17 28z z z


For this polynomial note that we can factor a 3z out of each term to get,

( )5 4 3 3 23 17 28 3 17 28z z z z z z =

Step 2Now, notice that the second factor is a quadratic and we know how to factor these. So, it looks like theform of the factoring should be,

( )( )5 4 3 33 17 28 3z z z z z z = + +

Step 3Finally once we write down the factors of the -28 we can see that the factoring of this polynomial is,

( )( )5 4 3 33 17 28 3 4 7z z z z z z = +

18. Factor the following polynomial.14 62 512x x


For this polynomial note that we can factor a 62x out of each term to get,

( )14 6 6 82 512 2 256x x x x =

Step 2Now, notice that the second factor is a difference of perfect squares and so we can further factor this as,

( ) ( )14 6 6 4 42 512 2 16 16x x x x x = +

Step 3Next, we can see that the third term is once again a difference of perfect squares and so can also befactored. After doing that the factoring of this polynomial is,

( ) ( )( )14 6 6 4 2 22 512 2 16 4 4x x x x x x = + +



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Step 4Finally we can see that we can do one more factoring on the last factor.

( )( )( )( )14 6 6 4 22 512 2 16 4 2 2x x x x x x x = + + +

Do not get too excited about polynomials that have lots of factoring in them. They will happen onoccasion so dont worry about it when they do.

Rational Expressions

1. Reduce the following rational expression to lowest terms.

2

2

6 7

10 21

x x

x x

+

Step 1First, we need to factor the numerator and denominator as much as we can. Doing that gives,

( ) ( )( ) ( )

2

2

7 16 7

10 21 7 3

x xx x

x x x x

+ =

+

Step 2Now all we need to do is cancel all the factors that we can in order to reduce the rational expression tolowest terms.

2

2

6 7 1

10 21 3

x x x

x x x

+=

+


2

26 99x x

x+ +

Step 1First, we need to factor the numerator and denominator as much as we can. Doing that gives,

( )

( )( )

22

2

36 9

9 3 3

xx x

x x x

++ +=

+



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2

2

6 9 3

9 3

x x x

x x

+ + +=


2

2

2 28

20

x x

x x

Step 1

First, we need to factor the numerator and denominator as much as we can. Doing that gives,

( )( ) ( )

( )( )

2 2

2 2

2 7 42 28 2 28

20 5 420

x xx x x x

x x x xx x

+ = =

+ +

Notice that in order to make factoring the denominator somewhat easier we first factored a minus sign outof the denominator.


2

22 28 2 720 5x x x

x x x +=

+

Recall that the minus sign in the denominator can be put out in front of the rational expression if wechoose to put it there (as we did here).

4. Perform the indicated operation in the following expression and reduce the answer to lowest terms.2 2

2 2

5 24 4 4

6 8 3

x x x x

x x x x

+ + +

+ +

Step 1So, we first need to factor each of the polynomials as much as possible.

( )( )

( )( )

( )

( )

( )

( )

( )2

8 3 2 8 2

4 2 3 4

x x x x x

x x x x x x

+ + + +=

+ + +



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Step 2Finally, just multiply the two terms together. Doing this gives,

( ) ( )( )

2 2

2 2

8 25 24 4 4

6 8 3 4

x xx x x x

x x x x x x

+ ++ + +=

+ + +


2 2

49 42

2 3 5 7 6

x x x

x x x x

+ +

Step 1So, we first need to do is convert this into a product.

2 2 2 2

2 2 2 2

49 42 49 7 6

2 3 5 7 6 2 3 5 42

x x x x x x

x x x x x x x x

+ + = + +

Make sure that you dont do the factoring and canceling until youve converted the division to a product.

Step 2Now we can factor each of the terms as much as possible to get,

( )( )( )( )

( ) ( )( )( )

2 2

2 2

7 7 1 649 42

2 3 5 7 6 2 5 1 7 6

x x x xx x x

x x x x x x x x

+ + + =

+ + + +

Step 3Finally cancel as much as possible to reduce to lowest terms and do the product.

2 2

2 2

49 42 7

2 3 5 7 6 2 5

x x x x

x x x x x

+ =

+ +


2 2

2 8 9 20

2 8 24 11 30

x x x x

x x x x

+

+

Step 1So, we first need to do is convert this into a product.

2 2 2 2

2 2 2 2

2 8 9 20 2 8 11 30

2 8 24 11 30 2 8 24 9 20

x x x x x x x x

x x x x x x x x

+ + =

+ +



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Make sure that you dont do the factoring and canceling until youve converted the division to a product.

Step 2Now we can factor each of the terms as much as possible to get,

( ) ( )( )( )

( ) ( )( ) ( )

2 2

2 2

4 2 5 62 8 9 20

2 8 24 11 30 2 2 6 5 4

x x x xx x x x

x x x x x x x x

+ +

= + +

Step 3Finally cancel as much as possible to reduce to lowest terms and do the product.

2 2

2 2

2 8 9 20 1

2 8 24 11 30 2

x x x x

x x x x

+ =

+

Dont worry if all the variables end up cancelling out after you are done reducing to lowest terms. It willhappen on occasion so dont worry about it when it does.

7. Perform the indicated operation in the following expression and reduce the answer to lowest terms.

2

3

14

11 10

xx

x x

++

+ +

Step 1This is just a division and so lets first convert it to a product.

2

2

33 11 101

4 1 4

11 10

x xxx x x

x x

+ ++ =+ + +

+ +

Step 2Now we can factor each of the second term as much as possible to get,

( ) ( )

2

31 1031

4 1 411 10

x xx

x x xx x

+ ++ =+

+ ++ +

Step 3Now cancel as much as possible to reduce to lowest terms and do the product.



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( )

2

33 101

4 4

11 10

xxx x

x x

++ =+ +

+ +

8. Perform the indicated operation in the following expression.

3

4 2 7

x

x x+

+

Step 1We first need the least common denominator for this rational expression.

( ) ( )lcd : 4 2 7x x +

Step 2Now multiply each term by an appropriate quantity to get the least common denominator into thedenominator of each term.

( )( ) ( )

( )( ) ( )

3 2 7 43

4 2 7 4 2 7 2 7 4

x x xx

x x x x x x

+ + = +

+ + +

Step 3All we need to do now is do the addition and simplify the numerator of the result.

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 23 2 7 43 6 21 4 2 214 2 7 4 2 7 4 2 7 4 2 7

x x xx x x x x xx x x x x x x x

+ + + + + ++ = = = + + + +


2 4

2 1 2

3 9 4x x x +

+


( )4lcd : 9 4x x +




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College Algebra

( )( )( )( )

( )( )

( )( )( )

2 4

2 4 42 2 4

2 3 4 2 91 42 1 2

3 9 4 9 43 3 4 4 9

x x xx

x x x x xx x x x x

+ + + = +

+ ++ +

Step 3All we need to do now is do the subtraction and addition then simplify the numerator of the result.

( )( ) ( )

3 2 4 4 3 2

2 4 4 4

6 24 4 182 1 2 18 6 24 4

3 9 4 9 4 9 4

x x x x x x x x

x x x x x x x

+ + + + + + = =

+ + +


2

8

12 36 6

x x

x x x

+ + +

Step 1We first need the least common denominator for this rational expression. However, before we get thatwell need to factor the denominator of the first term. Doing this gives,

( )22

8 8

12 36 6 66

x x x x

x x x xx

=

+ + + ++

Step 2The least common denominator is then,

( )2lcd : 6x +

Remember that we only take the highest power on each term in the denominator when setting up the leastcommon denominator.

Step 3Next, multiply each term by an appropriate quantity to get the least common denominator into thedenominator of each term.

( )

( )( )( ) ( )22

8 68

12 36 6 6 66

x xx x x

x x x x xx

+ =

+ + + + ++

Step 4Finally all we need to do is the subtraction and simplify the numerator of the result.

( ) ( )

( )

( )( ) ( )

2 2

2 2 22

2 488 68 48 3

12 36 6 6 6 6

x x xx x xx x x x

x x x x x x

+ + = = =

+ + + + + +



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2

2 1 113 42 6 7x x

x x x x

++ +

Step 1We first need the least common denominator for this rational expression. However, before we get thatwell need to factor the denominator of the first term. Doing this gives,

( ) ( )

2 2

2

1 1 1 1

13 42 6 7 6 7 6 7

x x x x

x x x x x x x x

+ ++ = +

+

Step 2

The least common denominator is then,

( ) ( )lcd : 6 7x x


Step 3Next, multiply each term by an appropriate quantity to get the least common denominator into thedenominator of each term.

( )( ) ( )( )( ) ( ) ( )( ) ( )

22

2

1 7 61 1 1

13 42 6 7 6 7 6 7 7 6

x x x xx x

x x x x x x x x x x

+ ++ = + +

Step 4Finally all we need to do is the addition and subtraction then simplify the numerator of the result.

( ) ( ) ( )( ) ( )

( ) ( ) ( ) ( )

22

2

2 3 2 3 2

1 1 7 61 1

13 42 6 7 6 7

1 6 7 6 7 6 6

6 7 6 7

x x x xx x

x x x x x x

x x x x x x x

x x x x

+ + ++ =

+

+ + + = =


( ) ( )3 2

10

3 8 3 8

x x

x x

++

+ +



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( )3

lcd : 3 8x +



( ) ( ) ( )

( )

( ) ( )3 2 3 2

3 810 10

3 8 3 8 3 8 3 8 3 8

x xx x x

x x x x x

++ ++ = +

+ + + + +

Step 3All we need to do now is do the addition and simplify the numerator of the result.

( ) ( ) ( ) ( )

2 2

3 2 3 3

10 10 3 8 3 9 10

3 8 3 8 3 8 3 8

x x x x x x x

x x x x

+ + + + + ++ = =

+ + + +

Complex Numbers

1. Perform the indicated operation and write your answer in standard form.

( )( )4 5 12 11i i +

Hint : You know how to do the operation with polynomials so you can do the operation here! Just recallthat you need to be careful to deal with any i2that might happen to show up in the process.

SolutionWe know how to multiply two polynomials and so we also know how to multiply two complex numbers.All we need to do is foil the two complex numbers to get,

( )( ) 2 24 5 12 11 48 44 60 55 48 16 55i i i i i i i + = + =

All we need to do to finish the problem is to recall that 2 1i = . Upon using this fact we can finish theproblem.



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( ) ( ) ( )4 5 12 11 48 16 55 1 103 16i i i i + = =


( ) ( )3 6 7i i

Hint : You know how to do the operation with polynomials so you can do the operation here!

SolutionWe know how to subtract two polynomials and so we also know how to subtract two complex numbers.

( ) ( )3 6 7 3 6 7 9 6i i i i i = + = +


( ) ( )1 4 16 9i i+ +

Hint : You know how to do the operation with polynomials so you can do the operation here!

SolutionWe know how to subtract two polynomials and so we also know how to subtract two complex numbers.

( ) ( )1 4 16 9 1 4 16 9 17 5i i i i i+ + = + + =


( )8 10 2i i+


SolutionWe know how to multiply two polynomials and so we also know how to multiply two complex numbers.All we need to do is distribute the 8ito get,

( ) 28 10 2 80 16i i i i+ = +

All we need to do to finish the problem is to recall that2 1i = . Upon using this fact we can finish the

problem.

( ) ( )8 10 2 80 16 1 16 80i i i i+ = + = +



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( ) ( )3 9 1 10i i +



( ) ( ) 23 9 1 10 3 30 9 90i i i i i + =

All we need to do to finish the problem is to recall that

2

1i =

. Upon using this fact we can finish theproblem.

( )( ) ( )3 9 1 10 3 30 9 90 1 87 39i i i i i + = =


( ) ( )2 7 8 3i i+ +



( ) ( ) 22 7 8 3 16 6 56 21i i i i i+ + = + + +

All we need to do to finish the problem is to recall that 2 1i = . Upon using this fact we can finish theproblem.

( )( ) ( )2 7 8 3 16 6 56 21 1 5 62i i i i i+ + = + + + = +


7

2 10

i

i

+



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Hint : Recall that standard form does not allow any i's in the denominator.

Step 1Because standard form does not allow for is to be in the denominator well need to multiply the

numerator and denominator by the conjugate of the denominator, which is 2 10i .

Step 2Multiplying by the conjugate gives,

( ) ( )( ) ( )

7 2 107 2 10

2 10 2 10 2 10 2 10

i ii i

i i i i

=

+ +

Step 3Now all we need to do is do the multiplication in the numerator and denominator and put the result instandard form.

2

2

7 14 72 10 4 72 4 72 1 9

2 10 4 100 104 104 104 26 13

i i i ii i

i i

+ = = = = +


1 5

3

i

i

+



numerator and denominator by the conjugate of the denominator, which is 3i .


( ) ( )( ) ( )

1 5 31 5 3

3 3 3 3

i ii i

i i i i

++=


2

2

1 5 3 15 15 3 15 3 5 1

3 9 9 9 9 3 3

i i i ii i

i i

+ + += = = + = +



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6 7

8

i

i

+



numerator and denominator by the conjugate of the denominator, which is 8 i+ .


( ) ( )

( ) ( )

6 7 86 7 8

8 8 8 8

i ii i

i i i i

+ ++ +=

+ +


2

2

6 7 8 48 62 7 41 62 41 62

8 8 64 65 65 65

i i i i ii

i i i

+ + + + += = = +

+

Solving Equations and Inequalities

Solutions and Solution Sets

1. Is 6x= a solution to ( )2 5 3 1 22x x = + ?

SolutionThere really isnt all that much to do for these kinds of problems. All we need to do is plug the givennumber into both sides of the equation and check to see if the right and left side are the same value.

Here is that work for this particular problem.



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( ) ( )?

2 6 5 3 1 6 22

7 7 OK

= +

=

So, we can see that the right and left sides are the same and so we know that 6x= is a solution to the

equation.

2. Is 7t= a solution to 2 3 10 4 8t t t+ = + ?



( ) ( ) ( )?2

7 3 7 10 4 8 7

60 60 OK

+ = +

=

So, we can see that the right and left sides are the same and so we know that 7t= is a solution to theequation.

3. Is 3t= a solution to 2 3 10 4 8t t t+ = + ?



( ) ( ) ( )?

23 3 3 10 4 8 3

10 20 NOT OK

+ = +

=

So, we can see that the right and left sides are the not the same and so we know that 3t= is not a

solution to the equation.

4. Is 2w= a solution to2 8 12

02

w w

w

+ +=

+?

Solution



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There really isnt all that much to do for these kinds of problems. All we need to do is plug the givennumber into both sides of the equation and check to see if the right and left side are the same value.

Note that for this problem we dont even really need to plug the value into the equation. We can see by a

quick inspection that if we were to plug 2w= into this equation we would have division by zero andwe know that is not allowed.

Therefore, 2w= is not a solution to this equation.

Be very careful with this kind of problem. If we had plugged 2w= into the equation wed have gottenzero in the numerator as well and we might be tempted to say that it is a solution to the equation. Wedbe wrong however. Regardless of the value of the numerator, we would still have division by zero and

that is just not allowed and so 2w= will not be a solution to this equation.

5. Is 4z= a solution to 2 26 3z z z + ?

SolutionThere really isnt all that much to do for these kinds of problems. All we need to do is plug the givennumber into both sides of the inequality and check to see if the inequality is true. In this case that willmean checking to see if the left side is larger than or equal to the right side.


( ) ( ) ( )?

2 26 4 4 4 3

8

+

19 NOT OK

So, we can see that the left side is neither larger than nor equal to the right side and so 4z= is not asolution to this inequality.

6. Is 0y= a solution to ( ) ( ) ( )2 7 1 4 1 3 4 10y y y+ < + + + ?

SolutionThere really isnt all that much to do for these kinds of problems. All we need to do is plug the givennumber into both sides of the inequality and check to see if the inequality is true. In this case that willmean checking to see if the left side is less than the right side.


( ) ( ) ( )( )?

2 0 7 1 4 0 1 3 4 0 10

13 34 OK

+ < + + +

+ ?

SolutionThere really isnt all that much to do for these kinds of problems. All we need to do is plug the givennumber into both sides of the inequality and check to see if the inequality is true. In this case that willmean checking to see if the left side is greater than the right side.


( ) ( )?

21 1 3 1 1

4

+ > +

> 4 NOT OK

Be very careful with this type of problem! Four is not greater than 4 (its equal to 4 big difference here)and so, we can see that 1x= is not a solution to this inequality.

Contrast the inequality in this problem with,

( )2

1 3 1x x+ +

While 1x= is not a solution to the inequality in the problem statement it is a solution to this inequalitysince 4 is in fact greater than or equal to4. The presence of the equal sign in the inequality can make allthe difference in the world and we really need to be on the lookout for it. It is easy to miss when its thereand it is easy to sometimes assume it is there when in fact it isnt.

Linear Equations

1. Solve the following equation and check your answer.

( )4 7 2 3 2x x x = +

Step 1First we need to clear out the parenthesis on the left side and then simplify the left side.

( )4 7 2 3 2

4 14 7 3 2

11 14 3 2

x x x

x x x

x x

= +

+ = +

= +



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Step 2

Now we can subtract 3xand add 14 to both sides to get all thexs on one side and the termswithout anxon the other side.

11 14 3 2

8 16

x x

x

= +

=

Step 3Finally, all we need to do is divide both sides by the coefficient of thex(i.e.the 8) to get the solution of

2x= .

Step 4Now all we need to do is check our answer from Step 3 and verify that it is a solution to the equation. Itis important when doing this step to verify by plugging the solution from Step 3 into the equation given inthe problem statement.

Here is the verification work.

( ) ( ) ( )?

4 2 7 2 2 3 2 2

8 8 OK

= +

=

So, we can see that our solution from Step 3 is in fact the solution to the equation.


( ) ( )2 3 10 6 32 3w w+ =

Step 1First we need to clear out the parenthesis on each side and then simplify each side.

( ) ( )2 3 10 6 32 3

2 6 10 192 18

2 4 192 18

w w

w w

w w

+ =

+ =

=

Step 2

Now we can add 18wand 4 to both sides to get all the ws on one side and the terms without an

won the other side.

2 4 192 18

20 196

w w

w

=

=

Step 3



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Finally, all we need to do is divide both sides by the coefficient of the w(i.e.the 20) to get the solution of

196 49

20 5w= = .

Dont get excited about solutions that are fractions. They happen more often than people tend to realize.



?

?

49 492 3 10 6 32 3

5 5

64 132 10 6

5 578 78

OK5 5

+ =

=

=



4 2 3 5

3 4 6

z z=

Step 1The first step here is to multiply both sides by the LCD, which happens to be 12 for this problem.

( ) ( ) ( )

4 2 3 512 12

3 4 6

4 2 3 512 12 12

3 4 6

4 4 2 3 3 2 5

z z

z z

z z

=

=

=

Step 2

Now we need to find the solution and so all we need to do is go through the same process that weused in the first two practice problems. Here is that work.



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College Algebra

( ) ( ) ( )4 4 2 3 3 2 5

16 8 9 10

2 7

7

2

z z

z z

z

z

=

=

=

=



( ) ( )7 7?2 2

35? 2

?

4 2 53

3 4 6

4 7 33 4 6

11 3 35

3 4 12

11 11OK

3 3

=

+ =

= +

=


Note that the verification work can often be quite messy so dont get excited about it when it does.Verification is an important step to always remember for these kinds of problems. You should always

know if you got the answer correct before you check the answers and/or your instructor grades theproblem!


2

4 1

25 5

t

t t=

Hint : Do not forget to watch out for values of tthat well need to avoid!

Step 1Lets first factor the denominator on the left side so we can identify the LCD. While we are at it we willalso factor a minus out of the denominator on the right side.



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College Algebra

( ) ( ) ( )

( ) ( )

2

4 1

25 5

4 1

5 5 5

4 15 5 5

t

t t

t

t t t

tt t t

=

= +

= +

So, after factoring the left side and factoring the minus sign out of the denominator on the right side wecan quickly see that the LCD for this equation is,

( )( )5 5t t +

From this we can also see that well need to avoid 5t= and 5t= . Remember that we have to avoiddivision by zero and we will clearly get division by zero with each of these values of t.

Step 2Next we need to do find the solution. To get the solution well need to multiply both sides by theLCD and the go through the same process we used in the first couple of practice problems. Hereis that work.

( )( )( ) ( )

( )( )

( )

4 15 5 5 5

5 5 5

4 5

4 5

5 5

1

tt t t t

t t t

t t

t t

t

t

+ = + +

= +

=

=

=

Step 3Finally we need to verify that our answer from Step 2 is in fact a solution.

The first thing to note is that it is not one of the values of tthat we need to avoid. Having determined thatwe know that we do have a potential solution (i.e.its not a value of twe need to avoid) all we need to dois plug the solution into the equation given in the problem statement.


( )( ) ( )

?

2

?

4 1 1

5 11 25

4 1

1 25 5 1

1 1OK

6 6

=

=

+

=



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College Algebra



3 4 72

1 1

y

y y

+ = +

Hint : Do not forget to watch out for values ofythat well need to avoid!

Step 1

First we can see that the LCD for this equation is,

1y

From this we can also see that well need to avoid 1y= . Remember that we have to avoid division byzero and we will clearly get division by zero with this value ofy.

Step 2

Next we need to do find the solution. To get the solution well need to multiply both sides by theLCD and the go through the same process we used in the first couple of practice problems. Hereis that work.

( ) ( )

( )

3 4 71 2 1

1 1

3 4 2 1 7

3 4 2 5

1

yy y

y y

y y

y y

y

+ = +

+ = +

+ = +

=

Step 3Finally we need to verify that our answer from Step 2 is in fact a solution and in this case there isnt a lotof work to that process. We can see that our potential solution from Step 2 is in fact the value ofythat weneed to avoid and so this equation has no solution.

We

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