ALGEBRA

21.16 Explain the use of symbols to represent mathematical values.

Basics of AlgebraAlgebra is a division of mathematics designed to help solve certain types of problems quicker and easier. Algebra is based on the concept of unknown values called variables, unlike arithmetic which is based entirely on known number values.

This lesson introduces an important algebraic concept known as the Equation. The idea is that an equation represents a scale such as the one shown on the right. Instead of keeping the scale balanced with weights, numbers, or constants are used. These numbers are called constants because they constantly have the same value. For example the number 47 always represents 47 units or 47 multiplied by an unknown number. It never represents another value.

The equation may also be balanced by a device called a variable. A variable is an an unknown number represented by any letter in the alphabet (often x). The value of each variable must remain the same in each problem.

Several symbols are used to relate all of the variables and constants together. These symbols are listed and explained below.

??? Multiply* Multiply/ Divide+ Add or Positive- Subtract or Negative

( ) Calculate what is inside of the parentheses first. (also called grouping symbols)

Basics of the Equation

The diagram on the right shows a basic equation. This equation is similar to problems which you may have done in ordinary mathematics such as:

__ + 16 = 30

You could easily guess that __ equals 14 or do 30 - 16 to find that __ equals 14.

In this problem __ stood for an unknown number; in an equation we use variables, or any letter in the alphabet.

When written algebraically the problem would be:x + 16 = 30

and the answer should be written:x = 14

Solving Equations

These equations can be solved relatively easy and without any formal method. But, as you use equations to solve more complex problems, you will want an easier way to solve them.

Pretend you have a scale like the one shown. On the right side there are 45 pennies and on the left side are 23 pennies and an unknown amount of pennies. The scale is balanced, therefore, we know that there must be an equal amount of weight on each side.

As long as the same operation (addition, subtraction, multiplication, etc.) is done to both sides of the scale, it will remain balanced. To find the unknown amount of pennies of the left side, remove 23 pennies from each side of the scale. This action keeps the scale balanced and isolates the unknown amount. Since the weight(amount of pennies) on both sides of the scale are still equal and the unknown amount is alone, we now know that the unknown amount of pennies on the left side is the same as the remaining amount (22 pennies) on the right side.

Solving Equations

Because an equation represents a scale, it can also be manipulated like one. The diagram below shows a simple equation and the steps to solving it.

Initial Equation / Problem x + 23 = 45

Subtract 23 from each side x + 23 - 23 = 45 - 23

Result / Answer x = 22

Solving Equations

Take a look at the equation below. As you can see, after the variable is subtracted from the left and the constants are subtracted from the right, you are still left with 2x on one side.

Initial Equation / Problem x + 23 = 3x + 45

Subtract x from each side x - x + 23 = 3x - x + 45

Result 23 = 2x + 45

Subtract 45 from each side 23 - 45 = 2x + 45 - 45

Result -22 = 2x

Switch the left and right sides of the equation

2x = -22

This means that the unknown number multiplied by two, equals -22. To find the value of x, use the process "dividing by the coefficient" described on the next page.

Identifying and Using Coefficients

The coefficient of a variable is the number which the variable is being multiplied by. In this equation, 2 is the coefficient of x because 2x is present in the equation. Some additional examples of coefficients:

Term Coefficient of x

2x 2

0.24x 0.24

x 1

-x -1

Note that in the last two examples, the following rules are applied

If the variable has no visible coefficient, then it has an implied coefficient of 1. If the variable only has a negative sign, then it has an implied coefficient of -1.

Continue to the next page to see how we use the coefficient of the variable x in the equation, 2, to find the value of x.

What is simplifying?In math class you simplified fractions to make them easier to understand and work with. For example, the first fraction, as shown below, can be simplified to a much smaller fraction with the same value.

173456 1------ has the same value as ---346912 2

For basic equations, like those presented in the Equation Basics lesson, your first step to solving the equation might be to subtract the variable on the left from both sides. But in some cases, you may not be able to tell if the variable is positive or negative because it has multiple signs in front of it,

5 - -2x = 6 + x

or because there are multiple terms with variables on the left.

5x + 4x + 7 = x + 15

To learn how to simplify multiple signs, and make the equations solvable,

Substitution LessonsIn algebra, letters such as x or y are used to represent values which are usually unknown. They can be used in equations or expressions to help solve a wide variety of problems. In many cases you may know the value of a variable. This is the case with the problem below:

b = 3, c = 185b - 2c + c / b

Since the values of both b and c are known, a numeric value for the expression can be found through the process of substitution.

When a school teacher becomes ill, a substitute teacher usually fills in for a few days. The substitute teacher temporarily takes the place of the usual teacher so that classroom activities may continue. Similarly, when using the method of substitution in algebra, a variable such as x or y is replaced with its value. The expression can then be simplified even further.

In this problem we replace the variables b and c since their values are given. Everywhere in the problem where the variable b is present, it is substituted with 3 in parentheses. Everywhere c appears, it is substituted with 18 in parentheses. It is important that the substituted number is placed in parentheses so that negative values are handled properly.

5(3) - 2(18) + (18) / (3)

Now, this expression can be simplified like any other.

The insides of all of the parentheses are simplified, and there are no exponents in the expression, so you can skip to simplifying multiplication and division in the order they appear.

15 - 36 + 6

Now simplify addition and subtraction in the order they appear (combine like terms):

-15

Proceed to the next page to begin simplifying a more complex expression.

Substitution

Simplifying Before Substituting

As you can see, the values of b and c are different for this problem:

b = -3, c = 2b2c + bc2 + 2b ?? bc

On the previous page, the expression could not be simplified before its variables, b and c, were substituted with their values.

Whenever possible, an expression should be simplified before substitution is applied, as it will often save time. Begin by simplifying multiplication: "2b ?? bc" becomes "2b2c".

b2c + bc2 + 2b2c

Now combine like terms: b2c and 2b2c are combined into 3b2c.

3b2c + bc2

The problem is now simplified to the following:

b = -3, c = 23b2c + bc2

We will begin the substition in a moment, but first compare this expression to the expression we started with. Both expressions are equivalent because we simplified properly. But, by reducing the number of terms in the expression, the substitution step will be easier.

Now substitute b with (-3) and c with (2).

3(-3)2(2) + (-3)(2)2

Use the Order of Operations to simplify the expression. First simplify exponents.

3(9)(2) + (-3)(4)

Simplify multiplication.

3(18) + -1254 - 12

Combine like terms.

42

The next page briefly explains two variations of the types of substitution problems shown.

Substitution

If the value for only one of the variables is given

Look over the problem below.

b = 35b - 2c + c / b

As you can see, the expression has two variables b and c, but in this problem, only the value for b is given. Don't panic! Simply substitute the variable b with the number 3, don't worry about doing anything with the variable c.

5(3) - 2c + c / (3)

Then simplify as much as you can using the Order of Operations:

If the value for one of the variables is not a number

In this case, as in the problem below, you must substitute the variable with the expression given.

b = c + 15b - 2c + c / b

NOTE

This problem is one instance where it is absolutely necessary to enclose the expression or number that you are substituting a variable for in parentheses. Because of the Order of Operations 5(c + 1) is not the same as 5c + 1.

Substitute:

5(c + 1) - 2c + c / (c + 1)

Simplify:

5c + 5 - 2c + c / (c + 1)

Proceed to the next page for more substitution resources.

Factoring A Difference Between Two Squares LessonsTake a look at the problem and work below:

(x + 3)(x - 2)x2 - 2x + 3x -6x2 + x - 6

As you can see above, two binomials in parentheses, (x + 3) and (x - 2) are multiplied using the FOIL method. When the outer terms and inner terms are multiplied they result in -2x and 3x. Because these two resulting terms have different coefficients, when we combine like terms the result is another term, +x, which appears in the last line of the problem.

Now observe the following problem:

(x - 2)(x + 2)x2 + 2x - 2x - 4x2 - 4

In this case, the multiplication of the outer terms and of the inner terms resulted in 2x and -2x. The terms 2x and -2x have coefficients of 2 and -2. These coefficients are essentially the same number, but with with opposite signs (one number is positive and the other is negative). These two terms form a zero pair, meaning when they are combined, they cancel each other out. You can see that in the last step of the problem, where like terms were combined, the zero pair 2x and -2x canceled out. As a result, only two terms remained on the last line.

The result from the last problem is called a Difference Between Two Squares. A Difference Between Two Squares is an expression with two terms (also known as a binomial) in which both terms are perfect squares and one of the two terms is negative.

The pages that follow show how to factor a difference between two squares. The factoring process, which converts an expression like "x2 - 4" into "(x - 2)(x + 2)", is essentially the opposite of the multiplication process we used above.

Factoring A Difference Between Two Squares

Take a look at the problem (expression) below:

4x2 - 16

The first step at factoring this is to make sure that the expression is a difference between squares.

Question Answer and Reason

Are there only two terms? Yes. The first term is 4x2; the second term is -16.

Are both coefficients (4 and 16) perfect squares?Yes. Notice 2 times 2 equals 4, and 4 times 4 equals 16.

Are all of the variables in the expression raised to an even (2,4,6, ...) power?

Yes. There is only one variable, x, and it has a power of 2 which is even.

Does one term have a positive coefficient, and another term have a negative coefficient?

Yes. The coefficient 4 is positive, and the coefficient -16 is negative.

Because "Yes" was answered to each of the above questions, we know that the expression is a difference between two squares. Begin the factoring process by writing two sets of open parentheses:

( )( )

Now find the square root of 4x2, the first term, by finding the square root of 4 and then dividing each exponent by 2. The square root of 4 is 2. Half of the exponent 2 is 1, thus x2 becomes x1 or x. Thus, the square root of the entire term is 2x. Write this term on the left inside of each set of parentheses.

(2x )(2x )

We will now consider 16, the second term without the negative sign. We will apply the same process that we applied to 4x2. There are no variables in 16, so we simply find that the square root of 16 is 4. Now 4 is written on the right inside of each set of parentheses.

(2x 4)(2x 4)

Add a plus sign to the middle of the first set of parentheses, then add a minus sign to the middle of the second set of parentheses.

(2x + 4)(2x - 4)

The result is two parentheses which can be multiplied to get the original expression 4x2 - 16. To check that this answer is correct, you can apply the FOIL Method which was presented in an earlier lesson.

Proceed to the next page to see another example of factoring a Difference Between Two Squares.

Factoring A Difference Between Two Squares

Examine the problem / expression below:

-9 + j4

Again, the first step at factoring this expression is to verify that the expression is a Difference Between Two Squares.

Question Answer and Reason

Are there only two terms? Yes. The first term is -9; the second term is j4.

Are both coefficients (9 and 1) perfect squares?Yes. Notice 3 times 3 equals 9 and 1 times 1 equals 1.

Are all of the variables in the expression raised to an even (2,4,6, ...) power?

Yes. There is only one variable, j, and it has a power of 4 which is even.

Does one term have a positive coefficient, and another term have a negative coefficient?

Yes. The coefficient 1 is positive, the coefficient -9 is negative.

Unlike the last problem, the first term is negative. To make factoring this expression easier, simply switch the two terms so that the negative term is second.

-9 + j4

becomes

j4 - 9

Now, continue factoring as in the last problem. Write two sets of open parentheses:

( )( )

Find the square root of the first term, j4. Write the result, j2, on the left inside of each set of parentheses.

( j2 )( j2 )

Find the square root of the second term, 9. Write the result, 3, on the right inside of each set of parentheses.

( j2 3)( j2 3)

Now write a plus sign in the middle of the first set of parentheses and write a minus sign in the middle of the second set of parentheses.

( j2 + 3)( j2 - 3)

Proceed to the next page for links to calculators, worksheets, and other resources for this lesson.

Solve by Factoring LessonsSeveral previous lessons explain the techniques used to factor expressions. This lesson focuses on an imporatant application of those techniques ??? solving equations.

Why solve by factoring?

The most fundamental tools for solving equations are addition, subtraction, multiplication, and division. These methods work well for equations like x + 2 = 10 - 2x   and   2(x - 4) = 0.

But what about equations where the variable carries an exponent, like x2 + 3x = 8x - 6? This is where factoring comes in. We will use this equation in the first example.

The Solve by Factoring process will require four major steps:

1. Move all terms to one side of the equation, usually the left, using addition or subtraction.2. Factor the equation completely.

3. Set each factor equal to zero, and solve.4. List each solution from Step 3 as a solution to the original equation.

First Examplex2 + 3x = 8x - 6

Step 1

The first step is to move all terms to the left using addition and subtraction. First, we will subtract 8x from each side.

x2 + 3x - 8x = 8x - 8x - 6x2 - 5x = -6

Now, we will add 6 to each side.

x2 - 5x + 6 = -6 + 6x2 - 5x + 6 = 0.

With all terms on the left side, we proceed to Step 2.

Step 2

We identify the left as a trinomial, and factor it accordingly:

(x - 2)(x - 3) = 0

We now have two factors, (x - 2) and (x - 3).

Step 3

We now set each factor equal to zero. The result is two subproblems:

x - 2 = 0

and

x - 3 = 0

Solving the first subproblem, x - 2 = 0, gives x = 2. Solving the second subproblem, x - 3 = 0, gives x = 3.

Step 4

The final step is to combine the two previous solutions, x = 2 and x = 3, into one solution for the original problem.

x2 + 3x = 8x - 6x = 2, 3

Proceed to the next page for an explanation of the theory behind our method, and another example.

Solve by Factoring: Why does it work?

Examine the equation below:

ab = 0

If you let a = 3, then logivally b must equal 0. Similarly, if you let b = 10, then a must equal 0.

Now try letting a be some other non-zero number. You should observe that as long as a does not equal 0, bmust be equal to zero.

To state the observation more generally, "If ab = 0, then either a = 0 or b = 0." This is an important property of zero which we exploit when solving by factoring.

When the example was factored into (x - 2)(x - 3) = 0, this property was applied to determine that either (x - 2) must equal zero, or (x - 3) must equal zero. Therefore, we were able to create two equations and determine two solutions from this observation.

A Second Example5x3 = 45x

Step 1

Move all terms to the left side of the equation. We do this by subtracting 45x from each side.

5x3 - 45x = 45x - 45x5x3 - 45x = 0.

Step 2

The next step is to factor the left side completely. We first note that the two terms on the left have a greatest common factor of 5x.

5x(x2 - 9) = 0

Now, (x2 - 9) can be factored as a difference between two squares.

5x(x + 3)(x - 3) = 0

We are left with three factors: 5x, (x + 3), and (x - 3). As explained in the "Why does it work?" section, at least one of the three factors must be equal to zero.

Step 3

Create three subproblems by setting each factor equal to zero.

1. 5x = 02. x + 3 = 03. x - 3 = 0

Solving the first equation gives x = 0. Solving the second equation gives x = -3. And solving the third equation gives x= 3.

Step 4

The final solution is formed from the solutions to the three subproblems.

x = -3, 0, 3

Third Example3x4 - 288x2 - 1200 = 0

Steps 1 and 2

All three terms are already on the left side of the equation, so we may begin factoring. First, we factor out a greatest common factor of 3.

3(x4 - 96x2 - 400) = 0

Next, we factor a trinomial.

3(x2 + 4)(x2 - 100) = 0

Finally, we factor the binomial (x2 - 100) as a difference between two squares.

3(x2 + 4)(x + 10)(x - 10) = 0

Step 3

We proceed by setting each of the four factors equal to zero, resulting in four new equations.

1. 3 = 02. x2 + 4 = 03. x + 10 = 04. x - 10 = 0

The first equation is invalid, and does not yield a solution. The second equation cannot be solved using basic methods. (x2 + 4 = 0 actually has two imaginary number solutions, but we will save Imaginary Numbers for another lesson!) Equation 3 has a solution of x = -10, and Equation 4 has a solution of x = 10.

Step 4

We now include all the solutions we found in a single solution to the original problem:

x = -10, 10

This may be abbreviated as

x = ??10

Solve by Factoring LessonsSeveral previous lessons explain the techniques used to factor expressions. This lesson focuses on an imporatant application of those techniques ??? solving equations.

Why solve by factoring?

The most fundamental tools for solving equations are addition, subtraction, multiplication, and division. These methods work well for equations like x + 2 = 10 - 2x   and   2(x - 4) = 0.

But what about equations where the variable carries an exponent, like x2 + 3x = 8x - 6? This is where factoring comes in. We will use this equation in the first example.

The Solve by Factoring process will require four major steps:

1. Move all terms to one side of the equation, usually the left, using addition or subtraction.

2. Factor the equation completely.3. Set each factor equal to zero, and solve.4. List each solution from Step 3 as a solution to the original equation.

First Examplex2 + 3x = 8x - 6

Step 1

The first step is to move all terms to the left using addition and subtraction. First, we will subtract 8x from each side.

x2 + 3x - 8x = 8x - 8x - 6x2 - 5x = -6

Now, we will add 6 to each side.

x2 - 5x + 6 = -6 + 6x2 - 5x + 6 = 0.

With all terms on the left side, we proceed to Step 2.

Step 2

We identify the left as a trinomial, and factor it accordingly:

(x - 2)(x - 3) = 0

We now have two factors, (x - 2) and (x - 3).

Step 3

We now set each factor equal to zero. The result is two subproblems:

x - 2 = 0

and

x - 3 = 0

Solving the first subproblem, x - 2 = 0, gives x = 2. Solving the second subproblem, x - 3 = 0, gives x = 3.

Step 4

The final step is to combine the two previous solutions, x = 2 and x = 3, into one solution for the original problem.

x2 + 3x = 8x - 6x = 2, 3

Solve by Factoring: Why does it work?

Examine the equation below:

ab = 0

If you let a = 3, then logivally b must equal 0. Similarly, if you let b = 10, then a must equal 0.

Now try letting a be some other non-zero number. You should observe that as long as a does not equal 0, b must be equal to zero.

To state the observation more generally, "If ab = 0, then either a = 0 or b = 0." This is an important property of zero which we exploit when solving by factoring.

When the example was factored into (x - 2)(x - 3) = 0, this property was applied to determine that either (x - 2) must equal zero, or (x - 3) must equal zero. Therefore, we were able to create two equations and determine two solutions from this observation.

A Second Example5x3 = 45x

Step 1

Move all terms to the left side of the equation. We do this by subtracting 45x from each side.

5x3 - 45x = 45x - 45x5x3 - 45x = 0.

Step 2

The next step is to factor the left side completely. We first note that the two terms on the left have a greatest common factor of 5x.

5x(x2 - 9) = 0

Now, (x2 - 9) can be factored as a difference between two squares.

5x(x + 3)(x - 3) = 0

We are left with three factors: 5x, (x + 3), and (x - 3). As explained in the "Why does it work?" section, at least one of the three factors must be equal to zero.

Step 3

Create three subproblems by setting each factor equal to zero.

1. 5x = 02. x + 3 = 03. x - 3 = 0

Solving the first equation gives x = 0. Solving the second equation gives x = -3. And solving the third equation gives x= 3.

Step 4

The final solution is formed from the solutions to the three subproblems.

x = -3, 0, 3