CHAPTER 1.3

Solving Equations

Equivalent EquationsThe equations are equivalent If they

have the same solution(s)

Example:

Determine whether 4x = 12 and 10x = 30 are equivalent equations

Determine whether 3x = 4x and 3/x = 4/x are equivalent equations

ADDITION AND MULTIPLICATION PROPERTIES OF EQUALITY

Let a, b, and c represent algebraic expressionsAddition property of equality: If

a = b,

then a + c = b + c

Multiplication property of equality: If a = b,

then a(c) = b (c)

APPLYING THE ADDITION PROPERTIE OF EQUALITY

In each equation, the goal is to isolate the variable on one side of the equation. To accomplish this, we use the fact that the sum of a number and its opposite is zero and the difference of a number and itself is zero.

p – 4 +4 = 4 +4

To isolate p, add 4 to both sides (-4 +4 = 0).

p – 4 = 11

Simplifyp- + 0 = 15p = 15

CHECK

Let’s try: : 4.7 13.9Solve y

4.7 13.9

4.7 13.9

13.9 4.7

18.6

:

4.7 13.9

13

4.7 4.7

0

18.6

.9 13.9

y

y

y

y

Check

Using the addition principle, adding 4.7 to both sides

The solution of this equation is 18.6

Substituting 18.6 for y

TRUE

Applying the Multiplication Properties of Equality

12x = 60Tip: Recall that the product of a number and its reciprocal is 1. For example:

(112

12) 1

12x = 60 12 12

To obtain a coefficient of 1 for the x-term, divide both sides by 12

Simplify

x = 5

Check!

Example:

2 1

9 3q

Tip: When applying the multiplication or division properties of equality to obtain a coefficient of 1 for the variable term, we will generally use the following convention:1. If the coefficient of the variable term is

expressed as a fraction, we usually multiply both sides by its reciprocal.

2. If the coefficient of the variable term is an integer or decimal, we divide both sides by the coefficient itself.

29 9( ) ( )

2( )92

1

3q To obtain a coefficient of 1 for

the q-term, multiply by the reciprocal ofwhich is

2

9

9

2

Simplify. The product of a number and its reciprocal is 1.

3

2q

CHECK!

Let’s try: : 8 72Solve t8 72

8 72

72

89

:

8 72

1 1( )8 8

(9

7 72

)

2

t

t

t

t

Check

Using the multiplication principle, multiply by 1/8 to both sides

The solution of this equation is 9

Substituting 9 for t

TRUE

Steps to solve a linear Equation in One Variable

Simplify both sides of the equation. Clear parentheses Combine like termsUse the addition or subtraction property of equality to

collect the variable terms on one side of the equation.Use the addition or subtraction property of equality to

collect the constant terms on the other side of the equation.

Use the multiplication or division property of equality to make the coefficient of the fvariable term equal to 1.

Check your answer.

We will want to find those values of x that make the equation true by isolating the x (this means get the x all by itself on one side of the equal sign)

12542 xxSince the x is in more than one place and inside of parenthesis the first thing we’ll do is get rid of parenthesis by distributing.51082 xx Now let’s get all constants

(terms without x’s) on the right side. We’ll do this by adding 8 to both sides.

+ 8 + 8

+ 3

3102 xx We are ready to get all x terms on the left side by adding 10x to both sides.+ 10x + 10x

312 xNow get the x by itself by getting rid of the 12. 12x means 12 times x so we get rid of it by dividing both sides by 12.

12 12 4

1x

Let’s check this answer by substituting it into the original equation to see if we get a true statement.

12542 xx4

1

4

1 Distribute and multiply

Distribute

Get a common denominator

2

15

2

15

It checks!

?

12

158

2

1 ?

52

58

2

1 ?

2

10

2

5

2

16

2

1

?

Let’s try:3( 4) 7x x

3 12 7

3 12 7

12 4

1 1( )( 12) ( )44

3

4

3

3

x x

x xx x

x

x

x

Using the addition principle, adding (-3x) to both sides

Using the multiplication principle, multiply by 1/4 to both sides

TRUE

Conditional Equations, Identities, and ContradictionsConditional Equations

An equation that is true for some values of the variable but false for other values.

ContradictionsEquation with no solution

IdentitiesAn equation that has all real numbers as its solutions.

Classify the equation

3 3

2

3 8 5 7

3 8 7 5 7 7

3 5

3 5

2, 21

x x

x x x x

x

x

x

x or

Using the addition principleSimplifying

Divide both sides b -1.

There is one solution, -2. For other choices of x, the equation is false. This equation is conditional since it can be true or false, depending on the replacement for x.

Using the addition principleSimplifying

Classify the equation3 5 3( 2) 4

3 5 3 6 4

3 5 3 2

3 5 2

2

3 3

5

3

x x

x x

x x

x x xx

Using the distributive law

Combining like terms.

The equation is false regardless of what x is replaced with, so all real numbers are solutions. There is no solution. This equation is a contradiction.

Using the addition principle, adding -3x to both sides

Classify the equation2 7 7( 1) 5

2 7 7 7 5

2 7 2 7

x x x

x x x

x x

Using the distributive law

Combining like terms.

The equation is true regardless of what x is replaced with, so all real numbers are solutions. This equation is a Identity.

7 2 3 4

2 9 3(4 1) 1

x x x

x x

Let’s try: