§ 1.4

Solving Linear Equations

Blitzer, Intermediate Algebra, 5e – Slide #2 Section 1.4

Linear Equations

Definition of a Linear Equation

A linear equation in one variable x is an equation that can be written in the form ax + b = 0, where a and b are real numbers and a is not equal to 0.

An example of a linear equation in x is 4x + 2 = 6. Linear equations in x are first degree equations in the variable x.

Blitzer, Intermediate Algebra, 5e – Slide #3 Section 1.4

Properties of Equality

Property Definition

Addition Property of Equality

The same real number or algebraic expression may be added to both sides of an equation without changing the equation’s solution set.

Multiplication Property of Equality

The same nonzero real number may multiply both sides of an equation without changing the equation’s solution set.

Blitzer, Intermediate Algebra, 5e – Slide #4 Section 1.4

Solving Linear Equations

Solving a Linear Equation

1) Simplify the algebraic expressions on each side.

2) Collect all the variable terms on one side and all the numbers, or constant terms, on the other side

3) Isolate the variable and solve.

4) Check the proposed solution in the original equation.

Blitzer, Intermediate Algebra, 5e – Slide #5 Section 1.4

Solving Linear Equations

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve and check: 5 - 3x + 4x = 1 - 7x + 12.

1) Simplify the algebraic expressions on each side.

5 - 3x + 4x = 1 - 7x + 12

5 + x = 13 - 7x Combine like terms:

-3x + 4x = x

1 + 12 = 13

Blitzer, Intermediate Algebra, 5e – Slide #6 Section 1.4

Solving Linear Equations

2) Collect variable terms on one side and constant terms on the other side.

5 + x + 7x = 13 - 7x + 7x Add 7x to both sides

CONTINUECONTINUEDD

5 + 8x = 13 Simplify

5 – 5 + 8x = 13 - 5 Subtract 5 from both sides

8x = 8 Simplify

Blitzer, Intermediate Algebra, 5e – Slide #7 Section 1.4

Solving Linear Equations

3) Isolate the variable and solve.

CONTINUECONTINUEDD

8x 8 Divide both sides by 8 8 8

x = 1 Simplify

Blitzer, Intermediate Algebra, 5e – Slide #8 Section 1.4

Solving Linear Equations

4) Check the proposed solution in the original equation.

CONTINUECONTINUEDD

5 - 3x + 4x = 2 - 7x + 6 Original equation

5 – 3(1) + 4(1) 1 – 7(1) + 12 Replace x with 1?=

5 – 3 + 4 1 – 7 + 12 Multiply=?

2 + 4 – 6 + 12 Add or subtract from left to right

=?

Add6 = 6

Blitzer, Intermediate Algebra, 5e – Slide #9 Section 1.4

Solving Linear Equations

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve and check:

1) Simplify the algebraic expressions on each side.

Multiply both sides by the LCD: 30

25

2

3

12 xxx

25

2

3

12 xxx

230

5

2

3

1230

xxx

21

30

5

2

1

30

3

12

1

30 xxxDistributive Property

Blitzer, Intermediate Algebra, 5e – Slide #10 Section 1.4

Solving Linear Equations

Cancel

CONTINUECONTINUEDD

Multiply

Distribute

21

03

5

2

1

03

3

12

1

03 xxx

11

15

1

2

1

6

1

12

1

10 xxx

xxx 151261020

xxx 15261210

Blitzer, Intermediate Algebra, 5e – Slide #11 Section 1.4

Solving Linear Equations

CONTINUECONTINUEDD

Combine like terms14x + 2 = 15x

2) Collect variable terms on one side and constant terms on the other side.

2 = x

Subtract 14x from both sides14x – 14x + 2 = 15x – 14x

Simplify

3) Isolate the variable and solve.

Already done.

Blitzer, Intermediate Algebra, 5e – Slide #12 Section 1.4

Solving Linear Equations

CONTINUECONTINUEDD 4) Check the proposed solution in the original equation.

2

2

5

22

3

122

2

2

5

22

3

14

2

2

5

0

3

3

Replace x with 2

Simplify

25

2

3

12 xxx

Original Equation

Simplify

?

?

?

?

Blitzer, Intermediate Algebra, 5e – Slide #13 Section 1.4

Solving Linear Equations

CONTINUECONTINUEDD

1 - 0 = 1 Simplify

1 = 1 Simplify

Since the proposed x value of 2 made a true sentence of 1 = 1 when substituted into the original equation, then 2 is indeed a solution of the original equation.

Blitzer, Intermediate Algebra, 5e – Slide #14 Section 1.4

Categorizing an Equations

Type of Equations Definitions

Identity An equation that is true for all real numbers

Conditional An equation that is not an identity but is true for at least one real number

Inconsistent

(contradiction)

An equation that is not true for any real number

Blitzer, Intermediate Algebra, 5e – Slide #15 Section 1.4

Categorizing an Equation

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve and determine whether the equation is an identity, a conditional equation or an inconsistent equation.

5 + 4x = 9x + 5

5 + 4x = 9x + 5

5 - 5 + 4x = 9x + 5 - 5

4x = 9x

4x – 4x = 9x – 4x

Subtract 5 from both sides

Simplify

Subtract 4x from both sides

Blitzer, Intermediate Algebra, 5e – Slide #16 Section 1.4

Categorizing an Equation

0 = x

Divide both sides by 5

Simplify

The original equation is only true when x = 0. Therefore, it is a conditional equation.

CONTINUECONTINUEDD

0 = 5x

5

5

5

0 x

Simplify

Blitzer, Intermediate Algebra, 5e – Slide #17 Section 1.4

Categorizing an Equation

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve and determine whether the equation is an identity, a conditional equation or an inconsistent equation.

5 – (2x – 4) = 4(x +1) - 2x

5 – (2x – 4) = 4(x +1) - 2x

5 – 2x +4 = 4x + 4 -2x

9 - 2x = 4 - 2x

Distribute the -1 and the 4

Simplify

Since after simplification we see a contradiction, we know that the original equation is inconsistent and can never be true for any x.

9 = 4 Add 2x to both sides.

Blitzer, Intermediate Algebra, 5e – Slide #18 Section 1.4

Categorizing an Equation

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve and determine whether the equation is an identity, a conditional equation or an inconsistent equation.

3 + 2x = 3(x +1) - x

3 + 2x = 3(x +1) - x

3 + 2x = 3x + 3 - x

3 + 2x = 2x + 3

Distribute the 3

SimplifySince after simplification we can see that the left hand side (LHS) is equal to the RHS of the equation, this is an identity and is always true for all x.