Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1 1 Section 2.3 Solving Linear Equations...

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1

Section 2.3

Solving LinearEquations

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1


Objective #1 Solve linear equations.


1. Simplify the algebraic expression on each side.

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

3. Isolate the variable and solve.

4. Check the proposed solution in the original equation.

Linear Equations

A step-by-step procedure for Solving Linear Equations:


Solve and check: 7x = 3(x + 8) 7x = 3(x + 8) 7x = 3x + 24 Use the Distributive Property.7x – 3x = 3x + 24 – 3x Subtract 3x from both sides. 4x = 24 Simplify. x = 6 Solve.

7(6) =3(6 + 8) Check the solution in the original equation.

42 = 3(14) 42 = 42 The true statement 42 = 42

verifies 6 is the solution.

Solving Linear Equations

EXAMPLEEXAMPLE


Objective #1: Examples

1a. Solve: 7 25 3 16 2 3x x x

Simplify the algebraic expression on each side. Collect variable terms on one side and constant terms on the other side.

Isolate the variable and solve. The solution set is 6 .

7 25 3 16 2 3

4 25 13 2

x x x

x x

4 25 2 13 2 2

2 25 13

2 25 25 13 25

2 12

x x x x

x

x

x

2 122 2

6

x

x



1a. Solve: 7 25 3 16 2 3x x x

Simplify the algebraic expression on each side. Collect variable terms on one side and constant terms on the other side.


7 25 3 16 2 3

4 25 13 2

x x x

x x

4 25 2 13 2 2

2 25 13

2 25 25 13 25

2 12

x x x x

x

x

x

2 122 2

6

x

x



1b. Solve: 4(2 1) 29 3(2 5)x x

Simplify the algebraic expression on each side.

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


4(2 1) 29 3(2 5)

8 4 29 6 15

8 25 6 15

x x

x x

x x

8 6 25 6 6 15

2 25 15

2 25 25 15 25

2 10

x x x x

x

x

x

2 102 2

5

x

x


Objective #2 Solve linear equations containing fractions.


Solving a linear equation involving fractions

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

230

5

2

3

1230

xxx

21

30

5

2

1

30

3

12

1

30 xxxDistributive Property

25

2

3

12 xxx



Cancel

CONTINUEDCONTINUED

Multiply

Distribute

30 2 1 30 2 30

1 3 1 5 1 2

x x x

11

15

1

2

1

6

1

12

1

10 xxx

xxx 151261020

xxx 15261210



CONTINUEDCONTINUED

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.



CONTINUEDCONTINUED

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.

?

?

?



CONTINUEDCONTINUED

Simplify

1 = 1 Simplify

Since the proposed x value of 2 makes the statement true,2 is indeed the solution of the original equation.



2. Solve: 2 5

4 3 6x x

Begin by multiplying both sides of the equation by 12, the least common denominator.

The solution set is 2 .

2 54 3 6

2 512 12

4 3 6

2 512 12 12

4 3 63 8 10

3 8 8 8 10

5 10

5 105 5

2

x x

x x

x x

x x

x x x x

x

x

x


Objective #3 Solve linear equations containing decimals.


3. Solve: 0.48 3 0.2( 6)x x

First apply the distributive property to remove the parentheses, and then multiply both sides by 100 to clear the decimals. The solution set is 15 .

0.48 3 0.2( 6)

0.48 3 0.2 1.2

100(0.48 3) 100(0.2 1.2)

48 300 20 120

48 300 300 20 120 300

48 20 420

48 20 20 20 420

28 420

28 42028 28

15

x x

x x

x x

x x

x x

x x

x x x x

x

x

x



Objective #4 Identify equations with no solution or infinitely

many solutions.


Some equations are not true for any real number. Such equations are called inconsistent equations or contradictions.

Example:Solve: 2x = 2(x +3) 2x = 2x + 6 Distributive Property.

2x – 2x = 2x + 6 – 2x Subtract 2x from both sides.

0 = 6 Simplify. False, 0 ≠ 6.

Inconsistent, no solution.

Linear equations having no solutions


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


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

4x = 9x

4x – 4x = 9x – 4x

Subtract 5 from both sides

Simplify

Subtract 4x from both sides



0 = x

Divide both sides by 5

Simplify

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

CONTINUEDCONTINUED

0 = 5x

5

5

5

0 x

Simplify



EXAMPLEEXAMPLE

SOLUTIONSOLUTION


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.



EXAMPLEEXAMPLE

SOLUTIONSOLUTION


3 + 2x = 3(x +1) – x

3 + 2x = 3(x +1) – x

3 + 2x = 3x + 3 – x

3 + 2x = 2x + 3

Distribute the 3.

Simplify.Since 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.



4a. Solve: 3 7 3( 1)x x

3 7 3( 1)

3 7 3 3

3 3 7 3 3 3

7 3

x x

x x

x x x x

The original equation is equivalent to the false statement 7 3. Thus, the equation has no solution.

The solution set is .



4b. Solve: 3( 1) 9 8 6 5x x x

3( 1) 9 8 6 5

3 3 9 3 6

3 6 3 6

3 3 6 3 3 6

6 6

x x x

x x

x x

x x x x

The original equation is equivalent to 6 6, which is true for every

value of x. The equation’s solution is all real numbers or is a real number .x x


Objective #5 Solve applied problems using formulas.


The formula

describes the pressure of sea water, p, in pounds per square foot, at a depth d feet below the surface.

At what depth is the pressure 30 pounds per square foot?

Substitute the given pressure into the equation for p,

p = 35. The equation becomes:

11

515

dp

11

51530

d

Linear Equations – An application


The depth where the pressure is 30 pounds per square foot is 33 feet.

530 15

115

30 11 15 11 Multiply both sides by 11, the LCD. 11

330 165 5d Simplify.

330 165 165 5d 165 Subtract 165 from both sides.

165 5 Simpl

d

d

d

ify.

165 5 Divide both sides by 5.

5 5d 33 Solve.

d

Linear Equations

CONTINUEDCONTINUED


The depth where the pressure is 30 pounds per square foot is 33 feet.

530 15

115

30 11 15 11 Multiply both sides by 11, the LCD. 11

330 165 5d Simplify.

330 165 165 5d 165 Subtract 165 from both sides.

165 5 Simpl

d

d

d

ify.

165 5 Divide both sides by 5.

5 533 Solve.

d

d

Linear Equations

CONTINUEDCONTINUED



5. It has been shown that persons with a low sense of humor have higher levels of depression in response to negative life events than those with a high sense of humor. This can be modeled by the following formulas:

Low-Humor Group: 10 539 9

D x

High-Humor Group: 1 269 9

D x

If the low-humor group averages a level of depression of 10 in

response to a negative life event, what is the intensity of that event?

Where x represents the intensity of a negative life event (from a low of 1 to a high of 10) and D is the level of depression in response to that event.



The formula indicates that if the low-humor group averages a level of depression of 10 in response to a negative life event, the intensity of that event is 3.7.

10 539 9

10 5310

9 910 53

9 10 99 9

90 10 53

90 53 10 53 53

37 10

37 1010 103.7

3.7

D x

x

x

x

x

x

x

x

x

CONTINUEDCONTINUED

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1 1 Section 2.3 Solving Linear Equations...

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Transcript of Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1 1 Section 2.3 Solving Linear Equations...