1.7_Linear and Absolute Value Inequalities

7/27/2019 1.7_Linear and Absolute Value Inequalities

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College Algebra: Section 1.7

Equations and InequalitiesInvolving Absolute Value

Objectives of this Section

Solve Equations Involving Absolute Value

Solve Inequalities Involving Absolute Value


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Equations Involving Absolute Value

If the absolute value of an expression equals

some positive numbera, then the expression

itself equals eithera or -a. Thus,u a u a u a is equivalent to or


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Solve: 2 3 11x

2 3 11x

2 3 11 2 3 11x x or

2 14x x 7

2 8x x 4

Solution set: {-4, 7}


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u a a u a

u a a u a

is equivalent to

is equivalent to

Theorem

In other words, |u| < a is equivalent

to -a < u and u < a.


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Ex. Solve y + 21 7

- 21 -21y -14

(-14) + 21 7

7 7

1. Draw the river

2. Subtract 21 fromboth sides

3. Simplify4. Check your answer

5. Graph the solution

-14 -13-15


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Ex. Solve 8y + 3 > 9y - 14

o17 1816

- 8y - 8y3 > y - 14

+ 14 + 1417 > y

y

< 178(16) + 3 > 9(16)14131 > 130

1. Draw the river2. Subtract 8y from both

sides

3. Simplify

4. Add 14 to both sides5. Simplify

6. Rewrite inequality

with the variable first

7. Check your answer8. Graph the solution


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Ex. Solve 8y + 3 > 9y - 14

o17 1816

- 8y - 8y3 > y - 14

+ 14 + 1417 > y

y

< 178(16) + 3 = 9(16)14131 > 130

Big Tip!!!

At the end of solving

inequality, always put the

variable at the LEFT hand

side. Then arrow of the

inequality sign tells you the

correct graph

y < 17

The graph should toward to

the left.


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To Solve the Absolute Value Inequalities

1. Isolate the absolute value expression.

2. Make sure the absolute value inequality can be defined.

3. For any defined absolute value inequality with template:|X| a, where a> 0 and = , ,

write the corresponding compound inequalities by following the

rules:

a) if |X| > a or |X| a, meaning greator, leaving

the jail

b) set jail boundaries as a and ac) write compound inequalities as

X a meaning stay left to the left

boundary or right to the right boundary

d) if |X| < a or |X| a, meaning less thand, going

to the jail

e) set jail boundaries as a and a

f) write compound inequalities as

a< X < a (and) meaning stay between the two

boundaries

4. Solve the converted compound inequalities.


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Solve | x3 | < 5

Absolute Value Inequality

(1) The absolute value expression is isolated(2) It is a well defined absolute value inequality

(3) It is a less thand inequality. (go to the jail)

a) set jail boundaries: 5, and 5

b) write compound inequalities: (stay in between the jailboundaries)

5 < x3 < 5

2 < x < 8

+3 +3 +3


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You try this!

Solve | x + 4 | < 1


(1) The absolute value expression is isolated

(2) It is a well defined absolute value inequality

(3) It is a less thand inequality. (go to the jail)


b) write compound inequalities: (stay in between the jail

boundaries)

1 < x + 4 < 1

5 < x


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Solve | 2x + 3 |3 2


(1) The absolute value expression is NOT isolated

| 2x + 3 |3 2

| 2x+ 3 | 5

(2) It is a well defined absolute value inequality(3) It is a greator inequality. (leaving the jail)


b) write compound inequalities: (stay left to left boundary

and right to the right boundary)

2x + 3 5 or 2x + 3 5

2x 8 or 2x 2

x 4 or x 1

3 3

+ 3 + 3

3 3


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You try this!

Solve | 4x3 | + 5 8


(1) The absolute value expression is NOT isolated

| 4x3| + 5 8

| 4x3 3

(2) It is a well defined absolute value inequality(3) It is a greator inequality. (leaving the jail)


b) write compound inequalities: (stay left to left boundary

and right to the right boundary)

4x3 3 or 4x3 3

4x 0 or 4x 6

x 0 or x 3/2

+ 3 + 3

5 5

+ 3 + 3


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Summary

1. Before solving absolute value equations or inequalities, you

MUST isolate the absolute value expression and check themare defined or not.

2. Remember telling yourself the jail story. It will help you set

up the correct equations or compound inequalities.

3. Solve the equations or compound inequalities and dont

forget the learned knowledge such as when multiplying or

dividing a negative number, you need flip the inequality

sign.


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Solve: 3 1 5x

3 1 5x

5 3 1 5x

4 3 6x

4

32x

Solution Set:

2,3

42

3

4| orxx


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u a u a u a

u a u a u a

is equivalent to or

is equivalent to or

Ifa is any positive number, then

Theorem


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Solve: 4 3 15x

4 3 15x

4 3 15x or 4 3 15x

4 18x

x 9

2

4 12x

x 3

),3[2

9,3

2

9|

orxorxx

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1.7_Linear and Absolute Value Inequalities