Chapter 1

Section 5Absolute Value Equations and Inequalities

ALGEBRA 2 LESSON 1-5ALGEBRA 2 LESSON 1-5

Solve each equation.

1. 5(x – 6) = 40 2. 5b = 2(3b – 8) 3. 2y + 6y = 15 – 2y + 8

4. 4x + 8 > 20

(For help, go to Lessons 1-3 and 1-4.)

Absolute Value Equations and InequalitiesAbsolute Value Equations and Inequalities

Solve each inequality.

6. 4(t – 1) < 3t + 55. 3a – 2 a + 6>–

1. 5(x – 6) = 40 2. 5b = 2(3b – 8)

= 5b = 6b – 16

x – 6 = 8 –b = –16x = 14 b = 16

Solutions

ALGEBRA 2 LESSON 1-5ALGEBRA 2 LESSON 1-5Absolute Value Equations and InequalitiesAbsolute Value Equations and Inequalities

5(x – 6)5

405

5. 3a – 2 a + 6 6. 4(t – 1) < 3t + 53a – a 6 + 2 4t – 4 < 3t + 5

2a 8 4t – 3t < 5 + 4a 4 t < 9

3. 2y + 6y = 15 – 2y + 8 4. 4x + 8 > 202y + 6y + 2y = 15 + 8 4x > 12

10y = 23 x > 3y = 2.3

>–>–>–>–

Absolute Value Absolute Value – the distance from zero a number is

on the number line – it is always positive Symbol : │x│ Definition:

If x is positive ( x > 0) then │x│ = x If x is negative ( x < 0) then │x │ = -x

Absolute Value Equations have a possibility of two solutions This is because the value inside the │ │ can equal either the

negative or the positive of the value on the other side of the equal sign

Always isolate the absolute value expression on one side of the equal sign before breaking the problem into two pieces.

Absolute Value Equations and InequalitiesAbsolute Value Equations and Inequalities

Solve |15 – 3x| = 6.

|15 – 3x| = 6

15 – 3x = 6 or15 – 3x = –6The value of 15 – 3x can

be 6 or –6 since |6| and |–6| both equal 6.

x = 3 or x = 7Divide each side of both

equations by –3.

–3x = –9 –3x = –21Subtract 15 from each

side of both equations.

Check: |15 – 3x| = 6 |15 – 3x| = 6|15 – 3(3)| 6 |15 – 3(7)| 6

|6| 6 |–6| 66 = 6 6 = 6

Try This Problem│3x + 2 │ = 7

3x + 2 = 7 or 3x + 2 = -7 3x = 5 or 3x = -9

x = 5/3 or x = -3

Check your answer by plugging it back into the equation.

.

77

725

72353

77

729

72)3(3

Absolute Value Equations and InequalitiesAbsolute Value Equations and Inequalities

Solve 4 – 2|x + 9| = –5.

4 – 2|x + 9| = –5

–2|x + 9| = –9 Add –4 to each side.

x = –4.5 or x = –13.5Subtract 9 from

each side of both equations. Check: 4 – 2 |x + 9| = –5 4 – 2|x + 9| = –5

4 – 2 |–4.5 + 9| –5 4 – 2 |–13.5 + 9| –5

4 – 2 |4.5| –5 4 – 2 |–4.5| –5

4 – 2 (4.5) –5 4 – 2 (4.5) –5

–5 = –5 –5 = –5

|x + 9| = Divide each side by –2.92

x + 9 = or x + 9 = – Rewrite as two equations.92

92

Try This Problem

Solve 2│3x - 1 │ + 5 = 33.

2 │3x - 1 │ = 28

│3x – 1│ = 14

3x – 1 = 14 or 3x – 1 = -14

3x = 15 or 3x = -14

x = 5 or x = -8/3

Check the solutions by plugging them into the original problem. They both work.

Extraneous Solutions

Extraneous solution – a solution of an equation derived from an original equation that is not a solution of the original equation

This is why we MUST check all answers to see if they work in the original problem.

Solve │2x + 5 │ = 3x + 4

2x + 5 = 3x + 4 5 = x + 4 1 = x

Check:2(1) + 5 = 3(1) + 4 2 + 5 = 3 + 4 7 = 7

2x + 5 = -(3x + 4)2x + 5 = -3x – 4 5x + 5 = -4 5x = -9 x = -9/5Check:│2(-9/5) + 5 │ = 3(-9/5) + 4

│-18/5 + 25/5 │ = -27/5 + 20/5

│7/5 │ = -7/5

│7/5 │ ≠ -7/5

The only solution is x = 1.

Try These Problems

Solve and check for extraneous solutions.

a) │2x + 3 │ = 3x + 2

2x + 3 = 3x + 2 or 2x + 3 = -3x – 2 3 = x + 2 or 5x + 3 = -2 1 = x or 5x = -5 1 = x or x = -1

Check:

The solution is x = 1

b) │x │ = x – 1

x = x – 1 or x = -x + 10 = -1 or 2x = 1 x = ½

Check:

There is No Solution.

Homework

Practice 1.5 #13-24 all

Absolute Value Inequalities

Let k represent a positive real number │x │ ≥ k is equivalent to x ≤ -k or x ≥ k

│x │ ≤ k is equivalent to -k ≤ x ≤ k

Remember to isolate the absolute value before rewriting the problem with two inequalities

Solve |2x – 5| > 3. Graph the solution.

|2x – 5| > 3

2x – 5 < –3 or 2x – 5 > 3 Rewrite as a compound inequality.

x < 1 or x > 4

2x < 2 2x > 8 Solve for x.

Try This Problem

Solve │2x - 3 │ > 7

2x – 3 > 7 or 2x – 3 < -7

2x > 10 or 2x < -4

x > 5 or x < -2

Solve –2|x + 1| + 5 –3. Graph the solution.>–

|x + 1| 4 Divide each side by –2 and reverse the inequality.

<–

–2|x + 1| + 5 –3>–

–2|x + 1| –8Isolate the absolute value expression. Subtract 5 from each side.

>–

–4 x + 1 4Rewrite as a compound inequality.

<– <–

–5 x 3Solve for x.

<– <–

Try This Problem

Solve |5z + 3| - 7 < 34. Graph the solution. |5z + 3| -7 < 34 |5z + 3| < 41-41 < 5z + 3 < 41-44 < 5z < 38-44/5 < z < 38/5

-8 4/5 < z < 7 3/5

Ranges in Measurement

Absolute value inequalities and compound inequalities can be used to specify an allowable range in measurement.

Tolerance – the difference between a measurement and its maximum and minimum allowable values Equals one half (½) the difference

between the max and min values

ToleranceFor example, if a manufacturing specification calls

for a dimension d of 10 cm with a tolerance of 0.1 cm, then the allowable difference between d and 10 is less than or equal to 0.1

|d - 10 | ≤ 0.1 absolute value inequality

d – 10 ≤ 0.1 and d – 10 ≥-0.1 equivalent compound inequality

-0.1 ≤ d – 10 ≤ 0.1 equivalent compound inequality

9.9 ≤ d ≤ 10.1 simplified compound inequality

The area A in square inches of a square photo is required to

satisfy 8.5 ≤ A ≤ 8.9. Write this requirement as an absolute

value inequality.

Write an inequality.–0.2 A – 8.7 0.2<– <–

Rewrite as an absolute value inequality.|A – 8.7| 0.2<–

Find the tolerance.8.9 – 8.5

2 =0.42 = 0.2

Find the average of the maximum andminimum values.

8.9 + 8.52 =

17.42 = 8.7

Try This ProblemThe specification for the circumference C in inches of a

basketball for junior high school is 27.75 ≤ C ≤ 30. Write the specification as an absolute value inequality.

375.2

75.

2

75.275.28

125.282

25.56

2

75.275.28

375.125.28 C

Find the tolerance.

Find the average from min and max values.

Write the absolute value inequality.

Homework

Practice 1.5 # 1 – 12, 25 – 30

Watch the units on the back page Make sure all units are inches,

centimeters, meters, etc…