# Absolute Value Equations - PowerPoint

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```					SOLVING ABSOLUTE-VALUE EQUATIONS

You can solve some absolute-value equations using mental math. For instance, you learned that the equation | x | 8 has two solutions: 8 and 8. To solve absolute-value equations, you can use the fact that the expression inside the absolute value symbols can be either positive or negative.

Solving an Absolute-Value Equation

Solve | x  2 |  5

SOLUTION
The expression x  2 can be equal to 5 or 5.

x  2 IS POSITIVE x  2 IS POSITIVE
|| x  2 ||  5 x2 5 x  2  5 x  2  5

x  2 IS NEGATIVE |x2|5 x  2  5 x  3

x7 x7

The equation has two solutions: 7 and –3.
CHECK

|72||5|5

| 3  2 |  | 5 |  5

Solving an Absolute-Value Equation

Solve | 2x  7 |  5  4 SOLUTION Isolate the absolute value expression on one side of the equation.

2x  7 IS POSITIVE IS POSITIVE

2x  7 IS NEGATIVE IS NEGATIVE | 2x  7 |  5  4 | 2x  7 |  9 2x  7  9 9 2x  2 TWO SOLUTIONS x  1

| 2x  7 |  5  4
| 2x  7 |  9 2x  7  +9 2x  16

x8

Solving an Absolute-Value Equation

Recall that | xis the distance between xxand 0. IfIf xx  x | is the distance between and 0. | | 8, 8, then any number between and 8 8 is a solution of the then any number between 88 andis a solution of the inequality.

8 7 6 5 4 3 2 1 8

0

1

2

3

4

5

6

7

You can use the following properties to solve absolute-value inequalities and equations.

SOLVING ABSOLUTE-VALUE INEQUALITIES

SOLVING ABSOLUTE-VALUE EQUATIONS AND INEQUALITIES
| ax  b |  c means ax  b  c and a x  b   c.

| a x  b | an absolute value b less than a number,  c. a x is  c and a x  b the When c means

inequalities are connected by and. When an absolute | a x  b |  cgreater thana x number, the inequalities  c. or a x  b are value is means a bc connected by or.
| ax  b |  c means means ax  b  c or a x  b   c.

| ax  b |  c

ax  b  c

or

a x  b   c.

Solving an Absolute-Value Inequality

Solve | x  4 | < 3

x  4 IS POSITIVE |x4|3 x  4  3 x7

x  4 IS NEGATIVE |x4|3 x  4  3  x1

Reverse inequality less than The solution is all real numbers greater than 1 and symbol. 7

This can be written as 1  x  7.

Solving an Absolute-Value Inequality

Solve | 2x  1 | 3  6 and graph the 2x + 1 IS POSITIVE 2x + 1 IS NEGATIVE solution.

2x + 1 IS POSITIVE 2x | 1 |  6 | 2x|  1   3  9 2x  1 9 | 2x  1 |+9

| 2x  1 |  3  6

2x + 1 IS NEGATIVE 2x | 1 |  9 | 2x|  1  3  6

| 2x  1 | 3  6

2x  1  9 | 2x  1 |  9 2x  10 2x  8 2x  1  9  2x  1  +9 x is x  5 The solution 4all real numbers greater10 or 2x  than 2x  8 equal x4 x  5 to 4 or less than or equal to  5. This can be written as the compound inequality x   5 or x  4.  Reverse 5 inequality symbol. 4.
6 5 4 3 2 1 0 1 2 3 4 5 6

Writing an Absolute-Value Inequality

You work in the quality control department of a manufacturing company. The diameter of a drill bit must be between 0.62 and 0.63 inch. a. Write an absolute-value inequality
to represent this requirement.

b. Does a bit with a diameter of 0.623
inch meet the requirement?

Writing an Absolute-Value Inequality

The diameter of a drill bit must be between 0.62 and 0.63 inch.
a. Write an absolute-value inequality to represent this requirement.

Let d represent the diameter (in inches) of the drill bit.
Write a compound inequality. Find the halfway point. Subtract 0.625 from

0.62  d  0.63 0.625

0.62  0.625  d  0.625  0.63  0.625

each part of the 0.005  d  0.625  0.005 compound inequality. an absolute-value inequality. d  0.625 |  0.005 Rewrite as | This inequality can be read as “the actual diameter must differ from 0.625 inch by no more than 0.005 inch.”

Writing an Absolute-Value Inequality

The diameter of a drill bit must be between 0.62 and 0.63 inch. b. Does a bit with a diameter of 0.623 meet the requirement?

| d  0.625 |  0.005 | 0.623  0.625 |  0.005 | 0.002 |  0.005
0.002  0.005
Because 0.002  0.005, the bit does meet the requirement.

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