Type

7 7

13 ( 13) 13

Example

Solution

Example Find the solution set:

a) |x| = 6; b) |x| = 0; c) |x| = –2

a) We interpret |x| = 6 to mean that the number x is 6 units from zero on a number line. Thus the solution set is {–6, 6}.

6-6

b) We interpret |x| = 0 to mean that x is 0 units from zero on a number line. The only number that satisfies this is zero itself. Thus the solution set is {0}.

.c) Since distance is always nonnegative, |x| = –2 has no solution.

Thus, the solution set is

Example Find the solution set:

a) |2x +1| = 5; b) |3 – 4x| = –10

The solution set is {–3, 2}.

x = –3 or x = 2

2x = –6 or 2x = 4

Substituting

a) We use the absolute-value principle, knowing that 2x + 1 must be either 5 or –5:

|2x +1| = 5

|X| = p

2x +1 = –5 or 2x +1 = 5

Find the zeros using “CALC”.

The solution set is {–3, 2}.

b) The absolute-value principle reminds us that absolute value is always nonnegative. The equation |3 – 4x| = –10 has no solution. The solution set is .

Example Find the solution set:

a) |2x +1| = 5; b) |3 – 4x| = –10

No zeros!

The solution set is .

Solution

Example Given that f (x) = 3|x+5| – 4, find all x for which f (x) = 11

Since we are looking for f (x) = 11, we substitute:

Replacing f (x) with 3|x + 5| - 4

x +5 = –5 or x +5 = 5

x = –10 or x = 0

The solution set is {–10, 0}.

3|x+5| – 4 = 11

f (x) = 11

3|x+5| = 15

|x+5| = 5

Sometimes an equation has two absolute-value expressions. Consider |a| = |b|. This means that a and b are the same distance from zero.

If a and b are the same distance from zero, then either they are the same number or they are opposites.

Example To solve |3x – 5| = |8 + 4x| we would consider the two cases.

and solve each equation.

3x – 5 = 8 + 4x

This assumes these numbers are the same.

This assumes these numbers are opposites.

3x – 5 = –(8 + 4x) or

Solution

Solution

Example Solve |x| < 3. Then graph.

The solutions of |x| < 3 are all numbers whose distance from zero is less than 3. By substituting we see that numbers like –2, –1, –1/2, 0, 1/3, 1, and 2 are all solutions.

The solution set is {x| –3 < x < 3}.

In interval notation, the solution is (–3, 3). The graph is as follows:

-3 31 abs( ) 3 0y x

Solution

Example Solve |x| < 3. Then graph.

The solutions of |x| < 3 are all numbers whose distance from zero is less than 3. By substituting we see that numbers like –2, –1, –1/2, 0, 1/3, 1, and 2 are all solutions.

The solution set is {x| –3 < x < 3}.

In interval notation, the solution is (–3, 3). The graph is as follows:

-3 31 abs( ) 3 0y x

Solution

The solutions of are all numbers whose distance from zero great than or equal to 3 units. The solution set is

In interval notation, the solution is

The graph is as follows:

Example Solve 3. Then graph.x

3x

{ | 3 3}. x x or x

( , 3] [3, ).

-3 3

1 abs( ) 3 0y x

Example Solve.

4 3 7a.) 8

5

x

b.) 7 3 4 5 25a