Transcript Absolute Value Precalc Absolute Valuex

1.6 Absolute Value Equations
and Inequalities
Learning Target: I can write
and solve equations
involving absolute value
Solving Absolute Value Equations
Absolute value is denoted by the bars | |.
Absolute value represents the distance a
number is from 0. Thus, it is always
positive.
 |8| = 8 and |-8| = 8
Solving absolute value equations
• 1. Isolate the absolute value expression.
• 2. Set up two equations to solve.
• For the first equation, drop the absolute
value bars and solve the equation.
• For the second equation, drop the bars,
negate the opposite side, and solve the
equation.
• 3. Check the solutions.
6|5x + 2| = 312
• Isolate the absolute value expression by dividing by 6.
6|5x + 2| = 312
|5x + 2| = 52
• Set up two equations to solve.
5x + 2 = 52
5x = 50
x = 10
or
5x + 2 = -52
5x = -54
x = -10.8
•Check: 6|5x + 2| = 312
6|5(10)+2| = 312
6|52| = 312
312 = 312
6|5x + 2| = 312
6|5(-10.8)+ 2| = 312
6|-52| = 312
312 = 312
3|x + 2| -7 = 14
• Isolate the absolute value expression by adding 7 and dividing by 3.
3|x + 2| -7 = 14
3|x + 2| = 21
|x + 2| = 7
• Set up two equations to solve.
x+2=7
x=5
or
•Check: 3|x + 2| - 7 = 14
3|5 + 2| - 7 = 14
3|7| - 7 = 14
21 - 7 = 14
14 = 14
x + 2 = -7
x = -9
3|x + 2| -7 = 14
3|-9+ 2| -7 = 14
3|-7| -7 = 14
21 - 7 = 14
14 = 14
6|2x + 5| = 6x + 24
• Isolate the absolute value expression by dividing by 6.
6|2x + 5| = 6x + 24
|2x + 5| = x + 4
• Set up two equations to solve.
2x + 5 = x + 4
2x = x – 1
x = -1
2x + 5 = -(x + 4)
2x + 5 = -x - 4
3x + 5 = -4
3x = -9 so
x = -3
Check your work.
Extraneous Solutions
Absolute
value equation
problem solving process isn’t
perfect:
Very
important to check answers
of absolute value equations
One or more solutions may be
extraneous.
Extraneous Solutions
Extraneous:
a solution
derived from an original
equation that is not a
solution to the original
equation.
|3x + 2| = 4x + 5
|3x + 2| = 4x+5
Set up two equations to solve.
3x + 2 = 4x + 5
-x = 3
x = -3
or
•Check:
|3x + 2| = 4x + 5
|3(-3) + 2| = 4(-3) + 5
|-9 +2| = -12 + 5
|-7| ≠ -7
EXTRANEOUS
3x + 2 = -(4x + 5)
3x + 2 = -4x - 5
7x = -7
x = -1
|3x + 2| = 4x + 5
|3(-1)+ 2| = 4(-1) + 5
|-3+2| = -4 + 5
|-1| = 1
1= 1
|2x + 5| = 3x + 4
|2x + 5| = 3x+4
Set up two equations to solve.
2x + 5 = 3x + 4
-x = -1
x=1
•Check:
2x + 5 = -(3x + 4)
2x + 5 = -3x - 4
or
5x = -9
−𝟗
x=
𝟓
|2x + 5| = 3x + 4
|2(1) + 5| = 3(1) + 4
|2x + 5| = 3x + 4
−𝟗
−𝟗
|2( )+ 5| = 3( ) + 4
|2 +5| = 3 + 4
|7| = 7
| |≠
𝟓
𝟓
EXTRANEOUS
𝟓
𝟕
−𝟕
𝟓