Transcript Absolute Value Equations

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
IS POSITIVE
POSITIVE
x  2 IS NEGATIVE
|| xx  22 ||  55
xx  22  5
5
|x2|5
xx  77
x  2  5
x  3
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
IS POSITIVE
POSITIVE
2x  7 IS
IS NEGATIVE
NEGATIVE
| 2x  7 |  5  4
| 2x  7 |  5  4
| 2x  7 |  9
| 2x  7 |  9
2x  7  +9
2x  16
x8
2x  7  9
9
2x  2
TWO SOLUTIONS
x  1
Solving an Absolute-Value Equation
Recall that |xx is
| isthe
thedistance
distancebetween
betweenxxand
and0.0.IfIf x
| x 
| 8,8,then
then
any number between 8 and 8 is a solution of the inequality.
8 7 6 5 4 3 2 1
0
1
2
3
4
5
6
You can use the following properties to solve
absolute-value inequalities and equations.
7
8
SOLVING ABSOLUTE-VALUE INEQUALITIES
SOLVING ABSOLUTE-VALUE EQUATIONS AND INEQUALITIES
| ax  b |  c
means
ax  b  c
and
a x  b   c.
means valueais
| a xWhen
 b | an
c absolute
x less
b than
c
and
a x the
b   c.
a number,
inequalities are connected by and. When an absolute
means
| a xvalue
 b | iscgreater
a x  b  cthe inequalities
or
a x  bare  c.
than a number,
connected by or.
| ax  b |  c
means
ax  b  c
or
a x  b   c.
| ax  b |  c
means
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
symbol.
The solution is all real numbers greater than 1 and
less than
7.
This can be written as 1  x  7.
Solving an Absolute-Value Inequality
Solve
 1 | 3  6 and graph
2x + 1| 2x
IS POSITIVE
2x +the
1 ISsolution.
NEGATIVE
| 2x  1 |  3  6
| 2x  1 | 3  6
2x + 1 IS POSITIVE
| 2x|2x1 | 13|  69
2x + 1 IS NEGATIVE
1 |  69
| 2x|2x1 | 3
2x  1  9
| 2x  1 |  9
2x  10
2x  8
2x  1  9
2x  1  +9
x4
x  5
2x  10
2x  8
The solution is all real numbers greater than or 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
2x11|  +9
| 2x
9
inequality symbol.
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.
0.62  d  0.63
Find the halfway point.
0.625
Subtract 0.625 from
each part of the
compound inequality.
0.62  0.625  d  0.625  0.63  0.625
0.005  d  0.625  0.005
Rewrite as an absolute-value inequality.
| d  0.625 |  0.005
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.