Solving Absolute Value Equations.
Formal Definition of Absolute Value:
a, ifa  0
a, ifa  0
a 
For example,
x-3
x 3
if
(x-3) ≥ 0
When simplified,
x-3
x 3
=
-(x-3) if (x-3) < 0
if x ≥ 3
=
-x+3 if x< 3
Solving Equations with absolute value:
Algebraically :
Step 1. Isolate absolute value on the left side of the equation.
Step 2. Split the equation into two equations using the formal definition of Absolute value
Step 3. Solve each equation.
Step 4. Check both solutions and record your answer.
Example 1.Solve Absolute value equations algebraically:

Solve | 2x – 3 | – 4 = 3
Step 1. Isolate the absolute-value part on the left side.
| 2x – 3 | – 4 = 3
| 2x – 3 | = 7
Step 2. Clear the absolute-value bars by splitting the equation into its two equations, one for each
sign:
(2x – 3) = 7
2x – 3 = 7
2x = 10
x=5
or
or
or
or
–(2x – 3) = 7
–2x + 3 = 7
–2x = 4
x = –2
Answer: x = –2, 5.
Solving Absolute value equations graphically:
Step 1. Graph two equations,
intersections of the graphs.
y1 = | 2x – 3 | – 4 and y2 = 3 and look for the
Find the x-coordinates of the points of intersection. Those are your solutions.
Answer: x= -2, 5
Example 2.
Solve
Example 3.
x 3  5
x-3 =5
or
x= 8
Solve
5x  9  x  3
-(x-3) =5
5x-9 =x+3 or -(5x-9)=x+3
-x+3= 5
-x=2
X=-2
4x=12
-5x+9=x+3
x=3
-6x=-6
x=1
Check the solutions:
|8-3| =5 Ok
|-2-3| =5 Ok
Answer: x= -2, 5
Example 4.
Solve
Check the solutions:
|5(3)-9|=3+3
OK
|5(1)-9| =1+3
OK
Answer: x=1, 3
Example 5.
4  2 6  3x  8
Solve |x-3|= 3x+5
-2|6-3x|=-12
x-3=3x+5
6-3x =6
-3x=0
x=0
x=4
|6-3x| =6
or
–(6-3x)=6
-6+3x=6
3x=12
-2x =8
x=-4
-(x-3)=3x+5
-x+3=3x+5
-4x=2
x=-.5
Check the solutions:
4-2|6-3(0)|=-8 ok
4-2|6-3(4)|=-8 ok
Check the solutions:
|-4-3|=3(-4)+5
Not OK! Extraneous solution.
|-.5-3|=3(-.5)+5 Ok.
Answer: x=0, 4
Answer: x=-.5
Example 6.
Solve | x2– 4x – 5 | = 7
Clear the absolute-value bars by splitting the equation into two equations:
( x2 – 4x – 5 ) = 7
Solve the first
equation:
2
x – 4x – 5 = 7
x2 – 4x – 12 = 0
(x – 6)(x + 2) = 0
x = 6, x = –2
–(x2 – 4x – 5) = 7
or
Solve the second equation:
–x2 + 4x + 5 = 7
–x2 + 4x – 2 = 0
0 = x2 – 4x + 2
Applying the Quadratic Formula to the above, I get:
Answer:
To confirm this graphically, you can look for the intersections of y1 = | x2 – 4x – 5 | and y2 = 7: