Pre-Calculus 6.0
SOLVING ABSOLUTE VALUE EQUATIONS
VIA THE “DISTANCE” APPROACH
OBJECTIVE:
Using the basic concept of absolute value as distance from a central value, solve
absolute value equations in which a linear variable (e.g. x raised to the first power)
exists only within the absolute value.
Given an absolute value equation of the form
x−h =
c where h and c are constants,
the solution set contains those values that are c units away from h
x = { h − c, h + c } .
KEY STEPS:
1) If the variable has a coefficient other than 1, factor that coefficient out of
both terms in the expression within the absolute value.
2) Use properties of absolute value to transform the expression within the
absolute value to be of the form x − h . Recall the key properties of absolute
values as follows:
− a =a ;
a•b = a • b ;
a/b = a / b
3) Isolate the absolute value term x − h ; at this point x − h should be set
equal to some constant, c.
4) Identify the central value, h. Calculate the solution values as given by those
values that are c units less than h and c units more than h.
5) State solution using proper set notation: x ={ h − c , h + c }
3 4x − 8 + 5 =
35
Example 1:
1) Factor out coefficient of linear term:
3 4 ( x − 2) + 5 =
35
2) Use properties of absolute value to “extract the coefficient:
3• 4 • x − 2 + 5 =
35
12 • x − 2 + 5 =
35
3) Isolate the absolute value term:
12 • x − 2 + 5 =
35
−5 −5
1
1
30  
 12 • x − 2 =
 12 
 12 
5
x−2 =
2
4) Identify central value and solution values
Central value: h = 2
5  1 9
 5
Solution values: x = 2- , 2 +  = − , 
2  2 2
 2
:
Pre-Calculus 6.0
SOLVING ABSOLUTE VALUE EQUATIONS
VIA THE “DISTANCE” APPROACH
2
3 − 2x − 3 =
4
3
Example 2:
1) Rewrite and factor out coefficient of linear term:
2
4
− 2x + 3 − 3 =
3
2
3
( −2 )  x −  − 3 =4
3
2

2) Use properties of absolute value to “extract the linear coefficient from the absolute value:
2
3
• − 2 • x − − 3 =4
3
2
2
3
• 2 • x − − 3 =4
3
2
4
3
x − −3 =
4
3
2
3) Isolate the absolute value term:
4
3
x − −3 =
4
3
2
+3 +3
3
3 4
3
x− =
7 
 
2
4 3
4
3 21
x− =
2
4
4) Identify central value and solution values
Central value: h =
3
2
 3 21 3 21   15 27 
Solution values: x =  - , +  = − , 
2 4 2 4   4 4 
Example 3:
4
x + 7 + 12 =
4
5
1) Isolate the absolute value term:
4
x + 7 + 12 =4
5
− 12 − 12
54
5
−8  
  x+7 =
45
4
−10
x+7 =
2) Identify central value and solution values
Central value: h = −7 BUT... Distance from Central Value cannot be negative
(because Absolute Value cannot be negative) therefore there is no solution.