Big Ideas Math Integrated I, 2016
BI
Big Ideas Math Integrated I, 2016 View details
4. Solving Absolute Value Equations
Continue to next subchapter

Exercise 8 Page 30

Write two separate equations.

x=7 and x=- 3

Practice makes perfect
When the absolute values of two expressions are equal, either the expressions are equal or the opposite of the expressions are equal. Let's look at an example equation. |ax+b|=|cx+d| For this equation, there are two possible cases to consider. ax+b=cx+d or ax+b=- (cx+d) To solve the given absolute value equation we will write two equations — similar to those above — when we remove the absolute value.
|x+8|=|2x+1|

lc x+8 ≥ 0:x+8 = (2x+1) & (I) x+8 < 0:x+8 = - (2x+1) & (II)

lcx+8=2x+1 & (I) x+8=- (2x+1) & (II)
â–Ľ
Solve for x
lx+8=2x+1 x+8=- 2x -1

(I), (II):LHS-8=RHS-8

lx=2x-7 x=- 2x -9
l- x=- 7 x=- 2x -9
lx=7 x=- 2x -9
lx=7 3x=- 9
lx_1=7 x_2=- 3
After solving an absolute value equation, it is necessary to check for extraneous solutions. To do this, we substitute the found solutions into the given equation and determine if a true statement is made.
|x+8|=|2x+1|
| 7+8|? =|2( 7)+1|
â–Ľ
Simplify equation
|7+8|? =|14+1|
|15|? =|15|
15=15
x=7 is not extraneous.
|x+8|=|2x+1|
|{\color{#0000FF}{\text{-} 3}}+8|\stackrel{?}=|2({\color{#0000FF}{\text{-} 3}})+1|
â–Ľ
Simplify equation
|\text{-} 3+8|\stackrel{?}=|\text{-} 6+1|
|5|\stackrel{?}=|\text{-} 5|
5\stackrel{?}=|\text{-} 5|
5=5
x=- 3 is not extraneous.