Big Ideas Math Algebra 1, 2015
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Big Ideas Math Algebra 1, 2015 View details
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Exercise 8 Page 47

Write two separate equations.

x=8 and x=-4.

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.
|20+2x|=|4x+4|

lc 20+2x ≥ 0:20+2x = (4x+4) & (I) 20+2x < 0:20+2x = - (4x+4) & (II)

lc20+2x=4x+4 & (I) 20+2x=- (4x+4) & (II)
Solve for x
l20+2x=4x+4 20+2x=-4x-4

(I), (II):LHS-20=RHS-20

l2x=4x-16 2x=-4x-24
l-2x=-16 2x=-4x-24
lx=8 2x=-4x-24
lx=8 6x=-24
lx= 8 x= -4
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.
|20+2x|=|4x+4|
|20+2( 8)|? =|4( 8)+4|
Simplify equation
|20+16|? =|32+4|
|36|? =|36|
36=36 ✓
Substituting 8 for x in the equation results in a true statement, so x=8 is not an extraneous solution.
|20+2x|=|4x+4|
|20+2({\color{#FF0000}{\text{-}4}})|\stackrel{?}=|4({\color{#FF0000}{\text{-}4}})+4|
Simplify equation
|20-8|\stackrel{?}=|\text{-}16+4|
|12|\stackrel{?}=|\text{-}12|
12\stackrel{?}=|\text{-}12|
12=12 ✓
Substituting -4 for x in the equation results in another true statement, so x=-4 is also not extraneous.