Big Ideas Math Integrated I, 2016
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Big Ideas Math Integrated I, 2016 View details
4. Solving Absolute Value Equations
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Exercise 13 Page 31

Break down the given absolute value equation two separate equations.

x=-1/5

Practice makes perfect
When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. |ax+b|=|cx+d| Although we can make 4 statements about this equation, there are actually only two possible cases to consider.
Statement Result
Both absolute values are positive. ax+b=cx+d
Both absolute values are negative. -(ax+b)=-(cx+d)
Only the left-hand side is negative. -(ax+b)=cx+d
Only the right-hand side is negative. ax+b=-(cx+d)
Because of the Properties of Equality, when the absolute values of two expressions are equal, either the expressions are equal or the opposites of the expressions are equal. Now let's consider these two cases for the given equation. Given Equation:& |5x-2|=|5x+4| First Equation:& 5x-2 = 5x+4 Second Equation:& 5x-2=- (5x+4) Let's solve for x in each of these equations simultaneously.
lc5x-2=5x+4 & (I) 5x-2=- (5x+4) & (II)
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Solve for x
l5x-2=5x+4 5x-2=- 5x- 4
l- 2≠ 4 5x-2=- 5x- 4
l- 2≠ 4 10x-2=- 4
l- 2≠ 4 10x=- 2
l- 2≠ 4 x=- 210
l- 2≠ 4 x=- 15
While solving the first equation, we end up with a contradiction, so the first equation has no solution. The solution to the second equation is x=- 15. After solving an absolute value equation, it is necessary to check for extraneous solutions. To do this, we substitute x=- 15 into the given equation and determine if it makes a true statement.
|5x-2|=|5x+4|
\left|5\left({\color{#0000FF}{\text{-}\dfrac{1}{5}}}\right)-2\right|\stackrel{?}=\left|5\left({\color{#0000FF}{\text{-}\dfrac{1}{5}}}\right)+4\right|
â–Ľ
Simplify
\left|\text{-}\dfrac{5}{5}-2\right|\stackrel{?}=\left|\text{-}\dfrac{5}{5}+4\right|
|\text{-} 1-2|\stackrel{?}=|\text{-} 1+4|
|\text{-} 3|\stackrel{?}=|3|
3\stackrel{?}=|3|
3=3
We end with a true statement, so x=- 15 is not an extraneous solution.