Big Ideas Math Algebra 1, 2015
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Big Ideas Math Algebra 1, 2015 View details
4. Solving Absolute Value Equations
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Exercise 42 Page 33

Break down the given absolute value equation two separate equations.

a=5/2 and a=1/2

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:& |6a-5|=4a First Equation:& 6a-5 = 4a Second Equation:& 6a-5=- 4a Let's solve for a in each of these equations simultaneously.
|6a-5|=4a

lc 6a-5 ≥ 0:6a-5 = 4a & (I) 6a-5 < 0:6a-5 = - 4a & (II)

lc6a-5=4a & (I) 6a-5=- 4a & (II)
â–Ľ
Solve for a
l-5=-2a -5=- 10a
l 52=a - 5=- 10a
l 52=a 510=a
la= 52 a= 510
la= 52 a= 12
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.
|6a-5|=4a
\left|6\cdot{\color{#0000FF}{\dfrac{5}{2}}}-5\right|\stackrel{?}=4\cdot{\color{#0000FF}{\dfrac{5}{2}}}
â–Ľ
Simplify
\left|\dfrac{30}{2}-5\right|\stackrel{?}=\dfrac{20}{2}
|15-5|\stackrel{?}=10
|10|\stackrel{?}=10
10=10
We end with a true statement, so a= 52 is not extraneous.
|6a-5|=4a
\left|6\cdot{\color{#009600}{\dfrac{1}{2}}}-5\right|\stackrel{?}=4\cdot{\color{#009600}{\dfrac{1}{2}}}
â–Ľ
Simplify
\left|\dfrac{6}{2}-5\right|\stackrel{?}=\dfrac{4}{2}
|3-5|\stackrel{?}=2
|\text{-} 2|\stackrel{?}=2
2=2
Again we end with a true statement, so a= 12 is also not extraneous.