Big Ideas Math Integrated I, 2016
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Big Ideas Math Integrated I, 2016 View details
5. Solving Compound Inequalities
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Exercise 38 Page 86

Break down the given absolute value equation into two separate equations.

Solutions: w=-26 and w=-2/3
Graph:

Numbers on Number Line
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:& |1/2w-6|=|w+7| First Equation:& 1/2w-6 = w+7 Second Equation:& 1/2w-6=- (w+7) These two equations can now be solved by conventional means.

Solving the Equations

Let's solve for w in each of these equations simultaneously.
lc1/2w-6=w+7 & (I) 1/2w-6=- (w+7) & (II)
â–Ľ
Solve for w
lc1/2w-6=w+7 & (I) 1/2w-6=- w-7 & (II)

(I), (II):LHS+6=RHS+6

lc1/2w=w+13 & (I) 1/2w=- w-1 & (II)
lc1/2w-w=13 & (I) 1/2w=- w-1 & (II)
lc-1/2w=13 & (I) 1/2w=- w-1 & (II)
lcw=-26 & (I) 1/2w=- w-1 & (II)
lcw=-26 & (I) 1/2w+w=-1 & (II)
lcw=-26 & (I) 3/2w=-1 & (II)
lcw=-26 & (I) w=-2/3 & (II)

Checking the Solutions

After solving an absolute value equation, it is necessary to check for extraneous solutions. To do this, we will substitute the found solutions into the given equation and determine if a true statement is made. We will start by checking w=-26.
|1/2w-6|=|w+7|
|1/2( -26)-6|? =| -26+7|
â–Ľ
Simplify equation
|-26/2-6|? =|-26+7|
|-13-6|? =|-26+7|
|-19|? =|-19|
19=19
w=-26 is not extraneous. Next, we will check w=- 23.
|1/2w-6|=|w+7|
|1/2( -2/3)-6|? =| -2/3+7|
â–Ľ
Simplify equation
|-2/6-6|? =|-2/3+7|
|-1/3-6|? =|-2/3+7|
|-1/3-18/3|? =|-2/3+21/3|
|-19/3|? =|19/3|
19/3? =|19/3|
19/3=19/3
w=- 23 is also not extraneous.

Graph

The solutions to the given equation are the points w=-26 and w=- 23. In order to graph the solutions, we will plot them on a number line.

Numbers on Number Line