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 5 Page 29

How many cases do you have after you remove the absolute value?

x=- 1 12 and x=- 5 12

Practice makes perfect
Before we can solve this equation, we need to isolate the absolute value expression using the Properties of Equality.
4|2x+7|=16
|2x+7|=4

Solving for x

An absolute value measures an expression's distance from a midpoint on a number line. |2x+7|= 4This equation means that the distance is 4, either in the positive direction or the negative direction. |2x+7|= 4 ⇒ l2x+7= 4 2x+7= - 4 To find the solutions to the absolute value equation, we need to solve both of these cases for x.
| 2x+7|=4

lc 2x+7 ≥ 0:2x+7 = 4 & (I) 2x+7 < 0:2x+7 = - 4 & (II)

lc2x+7=4 & (I) 2x+7=- 4 & (II)

(I), (II): LHS-7=RHS-7

l2x=- 3 2x=- 11

(I), (II): .LHS /2.=.RHS /2.

lx_1=- 32 x_2= - 112
Both - 32 and - 112 are solutions to the absolute value equation.

Checking Our Answers

When solving an absolute value equation, it is important to check for extraneous solutions. We can check our answers by substituting them back into the original equation. Let's start with x= - 32.
4|2x+7|=16
4|2( -3/2)+7|? =16
â–Ľ
Simplify
4|2(-3/2)+7|? =16
4|- 3+7|? =16
4|4|? =16
4(4)? =16
16=16
We end with a true statement, so x=- 32 is not an extraneous solution. We will check x= - 112 in the same way.
4|x+7|=16
4|2( -11/2)+7|? =16
â–Ľ
Simplify
4|2(-11/2)+7|? =16
4|- 11+7|? =16
4|- 4|? =16
4(4)? =16
16=16
Again we end up with a true statement. Therefore, x=- 32 and x=- 112 satisfy the original equation and neither of them is an extraneous solution.

Rewriting Our Answers

Although x=- 32 and x=- 112 are perfectly valid solutions, we can also rewrite them as mixed numbers. Let's start with x=- 32.
x=-3/2
x=-2+1/2
x=-(2/2+1/2)
x=-(1+1/2)
x=-1 12
Using a similar process for x=- 112, we will find that - 112=-5 12.