Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
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Exercise 17 Page 231

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

3, 7

Practice makes perfect
Before we can solve this equation, we need to isolate the absolute value expression using the Properties of Equality.
- 3|d-5|=- 6
|d-5|=- 6/- 3
|d-5|=6/3
|d-5|=2
An absolute value measures an expression's distance from a midpoint on a number line. |d-5|= 2 This equation means that the distance is 2, either in the positive direction or the negative direction. |d-5|= 2 ⇒ ld-5= 2 d-5= - 2 To find the solutions to the absolute value equation we need to solve both of these cases for d.
| d-5|=2

lc d-5 ≥ 0:d-5 = 2 & (I) d-5 < 0:d-5 = - 2 & (II)

lcd-5=2 & (I) d-5=- 2 & (II)

(I), (II): LHS+5=RHS+5

ld_1=7 d_2=3
Both 3 and 7 are solutions to the absolute value equation.