Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
1. Relations and Functions
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Exercise 45 Page 67

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

- 13, 15

Practice makes perfect
Before we can solve this equation, we need to isolate the absolute value expression using the Properties of Equality.
19+|x-1|=33
19+|x-1|-19=33-19
|x-1|=14
An absolute value measures an expression's distance from a midpoint on a number line. |x-1|= 14 This equation means that the distance is 14, either in the positive direction or the negative direction. |x-1|= 14 ⇒ lx-1= 14 x-1= - 14 To find the solutions to the absolute value equation, we need to solve both of these cases for x.
| x-1|=14

lc x-1 ≥ 0:x-1 = 14 & (I) x-1 < 0:x-1 = - 14 & (II)

lcx-1=14 & (I) x-1=- 14 & (II)

(I), (II): LHS+1=RHS+1

lx_1=15 x_2=- 13
Both - 13 and 15 are solutions to the absolute value equation.