Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
8. Unions and Intersections of Sets
Continue to next subchapter

Exercise 51 Page 220

Isolate the absolute value on the left-hand side. How many cases do you have after you remove it?

f=2 or f=8

Practice makes perfect
Before we can solve the given equation, we need to isolate the absolute value expression using the Properties of Equality. 4|f-5|=12 ⇒ |f-5|=3 An absolute value measures an expression's distance from a midpoint on a number line. |f-5|=3 This equation means that the distance between f and 5 is 3, either in the positive direction or the negative direction. |f-5|=3 ⇒ lf-5= 3 f-5= - 3 To find the solutions to the absolute value equation, we need to solve both of these cases for f.
| f-5|=3

lc f-5 ≥ 0:f-5 = 3 & (I) f-5 < 0:f-5 = - 3 & (II)

lcf-5=3 & (I) f-5=- 3 & (II)

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

lf_1=8 f_2=2