Exercise 25 Page 211

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

f=1.5 or f=-2

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Before we can solve this equation, we need to isolate the absolute value expression using the Properties of Equality.

|4f+1|-2=5

AddEqn

LHS+2=RHS+2

|4f+1|-2+2=5+2

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Add terms

|4f+1|=7

An absolute value measures an expression's distance from a midpoint on a number line.

|4f+1|= 7 This equation means that the distance is 7, either in the positive direction or the negative direction. |4f+1|= 7 ⇒ l4f+1= 7 4f+1= -7 To find the solutions to the absolute value equation, we need to solve both of these cases for f.

| 4f+1|=7

lc 4f+1 ≥ 0:4f+1 = 7 & (I) 4f+1 < 0:4f+1 = - 7 & (II)

lc4f+1=7 & (I) 4f+1=-7 & (II)

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

lc4f=6 & (I) 4f=-8 & (II)

(I), (II): .LHS /4.=.RHS /4.

lf_1=1.5 f_2=-2

Both f=1.5 and f=-2 are solutions to the absolute value equation.

Subchapter links

Lesson Check p.210

Practice and Problem-Solving Exercises p.211-213

Standardized Test Prep p.213

Mixed Review p.213

Exercise 25 Page 211

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