Exercise 19 Page 53

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

{- 1/5,1}

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

|5n-2|-6=- 3

LHS+6=RHS+6

|5n-2|= 3

An absolute value measures an expression's distance from a midpoint on a number line. |5n-2|= 3 This equation means that the distance is 3, either in the positive direction or the negative direction. |5n-2|= 3 ⇒ l5n-2= 3 5n-2= - 3 To find the solutions to the absolute value equation, we need to solve both of these cases for n.

| 5n-2|=3

lc 5n-2 ≥ 0:5n-2 = 3 & (I) 5n-2 < 0:5n-2 = - 3 & (II)

lc5n-2=3 & (I) 5n-2=- 3 & (II)

(I), (II): LHS+2=RHS+2

lc5n=5 & (I) 5n=- 1 & (II)

(I), (II): .LHS /5.=.RHS /5.

ln_1=1 n_2=- 15

Both 1 and - 15 are solutions to the absolute value equation.

Exercise 19 Page 53

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