Exercise 19 Page 32

An absolute value measures an expression's distance from a midpoint on a number line. |4b-5|= 19 This equation means that the distance is 19, either in the positive direction or the negative direction. |4b-5|= 19 ⇒ l4b-5= 19 4b-5= - 19 To find the solutions to the absolute value equation, we need to solve both of these cases for b.

| 4b-5|=19

lc 4b-5 ≥ 0:4b-5 = 19 & (I) 4b-5 < 0:4b-5 = - 19 & (II)

lc4b-5=19 & (I) 4b-5=- 19 & (II)

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

l4b=24 4b=- 14

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

lb=6 b= -144

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Write as Mixed Number

ReduceFrac

(II):a/b=.a /2./.b /2.

lb=6 b= -72

MoveNegNumToFrac

(II):Put minus sign in front of fraction

lb=6 b=- 72

Rewrite

(II):Rewrite 7 as 6+1

lb=6 b=- 6+12

WriteSumFrac

(II):Write as a sum of fractions

lb=6 b=-( 62+ 12)

CalcQuot

(II):Calculate quotient

lb=6 b=-(3+ 12)

AddTerms

Add terms

lb_1=6 b_2=- 3 12

Both 6 and - 3 12 are solutions to the absolute value equation. Let's graph these solutions on a number line.