Exercise 46 Page 46

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

1/2|3c+5|=6c+4

LHS * 2=RHS* 2

|3x+5|=2(6c+4)

Distribute 2

|3c+5|=12c+8

An absolute value measures an expression's distance from a midpoint on a number line. |3c+5|= 12c+8This equation means that the distance is 12c+8, either in the positive direction or the negative direction. |3c+5|= 12c+8 ⇒ l3c+5= 12c+8 3c+5= - (12c+8) To find the solutions to the absolute value equation, we need to solve both of these cases for c.

|3c+5|=12c+8

lc 3c+5 ≥ 0:3c+5 = (12c+8) & (I) 3c+5 < 0:3c+5 = - (12c+8) & (II)

lc3c+5=12c+8 & (I) 3c+5= -(12c+8) & (II)

Distr

(II): Distribute - 1

l3c+5=12c+8 3c+5= -12c-8

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

l3c=12c+3 3c= -12c-13

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Solve for c

SubEqn

(I): LHS-12c=RHS-12c

l- 9c=3 3c= -12c-13

AddEqn

(II): LHS+12c=RHS+12c

l-9c=3 15c= -13

DivEqn

(I): .LHS /(- 9).=.RHS /(- 9).

lc=- 39 15c= -13

ReduceFrac

(I): a/b=.a /3./.b /3.

lc=- 13 15c= -13

DivEqn

(II): .LHS /15.=.RHS /15.

lc= - 13 c= - 1315

Checking Our Answers

When solving an absolute value equation, it is important to check for extraneous solutions. We can check our answers by substituting them back into the original equation. Let's start with - 13.

|3c+5|=12c+8

Substitute

c= -1/3

|3( -1/3)+5|? =12( -1/3)+8

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Simplify

MultPosNeg

a(- b)=- a * b

|- 3*1/3+5|? =- 12*1/3+8

DenomMultFracToNumber

3 * a/3= a

|- 1+5|? =- 12*1/3+8