Exercise 57 Page 340

Practice makes perfect

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

|2x+5|= 21 This equation means that the distance is 21, either in the positive or the negative direction. |2x+5|= 21 ⇒ l2x+5= 21 2x+5= - 21 To find the solutions to the absolute value equation we need to solve both of these cases for x.

| 2x+5|=21

lc 2x+5 ≥ 0:2x+5 = 21 & (I) 2x+5 < 0:2x+5 = - 21 & (II)

l2x+5=21 2x+5=- 21

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

l2x=16 2x=- 26

(I), (II): .LHS /2.=.RHS /2.

lx_1=8 x_2=- 13

Both 8 and -13 are solutions to the absolute value equation.

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 x=8.

|2x+5|=21

Substitute

x= 8

|2( 8)+5|? =21

Multiply

|16+5|? =21

AddTerms

Add terms

|21|? =21

AbsPos

|21|=21

21=21 ✓

We will check x=-13 in the same way.

|2x+5|=21

Substitute

x= -13

|2( -13)+5|? =21

MultPosNeg

a(- b)=- a * b

|-26+5|? =21

AddTerms

Add terms

|-21|? =21

AbsNeg

|-21|=21

21=21 ✓

We see that both the x=8 and x=-13 satisfy the original equation.

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

2|3x-2|=24

DivEqn

.LHS /2.=.RHS /2.

|3x-2|=12

An absolute value measures an expression's distance from a midpoint on a number line. |3x-2|= 12This equation means that the distance is 12, either in the positive or the negative direction. |3x-2|= 12 ⇒ l3x-2= 12 3x-2= - 12 To find the solutions to the absolute value equation, we need to solve both of these cases for x.

| 3x-2|=12

lc 3x-2 ≥ 0:3x-2 = 12 & (I) 3x-2 < 0:3x-2 = - 12 & (II)

l3x-2=12 3x-2=- 12

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

l3x=14 3x=- 10

(I), (II): .LHS /3.=.RHS /3.

lx_1= 143 x_2=- 103

Both 143 and - 103 are solutions to the absolute value equation.

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 x= 143.

2| 3x-2|=24

Substitute

x= 14/3

2| 3( 14/3)-2|? =24

DenomMultFracToNumber

3 * a/3= a

2|14-2|? =24

SubTerm

Subtract term

2|12|? =24

AbsPos

|12|=12

2(12)? =24

Multiply

24=24 ✓

We will check x=- 103 in the same way.

2| 3x-2|=24

Substitute

x= -10/3

2| 3( -10/3)-2|? =24

MoveNegFracToNum

Put minus sign in numerator

2|3(-10/3)-2|? =24

DenomMultFracToNumber

3 * a/3= a

2|-10-2|? =24

SubTerm

Subtract term

2|-12|? =24

AbsNeg

|-12|=12

2(12)? =24

Multiply

24=24 ✓

We see that both the x= 143 and x=- 103 satisfy the original equation.

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

|4x+5|-8=31

AddEqn

LHS+8=RHS+8

|4x+5|=39

An absolute value measures an expression's distance from a midpoint on a number line. |4x+5|= 39This equation means that the distance is 39, either in the positive direction or the negative direction. |4x+5|= 39 ⇒ l4x+5= 39 4x+5= - 39 To find the solutions to the absolute value equation we need to solve both of these cases for x.

|4x+5 |=39

lc 4x+5 ≥ 0:4x+5 = 39 & (I) 4x+5 < 0:4x+5 = - 39 & (II)

l4x+5=39 4x+5=- 39

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

l4x=34 4x=- 44

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

lx= 344 x=- 11

ReduceFrac

a/b=.a /2./.b /2.

lx_1= 172 x_2=- 11

Both 172 and -11 are solutions to the absolute value equation.

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 x= 172.

|4x+5|-8=31

Substitute

x= 17/2

|4( 17/2)+5|-8? =31

MoveLeftFacToNum

a*b/c= a* b/c

|68/2+5|-8? =31

CalcQuot

Calculate quotient

|34+5|-8? =31

AddTerms

Add terms