Exercise 92 Page 292

Practice makes perfect

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

|9+3x|= 39 This equation means that the distance is 39, either in the positive direction or the negative direction. |9+3x|= 39 ⇒ 9+3x& = 39 9+3x& = - 39 To find the solutions to the absolute value equation, we need to solve both of these cases for x.

| 9+3x|=39

lc 9+3x ≥ 0:9+3x = 39 & (I) 9+3x < 0:9+3x = - 39 & (II)

lc9+3x=39 & (I) 9+3x=- 39 & (II)

(I), (II): LHS-9=RHS-9

l3x=30 3x=- 48

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

lx=10 x=- 16

To check our solutions, we will substitute them into the equation.

|9+3x|=39

Substitute

x= 10

|9+3( 10)|=39

Multiply

|9+30|=39

AddTerms

Add terms

|39|=39

AbsPos

|39|=39

39=39

Since the substitution resulted in an identity, x=10 is a correct solution.

|9+3x|=39

Substitute

x= - 16

|9+3( - 16)|=39

MultPosNeg

a(- b)=- a * b

|9-48|=39

SubTerm

Subtract term

|- 39|=39

AbsNeg

|-39|=39

39=39

Since the substitution resulted in an identity, x=- 16 is also a correct solution.

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

|2x+1|= 10 This equation means that the distance is 10, either in the positive direction or the negative direction. |2x+1|= 10 ⇒ 2x+1& = 10 2x+1& = - 10 To find the solutions to the absolute value equation, we need to solve both of these cases for x.

|2x+1|=10

lc 2x+1 ≥ 0:2x+1 = 10 & (I) 2x+1 < 0:2x+1 = - 10 & (II)

lc2x+1=10 & (I) 2x+1=- 10 & (II)

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

l2x=9 2x=- 11

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

lx=4.5 x=- 5.5

To check our solutions, we will substitute them into the equation.

|2x+1|=10

Substitute

x= 4.5

|2( 4.5)+1|=10

Multiply

|9+1|=10

AddTerms

Add terms

|10|=10

AbsPos

|10|=10

10=10

Since the substitution resulted in an identity, x=4.5 is a correct solution.

|2x+1|=10

Substitute

x= - 5.5

|2( - 5.5)+1|=10

Multiply

|- 11+1|=10

AddTerms

Add terms

|- 10|=10

AbsNeg

|-10|=10

10=10

Since the substitution resulted in an identity, x=- 5.5 is also a correct solution.

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

|- 3x+9|= 10 This equation means that the distance is 10, either in the positive direction or the negative direction. |- 3x+9|= 10 ⇒ - 3x+9& = 10 - 3x+9& = - 10 To find the solutions to the absolute value equation, we need to solve both of these cases for x.

|- 3x+9|=10

lc - 3x+9 ≥ 0:- 3x+9 = 10 & (I) - 3x+9 < 0:- 3x+9 = - 10 & (II)

l- 3x+9=10 - 3x+9=- 10

(I), (II): LHS-9=RHS-9

l- 3x=1 - 3x=- 19

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

l3x=-1 3x=19

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

lx_1=- 13 x_2= 193

To check our solutions, we will substitute them into the equation.

|- 3x+9|=10

Substitute

x= - 1/3

|- 3( - 1/3)+9|=10

MultNegNegOnePar

- a(- b)=a* b

|3* 1/3+9|=10

DenomMultFracToNumber

3 * a/3= a

|1+9|=10

AddTerms

Add terms

|10|=10

AbsPos

|10|=10

10=10

Since the substitution resulted in an identity, x=- 13 is a correct solution.

|- 3x+9|=10

Substitute

x= 19/3

|- 3( 19/3)+9|=10

DenomMultFracToNumber

3 * a/3= a

|- 19+9|=10

AddTerms

Add terms

|- 10|=10

AbsNeg

|-10|=10

10=10

Since the substitution resulted in an identity, x= 193 is also a correct solution.

d An absolute value measures an expression's distance from a midpoint on a number line. Since distance cannot be negative, the absolute value of a number cannot be negative.

|3.2x-4|= - 5.7 An absolute value cannot be equal to -5.7, so this equation has no solution.