Exercise 87 Page 40

a To solve the equation for x, we have to perform inverse operations until we have isolated x.

3x-1=4x+8-x

Subtract term

3x-1=3x+8

LHS-3x=RHS-3x

-1 ≠ 8

Trying to solve the equation, we ended up with a contradiction. This means there are no values of x that solves this equation. Therefore, the equation has no solution.

b Again, to solve the equation we have to perform inverse operations until x is isolated.

-10+5x=7x-4

SubEqn

LHS-7x=RHS-7x

-10-2x=-4

AddEqn

LHS+10=RHS+10

-2x=6

MultEqn

LHS * (- 1)=RHS* (- 1)

2x=-6

DivEqn

.LHS /2.=.RHS /2.

x=-3

We will check the solution by substituting it into the equation.

-10+5x=7x-4

Substitute

x= -3

-10+5( -3)? =7( -3)-4

MultPosNeg

a(- b)=- a * b

-10-15? =-21-4

SubTerms

Subtract terms

- 25=- 25

The solution is correct.

c Once more, we have to perform inverse operations until x is isolated on one of the equation's sides.

28-6x+4=30-3x

AddTerms

Add terms

32-6x=30-3x

AddEqn

LHS+3x=RHS+3x

32-3x=30

SubEqn

LHS-32=RHS-32

-3x=-2

MultEqn

LHS * (- 1)=RHS* (- 1)

3x=2

DivEqn

.LHS /3.=.RHS /3.

x=2/3

Let's check the solution by substituting it into the equation.

28-6x+4=30-3x

Substitute

x= 2/3

28-6( 2/3)+4? =30-3( 2/3)

MoveLeftFacToNum

a*b/c= a* b/c

28-12/3+4? =30-6/3

CalcQuot

Calculate quotient

28-4+4? =30-2

AddSubTerms

Add and subtract terms

28=28

The solution is correct.

d Let's isolate x.

4x-1=9x-1-5x

SubTerm

Subtract term

4x-1=4x-1

SubEqn

LHS-4x=RHS-4x

-1=-1

Attempting to solve the equation, we have reached an identity. Therefore the solutions to this equation are all real numbers. It does not matter which value of x you substitute into the equation. The left-hand side and right-hand side will always be equal.

Subchapter links

Exercises

Exercise 87 Page 40

Hints

Check the answer