Exercise 20 Page 143

To solve an equation, we should first gather all of the variable terms on one side and all of the constant terms on the other side using the Properties of Equality. In this case, the only variable term is already on the right-hand side of the equation, so we will start by subtracting 10 from both sides.

15 = - 5x+10

SubEqn

LHS-10=RHS-10

15-10 = - 5x+10-10

SubTerm

Subtract term

5=- 5x

DivEqn

.LHS /(- 5).=.RHS /(- 5).

5/- 5 = - 5x/- 5

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Simplify left-hand side

MoveNegDenomToFrac

Put minus sign in front of fraction

-5/5 = - 5x/- 5

DivNegNeg

- a/- b=a/b

-5/5 = 5x/5

CalcQuot

Calculate quotient

- 1 = x

RearrangeEqn

Rearrange equation

x= - 1

The solution to the equation is x= - 1. We can check our solution by substituting it into the original equation.

15 = - 5x+10

Substitute

x= - 1

15 ? = - 5( - 1)+10

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Simplify

MultNegNegOnePar

- a(- b)=a* b

15 ? = 5* 1+10

IdPropMult

Identity Property of Multiplication

15 ? = 5+10

AddTerms

Add terms

15=15

Since the left-hand side is equal to the right-hand side, our solution is correct.