Exercise 10 Page 26

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, using the Properties of Equality. In this case, we start by using the Distributive Property to simplify the right-hand side of the equation.

68=1/5(20x+50)+2

Distr

Distribute 1/5

68=4x+10+2

AddTerms

Add terms

68=4x+12

SubEqn

LHS-12=RHS-12

68-12=4x+12-12

SubTerm

Subtract term

56=4x

DivEqn

.LHS /4.=.RHS /4.

56/4=4x/4

CalcQuot

Calculate quotient

14=x

RearrangeEqn

Rearrange equation

x=14

Therefore, x=14 is the solution to the equation. We can check our answer by substituting it back into the original equation. If doing so results in an identity, our solution is correct. An identity is when both sides of the equation are exactly the same.

68=1/5(20x+50)+2

Substitute

x= 14

68 ? = 1/5(20( 14)+50)+2

Multiply

Multiply

68 ? =1/5(280+50)+2

AddTerms

Add terms

68 ? =1/5*330+2

MoveRightFacToNumOne

1/b* a = a/b

68 ? =330/5+2

CalcQuot

Calculate quotient

68 ? =66+2

AddTerms

Add terms

68=68

Because both sides are equal, we know that our answer is correct.