Exercise 14 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 both sides of the equation.

9(y-4)-7y=5(3y-2)

Distr

Distribute 9 & 5

9y-36-7y=15y-10

SubTerm

Subtract term

2y-36=15y-10

Now, we can continue solving for y using inverse operations. We need to move all the constants to the right-hand side and the variable terms to the left-hand side of the equation.

2y-36=15y-10

AddEqn

LHS+36=RHS+36

2y-36+36=15y-10+36

AddTerms

Add terms

2y=15y+26

SubEqn

LHS-15y=RHS-15y

2y-15y=15y+26-15y

SubTerm

Subtract term

- 13y=26

DivEqn

.LHS /(- 13).=.RHS /(- 13).

- 13y/- 13=26/- 13

CalcQuot

Calculate quotient

y=- 2

The solution to the equation is y=- 2.