Exercise 2 Page 105

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, we need to start by using the Distributive Property to simplify the left- and right-hand sides of the equation. Now we can continue to solve using the Properties of Equality.

5y-20=14y+7

AddEqn

LHS+20=RHS+20

5y-20+20=14y+7+20

AddTerms

Add terms

5y=14y+27

SubEqn

LHS-14y=RHS-14y

5y-14y=14y+27-14y

SubTerms

Subtract terms

- 9y=27

DivEqn

.LHS /(- 9).=.RHS /(- 9).

- 9y/- 9 = 27/- 9

CalcQuot

Calculate quotient

y=-3

The solution to the equation is y=- 3. We can check our solution by substituting it into the original equation.

5(y-4)=7(2y+1)

Substitute

y= - 3

5( - 3-4) ? = 7(2( - 3)+1)

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Simplify

MultPosNeg

a(- b)=- a * b

5(- 3-4) ? = 7(-6+1)

AddSubTerms

Add and subtract terms

5(- 7) ? = 7(- 5)

MultPosNeg

a(- b)=- a * b

- 35 = - 35 ✓

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