Exercise 11 Page 535

To solve an equation, we should first gather all of the variable terms on one side of the equation 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.

5(t^2-3t+2)=t(5t-2)

Distr

Distribute 5

5t^2-15t+10=t(5t-2)

Distr

Distribute t

5t^2-15t+10=5t^2-2t

Now we can continue to solve using the Properties of Equality.

5t^2-15t+10=5t^2-2t

SubEqn

LHS-5t^2=RHS-5t^2

5t^2-15t+10-5t^2=5t^2-2t-5t^2

SubTerms

Subtract terms

- 15t+10=- 2t

AddEqn

LHS+15t=RHS+15t

- 15t+10+15t=- 2t+15t

AddTerms

Add terms

10=13t

DivEqn

.LHS /13.=.RHS /13.

10/13=13t/13

CalcQuot

Calculate quotient

10/13=t

RearrangeEqn

Rearrange equation

t=10/13

The solution to the equation is t= 1013.