Exercise 9 Page 26

Gather all of the variable terms on one side of the equation and all of the constant terms on the other side.

k=- 9

Practice makes perfect

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 distributing -3 on the left-side of the equation.

2k-3(2k-3)=45

Distr

Distribute - 3

2k-6k+9=45

SubTerm

Subtract term

- 4k+9=45

SubEqn

LHS-9=RHS-9

- 4k+9-9=45-9

SubTerms

Subtract terms

- 4k=36

DivEqn

.LHS /(- 4).=.RHS /(- 4).

- 4k/- 4=36/- 4

CalcQuot

Calculate quotient

k=- 9

Therefore, k=- 9 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.

2k-3(2k-3)=45

Substitute

k= - 9

2(- 9)-3(2(- 9)-3) ? =45

▼

Simplify LHS

MultPosNeg

a(- b)=- a * b

- 18-3(-18-3) ? =45

SubTerm

Subtract term

- 18-3(-21) ? =45