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Core Connections: Course 3

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Chapter Closure

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Exercise 62 Page 211

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-hand sides of the equation.

3(2x-1)+7=- 44

Distribute 3

3(2x)-3(1)+7=- 44

Multiply

6x-3+7=- 44

Add terms

6x+4=- 44

Now we can continue to solve using the properties of equality.

6x+4=- 44

LHS-4=RHS-4

6x+4-4=- 44-4

Subtract term

6x= - 48

.LHS /6.=.RHS /6.

6x/6=- 48/6

Calculate quotient

x= - 8

The solution to the equation is x=- 8.

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.

6(2x-5)=- (x+4)

Distribute 6

6(2x)-6(5)=-(x+4)

Distribute - 1

6(2x)-6(5)=-1(x)+(- 1)(4)

Multiply

12x-30=- x-4

Now we can continue to solve using the properties of equality.

12x-30=- x-4

LHS+x=RHS+x

12x-30+x=- x-4+x

Add terms

13x-30=- 4

LHS+30=RHS+30

13x-30+30=- 4+30

Add terms

13x=26

.LHS /13.=.RHS /13.

13x/13=26/13

Calculate quotient

x=2

The solution to the equation is x=2.

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