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Core Connections: Course 3

CC

Core Connections: Course 3 View details

Chapter Closure

Exercise 124 Page 144

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

x=3

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 subtracting 4x from both sides of the equation to isolate the variable terms on the left-hand side of the equation.

6-x-3=4x-12

LHS-4x=RHS-4x

6-x-3-4x=4x-12-4x

CommutativePropAdd

Commutative Property of Addition

6-3-x-4x=4x-4x-12

Subtract terms

3-5x=- 12

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

3-5x=- 12

LHS-3=RHS-3

3-5x-3=- 12-3

CommutativePropAdd

Commutative Property of Addition

3-3-5x=- 12-3

Subtract terms

- 5x=- 15

.LHS /(- 5).=.RHS /(- 5).

- 5x/- 5=- 15/- 5

CancelCommonFac

Cancel out common factors

- 5x/- 5=- 15/- 5

Simplify quotient

x=- 15/- 5

- a/- b=a/b

x=15/5

Calculate quotient

x=3

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