Exercise 3 Page 157

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

- 3

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 subtracting w from both sides of the equation.

3w+2=w-4

SubEqn

LHS-w=RHS-w

3w+2-w=w-4-w

SubTerms

Subtract terms

2w+2=- 4

SubEqn

LHS-2=RHS-2

2w+2-2=- 4-2

SubTerms

Subtract terms

2w=- 6

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

2w=- 6

DivEqn

.LHS /2.=.RHS /2.

2w/2=- 6/2

CalcQuot

Calculate quotient

w=- 3

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

3w+2=w-4

Substitute

w= - 3

3({\color{#0000FF}{\text{-} 3}})+2\stackrel{?}={\color{#0000FF}{\text{-} 3}}-4

Multiply

\text{-} 9+2\stackrel{?}=\text{-} 3-4

AddSubTerms

Add and subtract terms

- 7=- 7

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

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Exercise 3 Page 157

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