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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

4. Solving Equations With Variables on Both Sides

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Exercise 24 Page 106

Gather all of the variable terms on one side of the equation and all of the constant terms on the other side. Start by removing the parentheses on the left-side of the equation.

g=1

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 side using the Properties of Equality. Notice that there are the parentheses on the left-side of the equation that we can remove.

(g+4)-3g=1+g

Remove parentheses

g+4-3g=1+g

CommutativePropAdd

Commutative Property of Addition

4+g-3g=1+g

Subtract term

4-2g=1+g

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

4-2g=1+g

LHS+2g=RHS+2g

4-2g+2g=1+g+2g

Add terms

4=1+3g

LHS-1=RHS-1

4-1=1+3g-1

Subtract terms

3=3g

.LHS /3.=.RHS /3.

3/3=3g/3

Calculate quotient

1=g

Rearrange equation

g=1

The solution to the equation is g=1. We can check our solution by substituting it into the original equation.

(g+4)-3g=1+g

g= 1

( 1+4)-3( 1)? =1+ 1

Add terms

5-3(1)? = 2

Identity Property of Multiplication

5-3? =2

Subtract terms

2=2 ✓

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

Subchapter links

Lesson Check p.105

Practice and Problem-Solving Exercises p.105-108

Standardized Test Prep p.108

Mixed Review p.108