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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

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Exercise 26 Page 161

Apply inverse operations to 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 gather all of the variable terms on one side of the equation and all of the constant terms on the other side using the Properties of Equality. To do so, we need to start with rearranging terms according to the Commutative Property of Addition and then combine like terms.

2a+5=9a-16

LHS-2a=RHS-2a

5=9a-16-2a

CommutativePropAdd

Commutative Property of Addition

5=9a-2a-16

Subtract terms

5=7a-16

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

5=7a-16

LHS+16=RHS+16

5+16=7a

Add terms

21=7a

.LHS /7.=.RHS /7.

21/7=a

Calculate quotient

3=a

Rearrange equation

a=3

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

2a+5=9a-16

a= 3

2( 3) +5 ? = 9( 3)-16

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Simplify

Multiply

6+5 ? = 27-16

Add and subtract terms

11= 11 ✓

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

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