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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

6. Ratios, Rates, and Conversions

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Exercise 53 Page 121

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 of the equation and all of the constant terms on the other side using the Properties of Equality. In this case, we will start by using the Multiplication Property of Equality.

3a+1/5=2

LHS * 5=RHS* 5

3a+1/5* 5=2* 5

FracMultDenomToNumber

a/5* 5 = a

3a+1=2* 5

Multiply

3a+1=10

LHS-1=RHS-1

3a+1-1=10-1

Subtract terms

3a=9

.LHS /3.=.RHS /3.

3a/3=9/3

Calculate quotient

a=3

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

3a+1/5=2

a= 3

3( 3)+1/5? =2

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Simplify

Multiply

9+1/5? =2

Add terms

10/5? =2

Calculate quotient

2=2

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

Subchapter links

Lesson Check p.119

Practice and Problem-Solving Exercises p.119-121

Standardized Test Prep p.121

Mixed Review p.121