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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

5. Literal Equations and Formulas

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Exercise 13 Page 112

Isolate y on the left-hand side of the equation, then substitute the values for x.

Equation: y=3x-9/5

x	y
- 1	-12/5
0	-9/5
1	-6/5

Practice makes perfect

Let's begin by solving the equation for y.

3x-5y=9

LHS-3x=RHS-3x

3x-3x-5y=9-3x

Subtract terms

- 5y=9-3x

.LHS /-5.=.RHS /-5.

-5y/-5=9-3x/-5

a/b=.a /-5./.b /-5.

y=9-3x/-5

- a/- b=a/b

y=-(9-3x)/5

DistrNegSignSwap

-(b-a)=a-b

y=3x-9/5

To find the value of y for each of the x-values, we substitute them into the equation and evaluate.

x	3x-9/5	y
- 1	3 ( -1)-9/5	-12/5
0	3( 0)-9/5	-9/5
1	3( 1)-9/5	-6/5

Subchapter links

Lesson Check p.112

Practice and Problem-Solving Exercises p.112-114

Standardized Test Prep p.114

Mixed Review p.114