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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 57 Page 114

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 need to start by using the Distributive Property to simplify the left-hand side of the equation.

2(y-4)=-4y+10

Distribute 2

2y-8=-4y+10

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

2y-8=-4y+10

LHS+4y=RHS+4y

2y-8+4y=-4y+10+4y

Add terms

6y-8=10

LHS+8=RHS+8

6y-8+8=10+8

Add terms

6y=18

.LHS /6.=.RHS /6.

6y/6=18/6

a/b=.a /6./.b /6.

y=3

The solution to the equation is y=3.

Subchapter links

Lesson Check p.112

Practice and Problem-Solving Exercises p.112-114

Standardized Test Prep p.114

Mixed Review p.114