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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

2. Solving Two-Step Equations

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Exercise 18 Page 91

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- and right-hand sides of the equation.

- 14 = - 5 + 3c

LHS+5=RHS+5

- 14 + 5 = - 5 + 5 + 3c

Add terms

- 9 = 3c

Rearrange equation

3c = - 9

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

3c = - 9

.LHS /3.=.RHS /3.

3c/3 = - 9/3

Calculate quotient

c = - 3

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

- 14 = - 5 + 3c

c= - 3

- 14 ? = - 5 + 3( - 3)

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Evaluate

Multiply

- 14 ? = - 5 + (- 9)

a+(- b)=a-b

- 14 ? = - 5 - 9

Subtract terms

- 14 = - 14

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

Subchapter links

Lesson Check p.91

Practice and Problem-Solving Exercises p.91-93

Standardized Test Prep p.93

Mixed Review p.93