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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 1 Page 91

Gather all the variable terms on one side of the equation and all the constant terms on the other side.

- 5

Practice makes perfect

To solve an equation, we should first gather all the variable terms on one side of the equation and all the constant terms on the other side. We can do this by using inverse operations and the Properties of Equality.

5x+12=- 13

LHS-12=RHS-12

5x+12-12=- 13-12

Subtract terms

5x=- 25

Now we can continue to solve using the Division Property of Equality.

5x=- 25

.LHS /5.=.RHS /5.

5x/5=- 25/5

MoveRightFacToNum

a/c* b = a* b/c

5/5x=- 25/5

MoveNegNumToFrac

Put minus sign in front of fraction

5/5x=- 25/5

Calculate quotient

1x = - 5

Identity Property of Multiplication

x=- 5

The solution to the equation is x=- 5. We can check our solution by substituting it into the original equation.

5x+12=- 13

x= - 5

5( - 5)+12? =- 13

▼

Evaluate left-hand side

a(- b)=- a * b

- 25+12? =- 13

Add terms

- 13=- 13 ✓

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