body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

Pearson Algebra 1 Common Core, 2011

PA

Pearson Algebra 1 Common Core, 2011 View details

2. Solving Two-Step Equations

Continue to next subchapter

Exercise 17 Page 91

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

- 27

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.

10 + h/3 = 1

LHS-10=RHS-10

10 - 10 + h/3 = 1 - 10

Subtract terms

h/3 = - 9

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

h/3 = - 9

LHS * 3=RHS* 3

3(h/3) = 3(- 9)

MoveLeftFacToNum

a*b/c= a* b/c

3h/3 = 3(- 9)

CancelCommonFac

Cancel out common factors

3h/3 = 3(- 9)

Simplify quotient

h = 3(- 9)

Multiply

h = - 27

The solution to the equation is h = - 27. We can check our solution by substituting it into the original equation.

10 + h/3 = 1

h= - 27

10 + (- 27/3) = 1

▼

Evaluate

MoveNegNumToFrac

Put minus sign in front of fraction

10 + (- 27/3) ? = 1

Calculate quotient

10 + (- 9) ? = 1

Add and subtract terms

1 = 1

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