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.

Glencoe Math: Course 2, Volume 2

GM

Glencoe Math: Course 2, Volume 2 View details

4. Solve Two-Step Equations

Continue to next subchapter

Exercise 2 Page 472

Start by using the Addition Property of Equality.

y = - 5

Practice makes perfect

To solve an equation, we should isolate the variable term. In this case, we will use the Addition Property of Equality and Division Property of Equality.

- 3y - 5 = 10

LHS+5=RHS+5

- 3y - 5 + 5 = 10 + 5

Add terms

- 3y = 15

.LHS /- 3.=.RHS /- 3.

- 3y/- 3 = 15/- 3

- a/- b=a/b

3y/3 = 15/- 3

MoveNegDenomToFrac

Put minus sign in front of fraction

3y/3 = - 15/3

Calculate quotient

y = - 5

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

- 3y - 5 = 10

y= - 5

- 3( - 5) - 5 ? = 10

MultNegNegOnePar

- a(- b)=a* b

15 - 5 ? = 10

Add terms

10 = 10

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

Subchapter links

Guided Practice p.472

Independent Practice p.473-474

H.O.T. Problems p.474

Standardized Test Practice p.474-476

Extra Practice p.475

Common Core Review p.476