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

7. Solve Inequalities by Multiplication or Division

Continue to next subchapter

Exercise 36 Page 512

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

k = 2

Practice makes perfect

To solve an equation, we should first gather all of the variable terms on one side and all of the constant terms on the other side using the Properties of Equality. In this case, we should use the Subtraction Property of Inequality and the Division Property of Equality to isolate the variable.

5k+6 = 16

LHS-6=RHS-6

5k = 10

.LHS /5.=.RHS /5.

k = 10/5

Calculate quotient

k = 2

The solution to the equation is k=2. We can check our solution by substituting it into the original equation.

5k+6 = 16

k= 2

5( 2) + 6 ? = 16

Multiply

10+6 ? = 16

Add terms

16=16 ✓

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

Subchapter links

Guided Practice p.508

Independent Practice p.509-510

H.O.T. Problems p.510

Standardized Test Practice p.510-512

Extra Practice p.511

Common Core Review p.512