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

6. Ratios, Rates, and Conversions

Continue to next subchapter

Exercise 51 Page 121

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

6

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 will start by using the Addition Property of Equality.

-2/3x-8=-12

LHS+8=RHS+8

-2/3x-8+8=-12+8

Add terms

-2/3x=-4

MoveNegFracToDenom

Put minus sign in denominator

2/-3x=-4

MoveRightFacToNum

a/c* b = a* b/c

2x/-3=-4

LHS * (-3)=RHS* (-3)

2x=-4(-3)

MultNegNegOnePar

- a(- b)=a* b

2x=12

.LHS /2.=.RHS /2.

2x/2=12/2

Calculate quotient

x=6

The solution to the equation is x=6. We can check our solution by substituting it into the original equation.

-2/3x-8=-12

x= 6

-2/3( 6)-8? =-12

MoveRightFacToNum

a/c* b = a* b/c

-2*6/3-8? =-12

Multiply

-12/3-8? =-12

Calculate quotient

-4-8? =-12

Subtract term

-12=-12

Since the left-hand side and right-hand side are equal, we know our solution is correct.

Subchapter links

Lesson Check p.119

Practice and Problem-Solving Exercises p.119-121

Standardized Test Prep p.121

Mixed Review p.121