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.

McGraw Hill Glencoe Algebra 1, 2012

MH

McGraw Hill Glencoe Algebra 1, 2012 View details

4. Solving Equations with the Variable on Each Side

Continue to next subchapter

Exercise 21 Page 100

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

a=1

Practice makes perfect

To solve an equation with a variable on both sides, we 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. Before we do that, we have products that can be simplified using the Distributive Property.

6(3a+1)-30=3(2a-4)

Distribute 6 & 3

18a+6-30=6a-12

Subtract term

18a-24=6a-12

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

18a-24=6a-12

LHS+24=RHS+24

18a-24+24=6a-12+24

Add terms

18a=6a+12

LHS-6a=RHS-6a

18a-6a=6a+12-6a

Subtract term

12a=12

.LHS /12.=.RHS /12.

12a/12=12/12

Calculate quotient

a=1

The solution to the equation is a=1. We can check our solution by substituting it into the original equation.

6(3a+1)-30=3(2a-4)

a= 1

6(3( 1)+1)-30 ? = 3(2( 1)-4)

▼

Simplify

a * 1=a

6(3+1)-30 ? = 3(2-4)

Add and subtract terms

6(4)-30 ? = 3(- 2)

a(- b)=- a * b

24-30 ? = - 6

Subtract term

- 6=- 6

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

Subchapter links

Check Your Understanding p.100

H.O.T. Problems p.101

Standardized Test Practice p.102

Spiral Review p.102

Skills Review p.102