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

4. Solving Equations With Variables on Both Sides

Continue to next subchapter

Exercise 22 Page 106

Gather all of the variable terms on one side of the equation and all of the constant terms on the other side. Start by using the Distributive Property.

b=7

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 need to start by using the Distributive Property to simplify the left-hand side of the equation.

8-(3+b)=b-9

Distribute -1

8-3-b=b-9

Subtract term

5-b=b-9

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

5-b=b-9

LHS+b=RHS+b

5-b+b=b-9+b

Add terms

5=2b-9

LHS+9=RHS+9

5+9=2b-9+9

Add terms

14=2b

.LHS /2.=.RHS /2.

14/2 = 2b/2

Calculate quotient

7=b

Rearrange equation

b=7

The solution to the equation is b=7. We can check our solution by substituting it into the original equation.

8-(3+b)=b-9

b= 7

8-(3+ 7) ? = 7-9

Add terms

8-10 ? =7-9

Subtract terms

- 2 = - 2 ✓

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

Subchapter links

Lesson Check p.105

Practice and Problem-Solving Exercises p.105-108

Standardized Test Prep p.108

Mixed Review p.108