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, 2015

PA

Pearson Algebra 1 Common Core, 2015 View details

Get Ready!

Continue to next subchapter

Exercise 1 Page 361

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

Identity

Practice makes perfect

To solve an equation, we need to start by using the Distributive Property to simplify the left- and right-hand sides of the equation.

3(2-2x)=-6(x-1)

Distribute 3

6-6x=-6(x-1)

Distribute - 6

6-6x=-6x+6

CommutativePropAdd

Commutative Property of Addition

6-6x=6-6x

Since the left-hand side is equal to the right-hand side, simplifying the equation resulted in an identity.

Subchapter links

Exercises p.361