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 2 Common Core, 2011

PA

Pearson Algebra 2 Common Core, 2011 View details

2. Standard Form of a Quadratic Function

Continue to next subchapter

Exercise 59 Page 208

If either of the variable terms would cancel out the corresponding variable term in the other equation, you can use the Elimination Method to solve the system.

Example Solution: Elimination Method, because there is no variable with a coefficient of $1 .$
Solution: $(7, - 1)$

Practice makes perfect

Since neither equation has a variable with a coefficient of

1,

the Substitution Method may not be the easiest. Instead, we will use the Elimination Method. To solve a system of linear equations this way, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation.

{3 x - 5 y = 26 - 2 x - 3 y = - 11 (I) (II)

Currently, none of the terms in this system will cancel out. Therefore, we need to find a common multiple between two variable like terms in the system. If we multiply (I) by

2

and multiply (II) by

3,

the

x -

terms will have opposite coefficients.

A shortcut to better grades

Download the app and create a free account to view full content

Subchapter links

Lesson Check p.206

Practice and Problem-Solving Exercises p.206-208

Standardized Test Prep p.208

Mixed Review p.208