Connecting...

This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Algebra 1
For more information, please see full course syllabus of Algebra 1
Algebra 1 Solving a System Using Elimination
Lecture Description
In this video we'll learn how to use the elimination method to solve a system of linear equations, and we'll learn how to recognize types of solutions. The goal with the elimination method is to combine the equations in a way so that we can solve for one variable at a time. In order to do this, sometimes we'll have to multiply by a constant. When multiplying by a constant, you may have to multiply one, or both of the equations to create a pair of terms that will cancel when added. We’ll also learn what happens when a system has exactly one solution, no solutions, or an infinite amount of solutions.
Bookmark & Share
Embed
Share this knowledge with your friends!
Copy & Paste this embed code into your website’s HTML
Please ensure that your website editor is in text mode when you paste the code.(In Wordpress, the mode button is on the top right corner.)
×
Since this lesson is not free, only the preview will appear on your website.
- - Allow users to view the embedded video in full-size.
Next Lecture
Previous Lecture
1 answer
Wed Apr 1, 2020 12:31 PM
Post by Jun Pang on March 31, 2020
On Example 2, you changed the x's from 4 and 3 to 12. Wouldn't it be easier to change the y's to 6, since 6 is a lower number than 12.
1 answer
Mon Jun 16, 2014 3:00 PM
Post by David Saver on June 10, 2014
Great Teaching!
3 answers
Last reply by: frank fan
Sun Nov 17, 2019 1:55 PM
Post by Henry Sith on February 10, 2014
I believe in one of the "Practice questions" there is an error, the stated question is: 12x + 3y = 4811x − 3y = 33
(BTW, Educator should look to make the questions easier to read)
Here is the error:
Step 1. 12x + 3y = 48
+11x − 3y = 33
Step 2. 22x = 81