Also in this series:

• Lesson 1: Introduction
• Lesson 2: Graphing
• Lesson 3: Elimination (current page)
• Lesson 4: Substitution

Welcome to Your Tutor Online Video Lessons. Today we continue our series on solving systems of linear equations. Today we will look at how to solve those equations by elimination.

For the elimination method you will simply add the two equations together in a way that will cancel out one of the variables. In our example here we will just go ahead and add the two equations together as we see them. The ys will cancel out. 3 plus 2 gives us 5 x. Negative 1 plus positive 1 cancels out; and 10 plus 20 is 30. Now that the ys are canceled out we can solve for x. In this example x is equal to 6. Then we want to take our solution, x = 6, go back to one of the original equations and plug it in. So we plugged 6 in for x and we will solve for y. 2 times 6 is 12, plus y, is equal to 20. Take away 12 from both sides and we are left with y = 8. Now we have both answers to our system of equations. X = 6 and y = 8.

Sometimes you need to change one of the equations before adding them together in order to get a variable to cancel out. In this example if we multiply the second equation by negative 2, the xs will cancel out. When you multiply an equation by a number, you need to multiply every single term by that number. So now we get -2x + 4y = 22. And we can just copy over the top equation exactly as we see it. 2x + 3y = 13. Now when we add these two equations together the xs will cancel out. Positive 2 plus negative 2 is zero. 4 and 3 is 7y, and 13 plus 22 is 35. We can divide this new equation by 7 on both sides and we will solve: Y = 5. Again we are going to take this same answer, plug it into one of our originals and we will get the answer for x. X -2 times 5 = -11. X - 10; -2 times 5 is -10 = -11. Add 10 to both sides and x is -1.

For our last example we will see that sometimes you have to multiple both equations by a factor in order to get something to cancel out. Here, none of the coefficients are multiples of the other. We need to multiply both equations in order to match up any of the coefficients. I am going to try to eliminate the x. In order to do that I will multiply the 2x by 5 and the 5x by -2 to get a 10 and a -10 to cancel out. 2x times 5 becomes 10x; 3y times 5 becomes 15y; and 16 times 5 is 80. For this equation we are multiplying by -2 so we get -10x; +8y; all that is equal to +12. Add both of these together. The 10s cancel out: +10-10 is zero. 15 + 8 = 23y. 80 + 12 = 92. We will divide both sides by 23 to get the y all by itself. And y = 4. Same process as before, we are going to take our answer for y and plug it into an original. 2x + 3y; y here is 4, is equal to 16. 2x + 12 = 16. Subtract 12 from both sides. 2x = 4; and finally 4 divided by 2. X = 2.

I hope you found this lesson useful. If you have any questions leave a comment on the blog at www.YourTutorOnline.com If you have any lesson suggestions send an email to podcast@yourtutoronline.com Thanks for watching, class dismissed.

This entry was posted on Saturday, October 25th, 2008 at 7:04 am and is filed under Algebra. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.