Your Tutor Online

Also in this series:

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

Welcome to Your Tutor Online video lessons. This is the last of a 4 part series on systems of equations. In this video we will look at how to solve systems of equations by substitution.

Substitution just means you’ll find out what one variable is equal to and you plug it into the other equation. In this first example we already see that x is the same thing as 2y minus 3. So that means where-ever we see an x in the other equation we can put in this for it instead. And that way we will only have ys to deal with and be able to solve the equation.

So I’ll begin to re-write the second equation: 3y plus 2 x, and instead of x we are going to put in what we know x is equal to. X is the same thing as 2y minus 3; is equal to 8.

Now that we only have one variable, we can solve for y. 3y and distribute the 2; plus 4y minus 6 is equal to 8. Combine like terms and get 7y minus 6 is equal to 8. Add 6 to both sides. 7y is equal to 14. And divide both sides by 7. We find out that y is equal to 2. We are going to take this answer we got for y and plug it back into either one of the original equations to solve for x.

I chose to plug this back into the first equation: x = 2y - 3. So we are going to take our y is equal to 2, put it where we see the y and go ahead and solve for x. 2 times 2 is 4. Subtract 3 and x is equal to 1. And now we know both solutions. X is equal to 1 and y is equal to 2.

Sometimes a variable will not already by isolated, or by itself, so you have to work with an equation to get one of the variables by itself. In this example the second equation could easily be turned into one where y is by itself if we just subtract 3x from both sides. So it becomes y = -3x + 5. And we can take that, just like we did before, and plug that into the other equation where-ever we see a y.

4x - 2; plug in the y; -3x +5 is equal to 10. Now all we have are x’s and we can solve for x. Distribute the -2; So we get plus 6x minus 10 is equal to 10. And we’ll continue this on another page. We’ll combine our like terms the 4x and the 6x leaves us with 10x and at the same time I’ll add 10 to both sides. 10 plus 10 is 20. We’ll divide both sides by 10 to get x by itself and have our answer for x is 2. Now that we know what x is we can go back to one of the original equations, plug in 2 where we see an x and solve for y.

Here I plugged in 2 into the first equation and we will solve for y. 4 times 2 is 8 minus 2y is equal to 10. We’ll subtract 8 from both sides and be left with a -2y on the left, a +2 on the right. Divide both sides by a -2 and y is equal to -1.

That’s all there is for solving systems of equations by substitution. They get a little more complex especially when the variables are not already by themselves. But follow this process and you’ll be able to solve any of them.

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.