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ELM: CSU Math Study Guide17 chapters | 147 lessons | 7 flashcard sets

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Lesson Transcript

Instructor:
*Jeff Calareso*

Jeff teaches high school English, math and other subjects. He has a master's degree in writing and literature.

When you have two variables and two equations, is solving the problem four times as hard? No! Not when you use the substitution or elimination method to solve systems of equations. We'll practice both in this lesson.

If there was always just one unknown thing or one problem at a time, life wouldn't be so tough. Let's say you get invited to a wedding and all you need to figure out is if you want the chicken, fish or vegetarian entrée. That's not so bad, right? But, life isn't that simple.

At one point or another, it's your own wedding. And then, there isn't just chicken or fish, there's this reception hall or that one, this best man or that one, this florist or, geez, when did this town get so many florists? And since when are there so many types of paper and different fonts for invitations?

Systems of equations can seem just as overwhelming. A **system of equations** is a group of two or more equations with the same variables. Multiple equations? Multiple variables? It's enough to make you want to elope. Fortunately, though, solving systems of equations is much more straightforward than it seems.

In this lesson, we're going to practice the two most common methods of solving systems of equations. First, there's the substitution method. The **substitution method** is when you solve one equation for either variable, then substitute the solution into the other equation.

Then there's the elimination method. The **elimination method** is when we add or subtract equations together to solve for a variable. Let's try out each method as we solve some equations.

Let's start with the substitution method. Here are two equations:

*y* - 2*x* = 1 and 5*x* - 2*y* = 3

Let's take the first one and solve for *y*. We add 2*x* to get *y* = 1 + 2*x*. Next, we substitute 1 + 2*x* for *y* in the second equation. So, we get 5*x* - 2(1 + 2*x*) = 3. Now we solve for *x*. First, we distribute the -2 and get -2 - 4*x*. Then, 5*x* - 4*x* is just *x*. We add 2 to both sides to get *x* = 5.

Now we have our *x* value. Let's plug that in to either equation and get a *y* value. Just pick the one you think will be easier. Let's use the first one. *y* - 2(5) = 1. *y* - 10 = 1. Add 10 and we get *y* = 11.

A good check is to put both variables back in both equations. If they don't work, you know you made a mistake somewhere. Let's try that here.

*y* - 2*x* = 1 becomes 11 - 2(5) = 1. That's 11 - 10 = 1. And yep, 1 = 1. And, 5*x* - 2*y* = 3 becomes 5(5) - 2(11) = 3. That's 25 - 22 = 3. Yep again! 3 = 3. We're good! And, no future in-laws were insulted in the solving of this problem.

Let's try one more. Here are two equations:

*y* = 3*x* - 4 and 2*y* - 5*x* = 2

In this one, we already have one solved for *y*, so let's just plug 3*x* - 4 in for *y* in the second equation. We get 2(3*x* - 4) - 5*x* = 2. 2 * 3*x* is 6*x* and 2 * 4 is 8. 6*x* - 5*x* is just *x*. Then, we add 8 to both sides, and we get *x* = 10.

Now, plug 10 in for *x* in that first equation. *y* = 3(10) - 4. *y* = 30 - 4, or 26.

Okay, let's check our work. We have *y* = 3*x* - 4, we get 26 = 3(10) - 4. That's 30 - 4, which is 26! And, 2*y* - 5*x* = 2 becomes 2(26) - 5(10) = 2. That's 52 - 50 = 2. Success!

Let's try some elimination practice. Here are two equations:

*x* + *y* = 5 and 5*y* - 1 = 2*x*

Think of this like making a sandwich. You need your ingredients to line up. If your tomato is over too far, it's going to slide right out. So, let's line everything up.

*x* + *y* = 5

5*y* - 1 = 2*x*

Great! Now we need to get coefficients balanced for one of the variables so they'll cancel out. This is like making sure your amounts of mayonnaise and mustard are in balance. Here, we can multiply the top equation by 2 so we get 2*x* on top to match the 2*x* on the bottom.

Now, let's subtract. The 2*x's* cancel out. We're left with -3*y* = 9. Divide by -3 and *y* = -3. Now, plug -3 in for *y* in the first equation. *x* - 3 = 5. So, *x* = 8. That's our sandwich! But wait. Let's check our work before we eat it.

In *x* + *y* = 5, we get 8 - 3 = 5. That works out. In 2*x* + 5*y* = 1, we get 2(8) + 5(-3) = 1. That's 16 - 15 = 1. Success!

Let's try another:

2*x* - *y* = 3

-4*x* + 3*y* = 1

Hey, look - our terms are neatly stacked. Thanks, sandwich gods! And, we can multiply the top equation by 2 to get rid of the *x* terms. So, it becomes 4*x* - 2*y* = 6. Here, we want to add. -2*y* + 3*y* is just *y*. And, 6 + 1 is 7. So, *y* = 7.

Now, plug 7 in for *y* in one of the equations. 2*x* - 7 = 3. 2*x* = 10. *x* = 5.

Let's check our answers. 2*x* - *y* = 3 becomes 2(5) - 7 = 3. That's 10 - 7 = 3, so that works. And, -4*x* + 3*y* = 1 becomes -4(5) + 3(7) = 1. That's -20 + 21 = 1, so that works too. We did it!

In summary, weddings are hard. Also, we practiced solving **systems of equations**. There are two common methods of solving systems of equations. The first is the **substitution method**. This involves solving one of the equations for one of the variables, then substituting the result for that variable in the other equation. The other method is the **elimination method**. This method involves stacking the equations and multiplying the terms in one so that when you add or subtract the equations, one of the variables disappears.

You should be able to solve systems of equations involving two unknown variables by both the substitution method and the elimination method after watching this video lesson.

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ELM: CSU Math Study Guide17 chapters | 147 lessons | 7 flashcard sets

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