Solving Two Equations with Two Unknowns – II

What do you know about solving two equations with two unknowns or variables? We began looking at this in the previous video, but let’s continue our exploration here.

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Transcript: Solving Two Equations with Two Unknowns

In the previous lesson, we discussed solving systems equation by substitution. And in that discussion, we noted that substitution in a way is kind of a limited value. In other words, if one of the coefficients is +1 or -1, than it’s very easy to substitute. But the general case, of course, will be we will have 2 equations and we will have 4 coefficients two for x and two for y and none of them will equal +1 or -1.

And if we try to solve the substitution in that case we are going to be introducing fractions to the equation. We are going to make things unnecessarily hard, again, substitution would work. If you like to fight your way through fractions, you could find the answer that way. But I think you’ll find that this method, this second method we’re gonna discuss is much easier.

The Elimination Method

In this lesson, we will discuss this method called elimination. So the elimination method, first and easy example. Pause the video and think about how you would solve this and then we’ll talk about it.

Okay, notice we are always allowed to add two equations. We can add an equation to an equation, that’s perfectly allowed.

Notice that we added these two equations, the +3y and the -3y would cancel. We would eliminate the 3ys. So this is a very easy example. We’re going to add these two equations. When we add, 7x + 2x = 9x. Positive 3y + (-3y) cancels, that gives us 0.

And 5 plus 13 gives us 18. So the sum of those two equations is the equation 9x = 18. Well, that’s very easy to solve. Divide by 9 and I get x = 2. Now we can plug that value into either equation and solve for y. Let’s plug into the second equation that looks a little friendlier.

Plug in x = 2, so we get 4- 3y = 13. Subtract 4, divide by- 3 and we get y = (-3). And so x = 2, y =- 3 is the solution to that system. That example was particularly easy because in the system as given, the coefficients of the same variable were equal and opposite in the two equations. So really, that was almost handed to us on a silver platter, that it was so convenient.

Ordinarily, it will not be the case, but we can make it the case. The strategy is to multiply both sides of one equation by one number, and both sides of the other equation by another number. For one of the variables, the two coefficients are equal and opposite, so that when we add they will cancel and you will eliminate that variable. Let’s look at the system we solved in the previous lesson.

Problem from Previous Lesson

So this is a system we solved with substitution and it works well for substitution because we have that one variable x just with a coefficient of 1. We are going to solve the same system now with elimination and we are going to solve it in two different ways.

So suppose first of all, we want to eliminate x. So the very first step in elimination is we have to choose the variable we want to eliminate.

We want to eliminate x, what that means is, a very convenient way to do that, would multiply the second equation by -2. So every single part of that second equation gets multiplied by -2 and then we will add the two equations. The -2x the +2x will cancel we will get -y = -7 multiply by -1, we get y = 7 then we could plug in and solve for x.

So that is one way to solve using the elimination method. We could start with those same equations and instead of eliminating x, we could also make the choice to eliminate y and that would be perfectly valid. If we want to eliminate y, then we could multiply the first equation by -2 and the second equation by positive 3. And the goal is to get the y’s to have equal and opposite coefficients.

So if I multiply the first one by -2 and the second by positive 3, then I get a -6y in one equation and a positive 6y in the other equation. That means that when I add them, they will cancel. I’ll solve for x, x = -3 and then I can plug in and find the value of y. With the elimination method, we begin by choosing which variable to eliminate.

And in general we could eliminate either one. If the two coefficients of one variable are the same, or if one is a multiple of the other than that variable is the better choice to eliminate. So sometimes we get that. For example, here I am going to say pause the video and decide which route you would take to solve this with elimination and then we will talk about it.

Okay, well, notice that the coefficients of y, +2 and -4, one is a multiple of the other. So that would be the better choice. It would not make sense if you eliminate x here. It’d be much more strategic to eliminate y. Because all we’d have to do is multiply that top equation by 2.

Multiply the top equation every term by 2. Then we have a +4y, a- 4y. We add, those cancel. We get 10x = 80, x = 8, Then we can plug that into the first equation, we can actually plug it into either one, it doesn’t matter. Plug it in to the first equation We get 2y = 3 or y = 3/2.

So x = 8, y = 3/2 is the solution to that.

Practice Problem

Here’s another practice problem. This is a system that we had in the last lesson. Remember at the end of the last lesson we had a system, and we said, substitution really didn’t work very well for this, so we didn’t solve it. So here it is again.

Now we’re gonna solve it with elimination. And in fact I’m gonna say pause the video, and try on your own to solve this with elimination. Okay, well let’s talk about this. I’m gonna choose to eliminate y. We could do either way but I’m gonna choose to eliminate y.

I’m gonna multiply the first equation by +2 and the second equation by -5. And what I’m trying to do is get to +10y and -10y. So when I do those multiplications these are the equations I get. I multiply every term in each equation. Then I add the two equations, I get -17x = -68. Well here it is good to know that 68 is 17 x 4.

So when we divide -68 by -17 we get 4. Now I can plug that into either equation. I will plug it into the first equation. And get 16 + 5y = 1. Subtract 16 from both sides, divide y -15, y = -3 and so the solution to that system is very simply x = 4, y = -3.

Practice Problem

Here’s another practice problem. Pause the video and then we’ll talk about this.

This is a tricky practice problem because this is a practice problem where it would be very easy to make the work for yourself 100 times more difficult than it has to be. As a general rule, if the test gives you a system and asks you to solve for the value of the expression, you do not need to solve for the values of the individual variables.

There will always be an elegant shortcut that leads directly to the value of the expression. Notice here that the coefficient of x is 3 bigger in the first equation than in the second equation, 5x, 5 is 3 bigger than 2. And notice that the coefficient of y, 2 is also 3 bigger than the coefficient in the other equation because 2 is 3 bigger than -1.

Well that’s very interesting. So that’s a suggestion right there, that we should subtract the equations. So I’m gonna multiply the second equation by -1. And then just add them. So when I do that lo and behold, I get 3x + 3y = 36, I divide by 3 and I get x + y = 12.

And that is the answer. Because the question was asking for the value of x + y. Elimination is the more convenient of the two, if the two coefficients of the same variable are equal or opposite or one is a multiple of the other. So certainly if we see that, that makes elimination easier, but in a general case, elimination is always easier than substitution.

If you are asked to find the value of an expression, you will almost always be able to find that without finding the values of the individual variables. And in that last problem it would have been a mistake to solve individually for the value of x and the value of y, all we needed was x + y.

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  • Mike MᶜGarry

    Mike served as a GMAT Expert at Magoosh, helping create hundreds of lesson videos and practice questions to help guide GMAT students to success. He was also featured as "member of the month" for over two years at GMAT Club. Mike holds an A.B. in Physics (graduating magna cum laude) and an M.T.S. in Religions of the World, both from Harvard. Beyond standardized testing, Mike has over 20 years of both private and public high school teaching experience specializing in math and physics. In his free time, Mike likes smashing foosballs into orbit, and despite having no obvious cranial deficiency, he insists on rooting for the NY Mets. Learn more about the GMAT through Mike's Youtube video explanations and resources like What is a Good GMAT Score? and the GMAT Diagnostic Test.

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