دو معادله، دو مجهول 2

Two Equations, Two Unknowns - II

In the previous lesson, we discussed solving systems of equations by substitution. And in that discussion we noted that substitution in a way, is kind of of limited value. In other words, if one of the coefficients is positive 1 or negative 1, then it’s very easy to substitute.

But the general case, of course will be, we’ll have two equations and we’ll have four coefficients, two for x and two for y, and none of them will equal positive 1 or negative 1. And so we try to solve with substitution in that case, we’re gonna be introducing fractions in the equation, we’re gonna 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. In this lesson, we will discuss this method elimination. So, the elimination method.

First an 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 positive 3y and the negative 3y would cancel. We would eliminate, the 3y’s. So this is a very easy example. We’re gonna add these two equations.

When we add, 7x plus 2x gives us 9x. Positive 3y plus negative 3y cancels, that gives us 0. And 5 plus 13 gives us 18. So the sum of those two equations is the equation 9x equals 18. Well that’s very easy to solve, divide by 9 again x equals 2. Now we can plug that value into either equation and solve for y.

Let’s plug it into the second equation, it looks a little friendlier. Plug in x equals 2, so we get 4 minus 3y equals 13. Subtract 4, divide by negative 3, and we get y equals negative 3. And so x equals positive 2, y equals negative 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, so that 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. 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’re gonna solve this same system now, with elimination.

And we’re gonna 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 se, be to multiply the second equation by negative 2.

So every single part of that second equation, gets multiplied by negative 2. Then we’ll add the two equations. The negative 2x, the positive 2x will cancel. We’ll get negative y equals negative 7. Multiply by negative 1, we get y equals 7. Then we can 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 negative 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 negative 2 and the second by positive 3, then I get a negative 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 equals negative 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, if the two coefficients of one variable are the same, or if one is a multiple of the other, then that variable’s the better choice to eliminate. So sometimes we get that. So for example here, I’m gonna say pause the video, and decide which route you would take to solve this with elimination, and then we’ll talk about it.

Okay, well notice that the coefficients of y, positive 2 and negative 4, one is a multiple of the other. So that would be the better choice, it would not make sense to eliminate x here. It would 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 plus 4y, a minus 4y. We add, those cancel, we get 10x equals 80, x equals 8. Then we can plug that into the first equation. We can actually plug it into either one, it doesn’t matter. Plug it into the first equation, we get 2y equals 3, or y equals 3 halves. So, x equals 8, y equals 3 halves, is the solution to that.

Here’s another practice problem. This is the 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 positive 2, and the second equation by negative 5.

And what I’m trying to do is get the plus 10y and minus 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 negative 17x equals negative 68. Well, here it’s good to know that 68 is 17 times 4. So when we divide negative 68 by negative 17, we get positive 4.

Now I can plug that into either equation. I’ll plug it into the first equation. And get 16 plus 5y equals 1. Subtract 16 from both sides, 5y equals negative 15, y equals negative 3. And so the solution to that system is very simply, x equals positive 4, y equals negative 3.

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’d be very easy to make the work for yourself a hundred 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, in the second equation, than in it is in the, 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 of the other equation. Because positive 2 is 3 bigger than negative 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 a negative 1, and then just add them.

So when I do that, lo and behold, I get 3x plus 3y equals 36. I divide by 3 and I get x plus y equals 12. And that is the answer. Because the question was asking for the value of x plus 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 plus y.