Int Algebra: Solving a System by Elimination

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07/01/2009

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Systems of Equations
Linear Systems in Two Variables
Solving a System With Elimination Page [1 of 2]
Solving two linear equations simultaneously by substitution is always a great way to go – solving for one of the variables, and then taking that result, plugging it into the second equation, and then solving for the other variable. There’s actually another method, which sometimes is even easier and preferable. That’s the method of elimination, where you take the two equations and start doing algebra, with both equations at the same time, and trying to get one of the variables to just drop out.
Let me try to illustrate this idea with the following example. Suppose I have 3x + 4y = -5, and, at the same time, I have x - 5y = -8. Now, what I’d like to do is solve this system of two linear equations and two variables at the same time. Now, what does that mean? Well, what could I possibly do? Notice that if I were to multiply one of the equations through on both sides by a number, it’s not going to change the value of the equation. It’s the exact same equation. If I take this, and multiply both sides by 2 here, and then 2 here, it’s the same exact equation. Can I think of a clever choice of multiplying one of these through so that it matches up, in some sense, with the top? Well, what if I just multiplied everything here through by 3. If I multiplied everything here through by 3, then what would I get? What I would see would be 3x - 15y = -24.
Now, let’s just write down that first equation right underneath this. I’m not going to do anything to the first equation, but you’ll notice that these terms are exactly the same, 3x and 3x. The other terms are not the same, I admit, but you have to admit these are. What I could do now is take these equations and do some arithmetic or algebra to the equations by taking the top equation and just subtracting off the bottom equation. What if I do that?
Well, if this is equal – these two things are equal, and these two things are equal. If I subtract 1, this side from that side, and that side from that side, this will still be equal. Right? If 7 = 7, and 3 = 3, then 7 - 3 will equal 7 - 3. It’s not going to change the value. But now look what happens – and I have to be careful that I subtract every single thing, not just the first term. 3x - 3x, they drop out. That’s the idea of the elimination, I lose the x’s. Now, what am I left with? I’m left with -15, but then I have a - +4, so that’s -19y. Then I have negative -24 - -5. Well, that’s +5, so that would equal -19.
Now, I can solve for y. You see how I eliminated the x? Now, I can see that y = 1. How would I find x? Well, I’d find x the same way that I did before. I would just plug in 1 into one of these equations and solve for x. For example, if I plug in 1 into this one – you can plug into either one. So now, I plug in x minus – I put 1 in here, so 5 x 1 is just 5 = -8. I see that x equals, if I bring the -5 over to the +5, -3. So, x = -3.
So, I just solved this system by elimination. I see that x = -3 and y = 1 is a solution. You can check, by the way. Plug in negative 3 for x and 1 for y. You’ll see -9 + 4 = -5. That checks. If you plug it back in here, you see negative -3 - 5 = -8. So, notice this is a solution to both of the equations at the same time. I did it by eliminating one of the variables.
Let’s try another one. How about 3x + 2y = 1 and 2 x + 3y = 1. Well, this one’s a trickier one to think about, because it’s not clear if I should multiply this by some number or that by some number, or something. There are a variety of things you could do. You could multiply this through by a particular fraction. That’s a good way of doing it. Another way is to multiply this by something and that by something else, and then have things line up.
For example, let’s suppose that I want to now get rid of the y’s. Notice what happens if I were to multiply this by 3. That would give me 6y, but then if I multiply this by 2, then I have 6y here. If I subtracted, it would drop out. What I’m going to do is multiply the top equation by 3 and the whole bottom equation by 2, not to change the value of the equation. It will change the look of the equation, but so what? Here, on the top, I would see 9x + 6y = 3. Notice that I multiplied everything through by 3. If I didn’t multiply this side by 3 here, it would change the equation.
Now, I’m going to multiply everything through by 2, and I see 4x + 6y = 2. Notice I have 6y and 6y. They match up perfectly. So, I can actually subtract these two from each other, and I’ll still get a true fact. I’m going to now eliminate – that’s why it’s called elimination, eliminate the 6y. So, if I take 9x and subtract 4x, I’m left with 5x, 6y - 6y = 0, and 3 - 2 = 1. So, I see that x = 51. So, there’s the x value.
Systems of Equations
Linear Systems in Two Variables
Solving a System With Elimination Page [2 of 2]
How could I find the y value? Just plug in that x value into either of these original equations. Either one will give you the same answer, and you can solve for y. For example, suppose I put into this bottom one here – if I put x = 51, I would see 51 • 2, so that’s 52 + 3y = 1. If I bring the 52 over to this side, I’d see 3y = 1 - 52, but 1 is 55, so 55 - 52 would be 53. If I divide both sides by 3, I’d see y equals – so now, I have to invert that 13 and multiply. Now, watch what happens here. You invert, it becomes 31, but 31 I can cancel, and I see 51.
So, that’s fun. The x and the y values turn out to be the same. If you plug them back in, you will see that, in fact, these are satisfied. Right? 53 + 52 = 1 and 52 + 53 = 1. So, in fact, I found the solution this time by using the elimination method, where I multiply equations through by appropriate numbers both sides, so that one of the variables lines up. If I subtract, they eliminate, and I can solve for the other.
Try some of these elimination problems on your own.