Int Algebra: Solving a System by Substitution

Great example for substituttion method!

12/31/2008

~ brittanie

Thanks for posting. I get it now :)

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Systems of Equations
Linear Systems in Two Variables
Solving a System by Substitution Page [1 of 2]
Now, one method for solving systems of linear equations is to actually just take one of the equations, and solve it for one of the variables. Then, take that answer, and plug it into the second equation. Therefore, you’d have just one equation with one variable that you can solve, and then find the other variable in the same way.
Let me try to show you exactly what I’m saying. This is what is known as solving a system by substitution. Now, suppose I give you the two lines, y = 4x - 3 and, and at the same time, y = 3x - 1. These are two different equations, two different lines. Now, I want to see if they crisscross. Well, they crisscross when the x’s and y’s are the same. What I could do is just ask myself, “Let’s see. If this y equals that, then I can actually substitute this value for y in the second equation.” You see how I’m substituting? I just say, “Well, if there’s a y here, and I know that y is supposed to equal that, I can plug that into here, and I notice I’ll have an equation just in terms of x’s.” So, that is actually the substitution method.
So, I’d have 4x - 3. That’s the “aka” y, y = 3x - 1. Now look, I could solve that pretty easily. I’ll bring the 3x over to here. It becomes a -3x, so I have an x. If I bring that -3 to here, it becomes a +3. With -1, it’s 2. So, I see x = 2. That was pretty easy. Then, I could find out what y equals just by plugging back into one of these things here. So, y would equal – it could be either one of these, 2(2) - 3, which would be 4 - 3, which is 1. So, if you do that––hold on a second, something is weird. Would that be really right? If I put a 2 in here––no, something’s wrong. Let’s see. If I bring this over to here, I see an x. If I bring this over to here, I see a +3 - 1 = 2, so that looks good. Oh, when I plug it in here––it should be a 4. So, that’s now going to be 8. 8 - 3 = 5. Watch me save this. There you go, fine. So, we have a 5, right? I just did that by checking.
Now, you can check, and you would have found this mistake by plugging the 2 into the other one and seeing 6 - 1 = 5. You see you get the answer, either way. The point is that x would equal 2, and y would equal 5. Those are the points that these two lines have in common, and you can check. Check that if you put in 2 here, you get 5 here. That would check. Check that if you get 2 in here, you get 5 here. So, that means that, in fact, these two lines crisscross right at 2,5. I solved them simultaneously, by substitution.
Let’s try this on a different one. 3x - 4y = 2 and 4x + 3y = 14. Well, again, the method is to solve for one of the variables, and then insert that into the second equation. Why don’t we just take the first equation here, and you can solve it for either x or y. It doesn’t make a difference. Why don’t we solve it for x? We’ll take this and solve for x. Again, I remind you, you could have taken this and solved it for y. You could have taken this one and solved it for x or solved it for y. So, if I solve this for x, I’ll bring this over to that side, and we see 3x = 2 + 4y. If I divide everything through by 3, I see that x = 2 + 4y, all divided by 3.
So, that first equation allowed me to figure out what x is, in terms of y. Now, what I could do is wherever I see an x in the second equation, I could insert that value right in for x. Now notice, after I do that, I’ll just have y’s. This will be an equation in y. What equation will it be? Well, it’s going to be 4 times this thing, so it’s going to be 34, because I have a third downstairs. 34(2 + 4y) + 3y = 14. Now, I can solve this by distributing, and I would see 38 + 316y + 3y. By the way, I see I’m going to need a common denominator, because I have a third down here. So, 3y is actually 39, if I multiply top and bottom there by 3. That equals 14. If I bring this 38 to the other side by subtracting it, what I would see is the following: That would go to this side here, but what would I have here? I would have 16 and 9, that’s 325. So, I’d have 325y, and that would equal 14 - 38. Now, how many thirds is 14? Well, if I multiply top and bottom here
Systems of Equations
Linear Systems in Two Variables
Solving a System by Substitution Page [2 of 2]
by 3, I would see a 2 and a 3, so it would be 342. I’d see 342, and then I subtract off that 38 on this side. Now, I can basically solve it pretty easily by doing a couple of things.
Well, first of all, I could actually just multiply everything through by 3. That’ll kill all the 3’s everywhere you see, so we’ll have 25y equals – and what’s 42 - 8, is that 36? Let’s see. Wait, is that right? No. If I subtract 8, what would I get? 34? Yes, I was right. So, 34. Therefore, y = 2534. So, that’s the y value. What’s the x value? Well, to find the x value, I take that y value, and I just put it right into here, and that will tell you what x equals. So, x would equal 2 + 4 x 2534, all divided by 3. Now, we can compute that. What would that equal? Well, I would have 2 plus – and here, I would see 4 x 34, so that’s 25136. 12 is the same thing as 2550, and 2550 + 25136 = 25186, all divided by 3. So, when you invert the 3 and multiply, the 3 goes on the bottom, and I see 75186. That’s the x value.
So, the answers would be – where do these things share a point in common? What’s the solution to this system at the same time? x = 75186. We already said y was going to be 2534. You can check those answers by plugging them in, and seeing, in fact, that both of these things are satisfied, but I’ll leave that to you.
We’ll take a look at more methods for solving two systems of linear equations simultaneously, next.