Solving a Multi-Variable Formula or Equation for One VariableDate: 07/24/2008 at 02:23:27 From: Tori Subject: Solving an equation with multiple variables I was given y^2 + 3yz - 8z - 4x = 0 and asked to solve for z. I haven't the slightest idea as to how to get rid of the y in 3yz. It seems like if you do manage to get rid of it, you have to put it back again later. I know you can factor out a y from the first half, and a -4 from the other half so it looks like this: y(y + 3z) - 4(2z + x) = 0. After that, I'm stuck. Date: 07/24/2008 at 16:54:39 From: Doctor Ian Subject: Re: Solving an equation with multiple variables Hi Tori, Suppose we start with something like xz + yz = 1 and we want to solve for z. We can use the distributive property to factor z out of the expression on the left, and then divide: z(x + y) = 1 z = 1/(x + y) Can you see how to apply this same reasoning to your problem? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 07/24/2008 at 23:44:08 From: Tori Subject: Solving an equation with multiple variables Here's what I did: y^2 + 3yz - 8z - 4x = 0 y(y + 3z) - 8z - 4x = 0 y(y + 3z) - 8z = 4x y + 3z - 8z = 4x/y -5z = (4x - y)/y 5z = (-4x + y)/y 5z = -4x z = (-4/5)x Date: 07/25/2008 at 08:55:09 From: Doctor Ian Subject: Re: Solving an equation with multiple variables Hi Tori, I'm glad you wrote back! It looks like there are a couple of things that we need to clear up. You wrote: >Here's what I did: > > y^2 + 3yz - 8z - 4x = 0 > y(y + 3z) - 8z - 4x = 0 > y(y + 3z) - 8z = 4x > y + 3z - 8z = 4x/y You ran into a problem here. Note that, when you divide an equation on both sides by y, you have to divide EVERYTHING on both sides by y, and you didn't do that. Here's how that step should have looked: y(y + 3z) - 8z = 4x (y + 3z) - 8z/y = 4x/y See the difference? > -5z = (4x - y)/y > 5z = (-4x + y)/y > 5z = -4x You did a similar thing here... So, it looks to me like there are a couple of things going on here. The first is that you're not quite clear on the idea of cancellation. That is, you're doing things like this, -4x + y ------- -> -4x y with variables, that you would never do with constants. For example, it would never occur to you to do this, 5 + 2 ----- -> 5 2 but that's essentially what you did, isn't it? The idea of cancellation is basically the same as the idea of equivalent fractions. That is, we can start with a fraction like 2/3, and multiply by 5/5, and get a fraction with the same value: 2 5 10 - * - = -- 3 5 15 Similarly, if we start with 10/15, we can break it into factors, and cancel some of those by separating them out: 10 2 * 5 2 5 2 2 -- = ----- = - * - = - * 1 = - 15 3 * 5 3 5 3 3 But this only works with MULTIPLICATION. If you try it with addition, you get nonsense: 10 7 + 3 7 -- = ------ = ... -- Wrong! 15 12 + 3 12 21 9 + 12 -- = ------ = ... = 9 Wrong! 12 12 And if we can't do something with numbers, we can't do it with variables--because variables are just numbers whose values we don't happen to know at the moment. They still have to obey all the same rules. This, by the way, is one of the reasons we spend so much time learning to factor polynomials. We can't do this, x^2 + 3x + 2 3x + 2 ------------ -> ------ Wrong! x^2 + 6x + 5 6x + 5 but we CAN do this, x^2 + 3x + 2 (x + 1)(x + 2) (x + 1) (x + 2) x + 2 ------------ = -------------- = ------- * ------- = ----- x^2 + 6x + 5 (x + 1)(x + 5) (x + 1) (x + 5) x + 5 Note that in this case, we have to stick a caveat on our final expression: x^2 + 3x + 2 x + 2 ------------ = ----- Where x is not -1 x^2 + 6x + 5 x + 5 This is because setting x to -1 in the ORIGINAL expression would cause a division by zero, which is undefined. Anyway, does this make sense? ***** The second thing is that it looks like you sort of lost sight of the goal early on. You want to solve for z, so you want to isolate the quantities involving z early on. For example, y^2 + 3yz - 8z - 4x = 0 y^2 + 3yz - 8z = 4x 3yz - 8z = 4x - y^2 Now we can factor out z, z(3y - 8) = 4x - y^2 and use division to cancel it on the left, z = (4x - y^2)/(3y - 8) Note how I'm using ()'s to make sure everything that needs to be divided, gets divided! My point is, factoring out a y, y^2 + 3yz - 8z - 4x = 0 y(y + 3z) - 8z - 4x = 0 isn't incorrect, but it's not especially helpful, either. Since you are trying to isolate z, the key is to see your original equation as [stuff] + [stuff]z - [stuff]z - [stuff] = 0 You don't really care what that other stuff is. It's just a distraction. Put any of the z terms on one side, and the non-z terms on the other. Factor the z out of the z terms, then divide both sides by the other factor to isolate the z. Does this make sense? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 07/26/2008 at 02:40:31 From: Tori Subject: Thank you (Solving an equation with multiple variables) Ah! That makes a lot more sense. I always seem to forget what you can and can't cancel out in fractions. I totally didn't see that you could factor out z's, too. Thank you for your help! |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/