Solving a System of Three Equations in Three VariablesDate: 06/23/2007 at 16:35:53 From: Christopher Subject: Can you help me with the formula for this equation My son is having problems with this equation and I have forgotten how to do this it has been so long. Can you help me to explain this to him? x + y + z = 10 x - 2y + z = -2 x - y - z = -12 Date: 06/24/2007 at 23:27:51 From: Doctor Peterson Subject: Re: Can you help me with the formula for this equation Hi, Christopher. I see that we don't seem to have an explanation of solving systems of three equations in our archives. I also see that this one has some tricks to it that might confuse your son even if he's fairly good at solving others. I'm not sure, therefore, whether I should show you the normal way or help your son with the specific issues in this problem. But since you wrote, not he, and you didn't show where he is having trouble, I'll choose to help you with the general methods. If necessary, he can write back and show what goes wrong. Let's take a different and simpler problem so we can focus on the method: x + 2y - z = -3 [1] x - y + z = 4 [2] x - 2y + 2z = 7 [3] This is known as a system of THREE equations in THREE variables. You may recall that you solve a system of TWO equations in TWO variables by combining them to eliminate one variable, so you reduce the system to ONE equation with ONE variable. With a three equation and three variable system, our strategy will again be to reduce the number of variables and equations--this time first reducing to two equations in two variables and then solving that by reducing to only one equation. We want to eliminate one variable. We can pick any of them; in this case I'll eliminate x because it has the coefficient 1 everywhere. Look at the first two equations: x + 2y - z = -3 [1] x - y + z = 4 [2] If we just subtract the second from the first (that is, add the negative of the second), we eliminate x: x + 2y - z = -3 [1] -x + y - z = -4 -[2] ----------------- 3y - 2z = -7 [4] I gave the new equation the name [4] so we can keep track of it later. We need another equation in the same two variables. We can combine the third equation with either of the first two and eliminate again-- making sure to eliminate x this time as well. I'll use the last two equations: x - y + z = 4 [2] -x + 2y - 2z = -7 -[3] ----------------- y - z = -3 [5] Now I've got two equations in two unknowns, which will both be true when all three of the original equations are true. Let's write them as a system: 3y - 2z = -7 [4] y - z = -3 [5] We can solve this by eliminating a variable. I'll eliminate y by multiplying [5] by -3 and adding it to [4]: 3y - 2z = -7 [4] -3y + 3z = 9 -3[5] ------------- z = 2 [6] I've found the value of z! Now how do I find x and y? Well, if you think about what we've done, we reduced the system to the two equations [4] and [5]; we've solved THAT system for z, but to finish solving it we have to find y. We go back to either [4] or [5] and plug in z = 2, then solve for y. I'll use equation [5]: y - z = -3 [5] y - 2 = -3 y = -1 Now we've finished solving this system, but we still need to find x, the first variable we eliminated. To find x, we have to go back to any of the original three equations and plug in both y = -1 and z = 2. I'll use [2]: x - y + z = 4 [2] x - -1 + 2 = 4 x + 3 = 4 x = 1 So the solution is x = 1, y = -1, z = 2. You should plug all three into each of the original three equations in order to make sure we didn't make a mistake anywhere. I want to take a moment to look back at the whole process. We started with three equations, [1], [2], and [3]; then we eliminated x to get two new equations, [4] and [5], with only two variables. Then we started as if this were a new problem, solving it by eliminating y and solving for z, then going back to one of the equations to get y. Finally we went back to the original system and found the remaining variable. I think of this process as like climbing Mt. Everest. We start at the bottom of the mountain (base camp) with a big ugly problem to solve. We then climb part way up and set up a high camp with a simpler problem to solve (only two variables). Then we climb to the top, solving for one variable. On the way up we discarded some equipment (the variables x and y). On the way back down, we pick them back up: y on our way back to the high camp, and x on the way to the bottom: [6] --> z ^ | | | / \ v [4] [5] --> y ^ ^ | | | | / \ / \ v [1] [2] [3] --> x In the process, we eliminated a variable three times, repeating the same process you use to solve two equations. Note that you could eliminate variables in any order--it doesn't have to be x then y then z, but whatever is easiest (or whatever you feel like). The important thing is that once you eliminate the first variable, you don't think about it again until you've finished solving the system of two equations completely; then you find the remaining variable by substituting in one of the original equations. If you have any further questions, feel free to write back. If you have trouble with the actual problem, show me your work (showing at least the equations [1] through [6] that you get, or whatever comes out oddly) and I'll help work out the kinks. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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