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anonymous
 3 years ago
I am still having so much trouble with systems of equations with 3 variables!! Can anyone help me out?
One of my problems is:
x+yz=1
x+y+z=3
3x2yz=4
anonymous
 3 years ago
I am still having so much trouble with systems of equations with 3 variables!! Can anyone help me out? One of my problems is: x+yz=1 x+y+z=3 3x2yz=4

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richyw
 3 years ago
Best ResponseYou've already chosen the best response.2alright. there are a few ways to do this. For a simple system like this (3 equations, 3 unknowns) probably the easiest way is to do it like this

richyw
 3 years ago
Best ResponseYou've already chosen the best response.2\[x+yz=1\]\[z+y+z=3\]\[3x2yz=4\]first solve one of the equations for one of the variables. I will solve the first one for \(z\)\[z=x+y+1\]alright, so now you have z in terms of x and y, so plug that value into another equation. I'll plug it into the second one.\[x+y+(x+y+1)=3\]now you can solve that one for one of the variables. I'll solve for x. \[2x=22y\]\[x=1y\]and now you can plug that into the third equation to find the value of x.

richyw
 3 years ago
Best ResponseYou've already chosen the best response.2I meant the value of y. \[3(1y)2y((1y)+y+1)=4\]\[33y2y2=4\]\[5y=5\]\[y=1\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How did you get x+y+1=z? I got x+y=z. Sorry, I don't mean to sound stupid but I never learned anything in Alg2. I'm good at Geometry concepts, not Alg2 and PreCalc :/

richyw
 3 years ago
Best ResponseYou've already chosen the best response.2now back substitute to find the other two values. \[x=1y=11=0\]\[z=x+y+1=1+0+1=2\]so the solution is \(x=0, \: y=1,\: z=2\)

richyw
 3 years ago
Best ResponseYou've already chosen the best response.2in the first equation we have \(x+yz=1\) and we want to isolate z right? so add z to both sides and we get \[x+yz+z=1+z\]\[x+y=z1\]now we want z by itself so we add 1 to each side and get \[x+y+1=z1+1\]\[x+y+1=z\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, okay, I see. That's not how I did it at all. See, what I tried to do was add the top two equations together, and then do the same for the bottom two. (I used the middle one 2 times) and then used the sum of both to add together and create a new equation to find the answer for a variable.

richyw
 3 years ago
Best ResponseYou've already chosen the best response.2that is an excellent method choice as well.

richyw
 3 years ago
Best ResponseYou've already chosen the best response.2the method I have shown you is probably one of two methods you will learn to solve these. I think it is often called "substitution". The method you described is much more powerful for larger systems of equations. If you take a course on linear algebra you will become very familier with it. I think in high school algebra they call the method "subtraction". You might find this method less confusing and I can explain that as well.

richyw
 3 years ago
Best ResponseYou've already chosen the best response.2most people use "substitution" for systems of equations 3x3 or smaller. and use "subtraction" for everything bigger than that. would you like me to show you how to solve this one using subtraction?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes please! and thank you

richyw
 3 years ago
Best ResponseYou've already chosen the best response.2alright. i'm just going to to it by hand and then take a pic to save time.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Alright, thank you very much.

richyw
 3 years ago
Best ResponseYou've already chosen the best response.2sorry I got a phone call. I'll actually just type this out.

richyw
 3 years ago
Best ResponseYou've already chosen the best response.2so for the subtraction method what you want to do is subtract one from the other. Now this is only going to help you if one of the variables is going to cancel out!. If they will not, you must multiply each side of the equation by a constant so that they do. but let's look at the first two equations. x+yz=1 x+y+z=3 ok so this is good, because we can immediately eliminate two of the variables by subtracting. So let's do that! x+yz=1 x+y+z=3  0+02z=4 now consider the first equation and the third x+yz=1 3x2yz=4 ok so here subtraction is not going to eliminate one of the variables. so we need to multiply one of these equations by something so that it does. I'll multiply the first equation by 3 (don't forget to do this on both sides of the = sign!) 3(x+yz)=3(1) so this gives us 3x+3y3z=3 3x2yz=4 ok now subtract again 3x+3y3z=3 3x2yz=4  0+5y2z=1 and we already know what 2 is, so this becomes 5y2(2)=1 and therefore y=1. Now you can solve for x really easy.

richyw
 3 years ago
Best ResponseYou've already chosen the best response.2so that was perhaps a bad example to show the subtraction method. There are also some problems you can run into when using this method if you don't use all of the equations (you will end up with statements like y=y, which is useless). There are also cases where the system has no solution. I don't want to confuse you with this, but if you would like to learn the best way to solve this using the subtraction method, I would encourage you you look up how to do it using an augmented matrix.

richyw
 3 years ago
Best ResponseYou've already chosen the best response.2haha that was way to much typing to explain this concept! some things are just better explained in person I guess. watch this video. http://www.youtube.com/watch?v=woqq3Sls1d8 it will clear up everything.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you SO very much for all of your help!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You started to do the addition/elimination method. I will show you that... x + y  z = 1 (equation 1) x + y + z = 3 (equation 2) 3x  2y  z =  4 (equation 3) Take the first 2 equations and add them trying to eliminate a variable. The best variable to eliminate in these two equations is z because they already cancel each other out. x + y  z =  1 x + y + z = 3  2x + 2y = 2 <===answer to first two equations Now take the last two equations and eliminate z ...you have to eliminate the same variable as you did in the first two equations. x + y + z = 3 3x  2y  z =  4  4x  y = 1 <===answer to last two equations Now take the answers from the previous equations and eliminate a variable. 2x + 2y = 2 > (2)2x + 2y = 2 4x  y =  1  I multiplied 1st equation by 2 to cancel out the x's 4x  4y =  4 (result of multiplying by a 2 4x  y =  1   5y =  5 y = 1 Now take either one of the equations you just used and substitute 1 in for y 2x + 2y = 2 2x + 2(1) = 2 2x + 2 = 2 2x = 22 2x = 0 x = 0/2 = 0 now that we know y = 1 and x = 0, we can substitute these in to any of the 1st three equations... x + y + z = 3 0 + 1 + z = 3 1 + z = 3 z = 3  1 z = 2 Now we can check with any of the 1st three equations... x + y  z =  1 0 + 1  2 = 1 1 = 1 (correct) check.. x + y + z = 3 0 + 1 + 2 = 3 3 = 3 (correct) check... 3x  2y  z =  4 3(0)  2(1)  2 = 4 0  2  2 =  4  4 =  4 (correct) I checked with all 3 equations and the answer is... x = 0, y = 1, z = 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you so much kelliegirl33!!
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