Need Help With Solving Equations with Three Variables

quaddingqueen04

New member
Joined
Oct 23, 2006
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5
An example of what I am doing would be this..

2x-y+2z=15
-x+y+z=3
3x-y+2z=18

My teacher had said that there are no fractions in the answer but I can't figure it out at all, and I have 8 of these problems. I have to solve for (x,y,z) with again no fractions. Please HELP!
 
I know that I can do this using the first two equations..

2x-y+2z=15
(2)-x+(2)y+(2)z=(2)3

So then it would be..

2x-y+2z=15
-2x+2y+2z=6

After you can cancle out the 2x and -2x left with this..

-y+2z=15
2y+2z=6
 
Then it would be...

y+4z=21

right?

So..

y=-4z+21


Then I have to plug it back into the other equation so with y in it so I can find out z or x?
 
Hello, quaddingqueen04!

\(\displaystyle \L >>\;Solve:\;\;\
\left\{ \begin{array}{l}
{\rm [1]2x - y + 2z = 15} \\
{\rm [2] - x + y + z = 3} \\
{\rm [3]3x - y + 2z = 18} \\
\end{array} \right.\)

Look for steps that will cancel terms out:

Multiply [1] by 2:\(\displaystyle \L \;4x\,-\,4y\,+\,4z=\,30\)

Multiply [2] by 4:\(\displaystyle \L \;\,-4x\,+\,4y\,+\,4z\,=\,12\)

Add [1] & [2]:\(\displaystyle \L \;8z\,=\,42\;\;z\,=\,\frac{21}{4}\)

What's this no fraction business? :shock:
 
I am not really sure, in all of the problems that she gave us on the worksheet, she said there are no fractions at all, but I don't understand how that can be. But she also said that you have to use two equations at a time and substitute and elimitate.
 
quaddingqueen04 said:
I am not really sure, in all of the problems that she gave us on the worksheet, she said there are no fractions at all, but I don't understand how that can be. But she also said that you have to use two equations at a time and substitute and elimitate.

Well if a system is consistent, no matter what way you solve it, it should get the same answer.
 
Yeah, I understand that much. Well thank you for the help, and I guess I will just do it to where I get fractions and deal with her tomorrow. Thanks again.
 
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