Question

solve by elimination method: x-3y+4z-2w=5 ; 2y+5z+w=2 ; y-3z=4

Answer 1

The 'elimination method means you get rid of variable until you have an equation you can solve.
First take this equation
y-3z=4
and rearrange it so it is y=??

Post you answer here

Answer 2

Hmm - I see a problem tho.

There are 4 variable here w,x,y &z

But you only have 3 equations.

I think you need at least as many independent equations as there are variable before you can solve a system like this.

Please check your question...

Answer 3

the question is correct, we just have to find the general solution since the system is 3 by 4 it will yield infinitely many solutions, so my problem is that i dont know which variables to assign new variables when doing the general solution.
Find the general sol now! Using the triangular system i got...
X-3y+4z-2w=5 ; 2y+5z+w=2 ; -11z-w=6

Answer 4

Not sure what you mean by a general solution.

Answer 5

Ok, thanks for trying hey

Answer 6

you have 1 free variable, right? let say it is z, and all others are depends on this free
as solving, you have:
\[\begin{cases} x-3y+4z-2w=5\\2y+5z+w=2\\y-3z=4\end{cases}\]
from the third one you have y = 4+3z
replace it to the second one you have 2(4+3z)+5z+w= 2 it gives you w =-6-11z
from the first one, you find x = 5+3y-4z+2w, plug all value of y, w in (which in term of z)
you have x = 5+3(4+3z)-4z+2(-6-11z) = 5-17z
Now, combine all, your solution is under the form
\[\left[\begin{matrix}x\\y\\z\\w\end{matrix}\right] =\left[\begin{matrix}5-17z\\4+3z\\z\\-6-11z\end{matrix}\right] =\left[\begin{matrix}5\\4\\0\\-6\end{matrix}\right] + z\left[\begin{matrix}-17\\3\\1\\-11\end{matrix}\right] \]
That is general form of the solution.
You can play with other free variable, let say you choose the free is y, then do the same, you will have another form of general solution.