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I need help solving w,x,y,z

w + x - 2y + z = 1
2w - x +  3y - 2z = 2
w + 3x - y + 4z = 7
3w + x - y + 2z = 5
Thank you! Though how do you work it out?

Solve them simultaneously. Honestly, it is very tedious but it's the only way i deem fit. Try subtracting 1) and 2)....multiply 1) by 2 and add to ..etc etc it's very tedious.

There are two general approaches.  You can solve it yourself, or you can use technology to do it for you.  If this is a "show your work" problem, then you're expected to use the first approach.

With in that, there are also two primary ways to proceed.

1. Substitution: Rewrite the first equation as w = 1 - x + 2y - z.  Substitute (1-x+2y-z) for each w in the remaining 3 equations and you've reduced the problem to 3 equations in 3 unknowns.  Solve that (by any method) and you have the x,y,z values needed to compute w from the formula above.

2. Elimination.  Take one equation with a w in it, preferably one where the coefficient is 1, and add or subtract multiples of that to each of the other three equations to make the w term vanish.  For the given system, I subtract 2, 1 and 3 times the first equation from the 2nd, 3rd and 4th equations, respectively, to get:

w + x - 2y + z = 1
-3x + 7y -4z = 0
2x + y + 3z = 6
-2x + 5y - z = 2

That leaves an equivalent system with only one equation that has a w term in it.  So w has been "eliminated".  Pick another equation and another variable and eliminate that.  I'd pick y in the 3rd equation or z in the 4th, since the coefficients of 1 and -1 are easy to work with, and no fraction will be created.  At least, not yet.

Eliminate all four variables in that way, and you'll be left with a system where each equation is a linear equation in just one variable, and is easily solved.

When the system is reduced to 3x3 or less, might want to use Cramer's rule.  That's especially easy on 2x2 systems.  It works for any size problem, but for 4x4 and larger, I find the other methods easier to use.

Make an augmented matrix:
|1  1 -2  1|1|
|2 -1 3 -2|2|
|1  3 -1  4|7|
|3  1 -1 2|5|

Multiply row 1 by -2 and add to row 2:
|1  1 -2  1|1|
|0 -3 -1 -4|0|
|1  3 -1  4|7|
|3  1 -1 2|5|

Multiply row 1 by -1 and add to row 3:
|1  1 -2  1|1|
|0 -3 -1 -4|0|
|0  2  1  3|6|
|3  1 -1 2|5|

Multiply row 1 by -3 and add to row 4:
|1  1 -2  1|1|
|0 -3 -1 -4|0|
|0  2  1  3|6|
|0  -2  5 -1|2|

Multiply row 2 by 2 and row 3 by 3 and put results in row 3:
|1  1 -2  1|1|
|0 -3 -1 -4|0|
|0  0  1  1|18|
|0  -2  5 -1|2|

Multiply row 2 by -2 and row 4 by 3 and store the sum in row 4:
|1  1 -2  1|1|
|0 -3 -1 -4|0|
|0  0  1  1|18|
|0  0  17 5|6|

Add row 3 to row 2 and store results in row 2:
|1  1 -2  1|1|
|0 -3 0 -3|18|
|0  0  1  1|18|
|0  0  17 5|6|

Divide row 2 by -3:
|1  1 -2  1|1|
|0  1 0 1|6|
|0  0  1  1|18|
|0  0  17 5|6|

Multiply row 3 by -17 and add to row 4:
|1  1 -2  1|1|
|0  1 0 1|6|
|0  0  1  1|18|
|0  0  0 -12|-300|

Divide row 4 by -12:
|1  1 -2  1|1|
|0  1 0 1|6|
|0  0  1  1|18|
|0  0  0 1|25|

Subtract row 4 from row 3:
|1  1 -2  1|1|
|0  1 0 1|6|
|0  0  1  0|-7|
|0  0  0 1|25|

Subtract row 4 from row 2:
|1  1 -2  1|1|
|0  1 0 0|-19|
|0  0 1 0|-7|
|0  0  0 1|25|

Subtract row 4 from row 1:
|1  1 -2  0|-24|
|0  1 0 0|-19|
|0  0 1 0|-7|
|0  0  0 1|25|

Multiply row 3 by -2 and add to row 1:
|1  1 0  0|-38|
|0  1 0 0|-19|
|0  0 1 0|-7|
|0  0  0 1|25|

Subtract row 2 from row 1:
|1  0 0  0|-19|
|0  1 0 0|-19|
|0  0 1 0|-7|
|0  0  0 1|25|

This means that w = -19, x = -19, y = -7, and z = 25