I am trying to figure out how to solve system of equations in 3 variables and the equations are..2x-y-3z=-1,2x-y+2=-9,x+2y-4z=17

First of all, Alba, the second equation must be 2*x*-*y*+*z*=^{-}9 knowing that all the equations must have three variables. Now, on with the procedure. First, it is preferable to label the three equations in the system.

[1] 2*x*-*y*-3*z*=^{-}1

[2] 2*x*-*y*+*z*=^{-}9

[3] *x*+2*y*-4*z*=17

Notice that equations [1] and [2] have the same two first terms. The idea here is to solve both equations for 2*x*-*y*.

2*x*-*y*-3*z*=^{-}1 is equivalent to 2*x*-*y*=3*z* - 1.

2*x*-*y*+*z*=^{-}9 is equivalent to 2*x*-*y*=^{-}*z
*- 9.

Since the left-hand side of both equivalent equations are equal, the right-hand side of both equations can be equalled so the value of
*z* can be found.

3*z* - 1 = ^{-}*z* - 9

3*z* + *z* = 1 - 9

4*z* = ^{-}8

*z* = ^{-}2

Now the value of *z* is found we have to find out what are the values of the
*x* and *y*. We'll consider equations [2] and [3]. To make this procedure easier, we'll substitute the value of
*z* in both equations.

In [2] when *z* = ^{-}2, 2*x*-*y*+(^{-}2)=^{-}9 or 2*x*-*y*=^{-}9+2 or 2*x*-*y*=^{-}7, which I will label it [4].

In [3] when *z* = ^{-}2, *x*+2*y*-4(^{-}2)=17 or
*x*+2*y*+8=17 or *x*+2*y*=17-8 or *x*+2*y*=9, which I will label it [5].

For the next step, I can eliminate any of the two variables in [4] and [5]. Randomly, I can eliminate
*x*. To eliminate *x*, multiply [5] by ^{-}2 and add it to [4].

[4]+(^{-}2)*[5] implies...

2*x*- *y*=^{-}7

+ -2*x*-4*y*=^{-}18

^{-}5*y*=^{-}25 or *y*=5

Having found the values of both *y* and *z*, use any of the three equations and substitute the corresponding values of both variables to find the value of
*x*. I will use [3] for the final step.

*x*+2(5)-4(^{-}2)=17

*x*+10+8=17

*x*+18=17

*x*=17-18

*x*=^{-}1

So the solution is (^{-}1, 5, ^{-}2).