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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

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First of all, Alba, the second equation must be 2x-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] 2x-y-3z=-1

[2] 2x-y+z=-9

[3] x+2y-4z=17

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

2x-y-3z=-1 is equivalent to 2x-y=3z - 1.

2x-y+z=-9 is equivalent to 2x-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.

3z - 1 = -z - 9

3z + z = 1 - 9

4z = -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, 2x-y+(-2)=-9 or 2x-y=-9+2 or 2x-y=-7, which I will label it [4].

In [3] when z = -2, x+2y-4(-2)=17 or x+2y+8=17 or x+2y=17-8 or x+2y=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...


+ -2x-4y=-18

-5y=-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.






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