system of equation with 3 equations using elimination

## x+4y-z=63x+2y+3z=162x-y+z=3

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# 1 Answer

I will restate your equations with spacing that is a bit easier to see

x + 4y -z = 6 first equation

3x +2y + 3z = 16 second equation

2x -y +z = 3 third equation

The first step is to add the third equation to first one this gives

3x +3y = 9 This eliminates z, so we look to do this again.

So triple the first equation and add to it the second equation. This gives

6x +14 y = 34 Now we have two new equations involving only x and y

Double the first new equation and subtract it from the second new equation. This gives

8y = 16 So y = 2. Substitute y = 2 into the first new equation to get x = 1

The substitute x = 1 and y = 2 into any of the original equations to get z = 3.