Question

solve the system
2x-2y+z=-1
x+3y+4z=11
3x+0y+2z=5

Answer 1

Eliminate the z variable so that you'll be left with a system of two equations in variables x and y.

Answer 2

You can do this easily by pairing the 1st and 3rd equations
2x-2y+z=-1
3x+0y+2z=5

Then by pairing the 2nd and 3rd equations:
x+3y+4z=11
3x+0y+2z=5

Now, working with the 1st and 3rd equations, Multiply the 1st equation by 2, then subtract:

4x - 4y + 2z = -2
3x +0y + 2z = 5

After subtraction you get:
x - 4y = -7

Working with the 2nd and 3rd equations, Multiply the 3rd equation by 2 then subtract:

x + 3y + 4z = 11
6x + 0y + 4z = 10

After subtraction you get:
-5x + 3y = 1

You're left with a system of two equations:
x - 4y = -7
-5x + 3y = 1

Which you might be able to solve.

Answer 3

For the system of 2 equations, you can multiply the first equation by 5 to get:
5x - 20y = -35
-5x + 3y = 1

Combining both equations together you get
-17y = -34
34 = 17y

Dividing both sides by 17 to get

2 = y

From there, you can do back substitution to find x and z.

Answer 4

The easiest way to solve this, however, is with matrices.

Answer 5

x - 4y = -7 means x = 4y - 7
-5x + 3y = 1 means x = (1 - 3y) / -5

Equal both resulting equations,
4y - 7 = 3/5y - 1/5

Rearrange, and y = 2.

If y = 2, x = 4(2) - 7 = 1.

Thefore,
y = 2
x = 1

Now, z, since 3x + 2z = 5,
3(1)+2z = 5
2z = 2, thefore z = 1.

x = 1, y = 2, and z = 1.

Answer 6

(I continued from @Hero 's first response)
And he is very right about using matrices to make this easier.