What is the solution of the system of equations?
3x+2y+z=7
5x+5y+4z=3
3x+2y+3z=1

The answer is x = 4
y = -1
z = -3

3x+2y+z=7
5x+5y+4z=3
3x+2y+3z=1

Let's take the first and the third equation, multiply the first one by (-1) and sum them up:
3x+2y+z=7
3x+2y+3z=1

-3x-2y-z=-7
3x+2y+3z=1

2z = -6
z = -6/2
z = -3

Substitute z in all equations:
3x+2y+z=7
5x+5y+4z=3
3x+2y+3z=1

3x + 2y - 3 = 7
5x + 5y - 12 = 3
3x + 2y - 9 = 1

3x + 2y = 10
5x + 5y = 15
3x + 2y = 10

The first and the third equations are the same, so one can be excluded:
3x + 2y = 10
5x + 5y = 15

Multiply the first equation by -5 and the second equation by 2 and sum them up:
-15x - 10y = -50
10x + 10y = 30

-5x = -20
x = -20/-5
x = 4

Substitute x:
3x + 2y = 10
12 + 2y = 10
2y = 10 - 12
2y = -2
y = -2/2
y = -1

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