3x+2x+z=6
x+y+3z=-5
4x+y-z=10
3x+2x+z=6
x+y+3z=-5
4x+y-z=10
Given: (1) 3x + 2y + z = 6
(2) x + y + 3z = -5
(3) 4x + y - z = 10
Since you are only looking to solve for x in this system of equations, the first step is to eliminate either of the other two variables. Let's try to eliminate y first by multiplying (2) by -2 and adding it to (1):
-2(x + y + 3z = -5) ==> -2x - 2y - 6z = 10
Add to (1):
-2x - 2y - 6z = 10
+ 3x + 2y + z = 6
____________________
x - 5z = 16
Now multiply the (3) by -2 and add it to (1):
-2(4x + y - z = 10) ==> -8x - 2y + 2z = -20
-8x - 2y + 2z = -20
+ 3x + 2y + z = 6
___________________
-5x + 3z = -14
Now we take the system of the two new equations we created and eliminate the z variable to solve for x:
(1*) x - 5z = 16
(2*) -5x + 3x = -14
To eliminate z, first multiply (1*) by 3 and multiply (2*) by 5:
(1*) 3(x - 5z = 16) ==> 3x - 15z = 48
(2*) 5(-5x + 3z = -14) ==> -25x + 15z = -70
Solve for x by combining these two equations:
3x - 15z = 48
+ -25x + 15z = -70
_____________________
-22x = -22
==> (-22x)/-22 = (-22)/-22
x = 1
3x+2y+z=6 ...... (1) <==Is this what you meant?
x+y+3z=-5 ....... (2)
4x+y-z=10 ...... (3)
(3)-(2): 3x - 4z = 15 ....... (4)
(1)-2*(2): x-5z = 16 ......(5)
(4)-3*(5): 11z = -33
z = -3
x = 16+5z = 1
y = z-4x = -1-4 = -5
Answer: (1, -5, -3)