3x+2x+z=6

x+y+3z=-5

4x+y-z=10

3x+2x+z=6

x+y+3z=-5

4x+y-z=10

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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)