Question 186539
2x+2y+3z=10		(1)
3x+y-z=0			(2)
x+y+2z=6			(3)

Multiply eqn (2) by 2 and subtract from eqn(1)
2x+2y+3z=10		(1)		
6x+2y - 2z=0		(2) x 2

- 4x + 5z = 10		(4)        (1) – (2) x 2
Multiply eqn (3) by 2 and subtract from eqn(1)

2x+2y+3z=10		(1)
2x+2y+4z=12		(5)         (3) x 2

-	z = - 2
So z = 2
Plugging in the value of z in eqn(4)
-	4x + 5(2) = 10;  -4x = 10-10 =0; so x = 0
Now plug-in the values of x (=0) ad z (=2) in eqn (3)

x + y + 2z = 6   		(3)
0 + y +2(2) =6
y + 4 =6;  so y = 6-4 = 2

So the answers are x =0,  y=2,   z = 2


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