Question 33348
x-3y+2z =-11 --- eq 1
2x-4y+3z =-15 --- eq 2
3x-5y-4z =5  --- eq 3

from eq 1 we get x = 3y - 2z -11
substituting the value of x in eq 2
2(3y - 2z -11) - 4y +3z = -15
6y - 4z -22 -4y +3z = -15
2y - z = 22 - 15 = 7 
2y - z = 7 ----eq 4
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substitute value of x again in eq 3
3(3y - 2z -11)-5y-4z = 5
9y - 6z -33-5y-4z = 5
4y-10z = 33 + 5 = 38
4y-10z = 38 --- eq 5
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2y - z = 7 ----eq 4
4y-10z = 38 --- eq 5
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multiply entire eq 4 with 2 
4y - 2z = 14  --- eq 4  (after multipling the entire equation with 4)

subtracting eq 5 and eq 4

4y-10z = 38
4y - 2z = 14
(- )  (+ )   (-)      changing signs
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  -8z = 24
z = -(24/8)= -3

substituting the value of Z in eq 4
2y - z = 7 
2y -(-3)=7
2y +3 = 7
2y = 7-3 = 4
y = 4/2 = 2

substitue the values of y and z in : x = 3y - 2z -11
x = 3(2)-2(-3)-11
x = 6 +6-11 = 12 -11 = 1

x = 1 ; y = 2 ; z = -3