Answer:
x =-2 and z =-4
Explanation:
We need to solve the following systems of equation
-3x-2y+4z = -16 eq(1)
10x+10y-5z = 30 eq(2)
5x+7y+8z = -21 eq(3)
Multiply eq(1) with 10 and eq(2) with 3
-30x-20y+40z = -160
30x+30y-15z = 90
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10y+25z = -70
Divide by 10
2y+5z = -14 eq(4)
Multiply eq(1) with 10 and eq(3) with 6
-30x-20y+40z = -160
30x+42y+48z = -126
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22y+88z = -286
Divide by 11
2y+8z = -26 eq(5)
Subtract eq(4) and eq(5)
2y+5z = -14
2y+8z = -26
- - +
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-3z = 12
z = 12/-3
z = -4
Putting value of z in eq(4)
2y+5z = -14
2y +5(-4) = -14
2y = -14 +20
2y = 6
y = 3
Putting value of z and y in eq(1)
-3x-2y+4z = -16
-3x-2(3)+4(-4) = -16
-3x -6 -16 = -16
-3x = -16+16+6
-3x = 6
x = 6/-3
x = -2