I:2x – y + z = 7
II:x + 2y – 5z = -1
III:x – y = 6
you can first use III and substitute x or y to eliminate it in I and II (in this case x):
III: x=6+y
-> substitute x in I and II:
I': 2*(6+y)-y+z=7
12+2y-y+z=7
y+z=-5
II':(6+y)+2y-5z=-1
3y+6-5z=-1
3y-5z=-7
then you can subtract II' from 3*I' to eliminate y:
3*I'=3y+3z=-15
3*I'-II':
3y+3z-(3y-5z)=-15-(-7)
8z=-8
z=-1
insert z in II' to calculate y:
3y-5z=-7
3y+5=-7
3y=-12
y=-4
insert y into III to calculate x:
x-(-4)=6
x+4=6
x=2
so the solution is
x=2
y=-4
z=-1
Answer:
Step-by-step explanation:
10
Sent a picture of the solution to the problem (s).
The answer is C, one over thirty-two
