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:
The answer is 100% without a doubt A
Step-by-step explanation:
Answer:
1/6
Step-by-step explanation:
Hope it helps plz give brainliest! :D
There are 3 terms in this expression. You can determine this answer by knowing that terms are either a single number or variable or numbers and variables multiplied together. Here's a picture example below:
Hope it helps! Mark as brainliest!
Answer:
7:55 am
Step-by-step explanation:
To determine the time needed to leave, add up all of the elapsed time and subtract that from the 8:50 departure time. The 10 minutes from walking to the platform puts us at 8:40 am, 30 minutes to reach the station puts us at 8:10 am, and the extra 15 minutes gives a time of 7:55 am.
