Where is the pic of the angles to measure ? I would help
You
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
First you subtract 3x because you are trying to get x alone so subtract 3x on both side so you would get 2y=-3x-4 and then divide the whole thing by 2 so you would get y=-3-4/2 and then you simplify and get y=-7/2 so y=-7/2 is your answer.
the student needed to place a ten outside the radical, as a perfect square of 100 composes 200. This way, a two should be left under the radical sign. Also, this would be about 14.1 because 14^2 is 196 and 15^2 is 225. The correct answer must be less than 225, more than 196, but closer to 196. Therefore, a .1 is added to the 14