To find the x-intercepts algebraically, we let y = 0 y=0 y=0 in the equation and then solve for values of x. In the same manner, to find for y-intercepts algebraically, we let x = 0 x=0 x=0 in the equation and then solve for y.
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
Her scores would be 98, 82, and 81. Their sum would be 261 then you would divide it be 3 and it would give you 87 then to find the median line the numbers up from greatest to least and 82 would be in the middle. To find the range subtract 81 from 98 and it gives you 17. Did this help?
It has to be an equation that adds up to an exponent of 3
