Just plot the values onto a normal number line
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
Step-by-step explanation:
Given points are (-2,5) and (3,-17)
to find midpoint, use this formula
( x1+x2/2 , y1+y2/2)
where points are a(x1,y1) and b(x2,y2)
coming to the question,
midpoint is :
[ (-2-3)/2 , (5-17)/2 ]
[ -5/2 , -12/2 ]
[ -2.5 , -6 ]
OPTION
IS CORRECT
Answer:
The ratio of sixth-grade students to fifth-grade students on the team was <u>7 : 8</u>.
Step-by-step explanation:
Given:
The girl's basketball team had 8 fifth-grade students and 7 sixth-grade students.
Now, to find the ratio of sixth-grade students to fifth-grade students on the team.
<em>Number of fifth-grade students = 8.</em>
<em>Number of sixth-grade students = 7.</em>
Now, to get the ratio of sixth-grade students to fifth-grade students on the team :


Therefore, the ratio of sixth-grade students to fifth-grade students on the team was 7 : 8.