Question

Solve each system by substitution.
-4x+3y-2z=7
2x-2y+3z=15
-x+2y-2z=-6
Medal will be given

Answer 1

Ok

-4x+3y-2z=7
3y = 7 + 4x +2z
y = 7/3 + 4/3x + 2/3 z (equation 1)

substitute y to 2x-2y+3z=15, then

2x-2y+3z=15
2x - 2(7/3 + 4/3x + 2/3 z) + 3z = 15
2x - 14/3 -8/3x - 4/3 z + 3z = 15
2x - 8/3x - 4/3z + 3z = 15 + 14/3
-2/3x + 5/3z = 59/3

both side times 3, then

-2x + 5z = 59
2x = 5z - 59
x = 5/2z - 59/2 (equation 2)

substitute x to -x+2y-2z=-6, then

-x+2y-2z=-6
(5/2z - 59/2) + 2y - 2z = -6
5/2z - 59/2 + 2y - 2z = -6
5/2z + 2y - 2z = -6+ 59/2
5/2z + 2y - 2z = 47/2
5/2z -2z+ 2y = 47/2
z/2 + 2y = 47/2

both side times 2, then
z + 4y = 47
z = 47 - 4y (equation 3)

substitute this z (equation 3) to x (equation 2), then

x = 5/2z - 59/2
x = 5/2 (47 - 4y ) - 59/2
x = 235/2 - 10y - 59/2
x = 176/2 - 10y (equation 4)

OF COURSE

substitute this x (equation 4) and z (equation 3) to y (equation 1), then
y = 7/3 + 4/3x + 2/3 z
y = 7/3 + 4/3 (176/2 - 10y) + 2/3 (47 - 4y)
y = 7/3 + 352/3 - 40/3y + 94/3 - 8/3y
y + 40/3y + 8/3y = 7/3 + 352/3 + 94/3
y (1 + 40/3 + 8/3) = 7/3 + 352/3 + 94/3

both side times 3
y (3 + 40 + 8) = 7 + 352 + 94
y (51) = 453
y = 453 /51

Answer 2

from equation 4 we get
x = 176/2 - 10y
x = 176/2 - 10 (453 /51)
x = 176/2 - 4530 /51

Answer 3

from equation 3 we get
z = 47 - 4y

z = 47 - 4(453 /51)
z = 47 - 1812/51

Answer 4

Thanks helped a lot

Answer 5

you are welcome :)