Question

how do you use eliminator method or substitution method for three linear questions?

like in this problem...
4x+3y+z=-5
2x-3y-z=23
5x+y+5z=0

Answer 1

add equations 1 and equation 2 to eliminate z and y

4x + 3x + z = -5
2x =3y - z = 23
--------------
6x = 18

then
x = 3

substitute into all equations and you have

12 + 3y + z = -5 3y + z = -17 (4)
6 - 3y - z = 23 -3y - z = 17 (5)
15 + y + 5z = 0 y + 5z = -15 (6)

Equation (5) + 3 x equation (6) to eliminate y and find z

-3y - z = 17
3y + 15z = -45
----------------
14z = - 28

z = -2

substitute x = 3 and z = -2 into the 1st equation to find y

12 + 3y - 2 = -5

3y = -15

y = -5

so the solution is x = 3, y = -5 and z = -2

hope this helps

Answer 2

thank you!