What is the solution to the following system?
a
x+y+z= 6
X-Y+Z= 8
X+y-Z=0

x= 4

Y= -1

Z= 3

step by step procedure:

let x+y+z=6, x-y+z=8, x+y-z=0 be equations 1, 2, and 3 respectively.

subtract equation 2 from equation 1 ; x+y+z-(x-y+z) = 6 - 8 y+y = -2 , 2y = -2, therefore y = -1substituting the value of y into equations 1, 2, and 3 . we get x+z=7---eqn 4, x+z=7---eqn 5 and x-z=1---eqn 6now add equation 6 to equation 5 ; x+z+(x-z)=7+1 ,x+x=7+1, 2x=8, therefore x=4put the value of x (x=4) back into equation 5; 4+z=7, z=7-4, therefore z=3.

you are welcome bro

The 3rd quadrant as 3x+g5ti=56i: 92d-543