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 = - 1 substituting 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 6 now add equation 6 to equation 5; x+z + (x-z) = 7+1, x+x=7+1, 2x=8, therefore x=4 put the value of x (x=4) back into equation 5; 4+z=7, z=7-4, therefore z=3.

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