Find the solution set of the systemx + 2y + z = 1;2x – 3y – w = 2 ;x ≥ 0 ; y ≥ 0 ; z ≥ 0 ; w≥ 0.

Sol. The given system is
x + 2y + z = 1 . . . . . . . . . . . . (1)
2x – 3y – ω ≥ 0
Multiplying eqn. (1) by (2) and subtracting from (2), we get
7y + 2z + ω = 0 ⇒ ω = - (7y + 2z)
Now if y, z > 0, ω < 0(not possible)
If y = 0, z = 0 then x = 1 and ω = 0.
∴ The only solution is x = 1, y = 0, z = 0, ω = 0.