Given : x+y+z=3 ...(i) 2x+2y–z=3...(ii) x+y–z=1...(iii) (i) + (ii) gives; 3(x+y)=6⇒x+y=2  ...(iv)(ii) – (iii) gives, x+y=2 ...(v) Since (iv) and (v) are same, the 3 given equations are not independent. Hence, in such cases we have infinite number of solutions.