3x - 2y + z = 2 .....(1)
2x + 3y - z = 5.....(2)
x + y + z = 6 .....(3)
Substituting eqn(6) in eqn(4), we get,
5(1)+y=7[y=2]...........(7)
Substituting eqn(6) and eqn (7) in eqn(3), we get,
1+2+z=6[z=3]...........(8)
Ans:
[x = 1]
[y = 2]
[z = 3]
Verification:
Substituting the values of x,y and z in eqn(2), we get,
LHS = 2(1)+3(2)−3
= 2+6−3=5
= RHS