Solve the given set of three equations to get the value

Solution

x + 2 z = 3
x+2y+3z=5 x + 2 y + 3 z = 5
3x−5z=−13 3 x − 5 z = − 13 From the first equation:
x=3−2z x = 3 − 2 z . Substitute x=3−2z x = 3 − 2 z  into the third equation:
3(3−2z)−5z=−13 3 ( 3 − 2 z ) − 5 z = − 13
9−6z−5z=−13 9 − 6 z − 5 z = − 13
−11z=−22 − 11 z = − 22
z=2 z = 2 . Now substitute z=2 z = 2  back into x=3−2z x = 3 − 2 z :
x=3−2(2)=−1 x = 3 − 2 ( 2 ) = − 1 . Finally, substitute x=−1 x = − 1  and z=2 z = 2  into the second equation:
−1+2y+3(2)=5 − 1 + 2 y + 3 ( 2 ) = 5
−1+2y+6=5 − 1 + 2 y + 6 = 5
2y=0 2 y = 0
y=0 y = 0 Hence, x = -1, y = 0 and z = 2