Question

Solve: $x+2y+z=7$,
$x+3z=11$ and
$2x-3y=1$.

Answer 1

We have,
$x+2y+z=7$ .(i)

$x+3z=11$ .(ii)

$2x-3y=1$ ..(iii)

From equation (i), we get

$z=7-x-2y$

Substituting $z=7-x-2y$ in equation (ii), we get

$x+3(7-x-2y)=11$

$\Rightarrow x+21-3x-6y=11$

$\Rightarrow -2x-6y=-10$ ..(iv)

Adding equations (iii) and (iv), we get
$-9y=-9\Rightarrow y=1$

Putting $y=1$ in equation (iii), we get $x=2$.

Putting $x=2,y=1$ in equation (i), we get

$2+2+z+7\Rightarrow z=3$

Hence, $x=2,y=1,z=3$.