Solve the following systems of equations:

Solution:

The given equations are:

$x-y+z=4 \ldots(i)$

$x+y+z=2 \quad \ldots($ ii $)$

$2 x+y-3 z=0 \quad \ldots($ iii $)$

First of all we find the value of $x$

$x=4+y-z$

Put the value of $x$ in equation $(i)$, we get

$4+y-z+y+z=2$

$\Rightarrow 2 y=-2$

 

$\Rightarrow y=-1$

Put the value of $x$ and $y$ in equation in $(i i i)$ we get

$2(4+y-z)+y-3 z=0$

$\Rightarrow 8-2-2 z-1-3 z=0$

$\Rightarrow-5 z=-5$

 

$\Rightarrow z=1$

Put the value of $y$ and $z$ in equation $(i)$, we get

$x-(-1)+1=4$

 

$\Rightarrow x=2$

Hence the value of $x=2, y=-1$ and $z=1$.