Solve the following systems of equations:

Solution:

The given equations are:

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

$x-2 y-2 z=9 \ldots(i i)$

 

$2 x+y+3 z=1 \ldots(i i i)$

First of all we find the value of $x$

$x=4+y-z$

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

$4+y-z-2 y-2 z=9$

 

$\Rightarrow-3 z-y=5 \ldots(i v)$

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

$2(4+y-z)+y+3 z=1$

$\Rightarrow 8+2 y-2 z+y+3 z=1$

 

$\Rightarrow 3 y+z=-7 \quad \ldots(v)$

Multiply equation $(i v)$ by 3 and add equations $(i v)$ and $(v)$, we get

Put the value of $z$ in equation $(v)$, we get

$3 y-1=-7$

$\Rightarrow 3 y=-6$

 

$\Rightarrow y=-2$

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

$x-(-2)-1=4$

 

$\Rightarrow x=3$

Hence the value of $x=3$, $y=-2$ and $z=-1$.