Question

Solve the following equations by method of reduction:
$x-y+z=4,2x+y-3z=0,x+y+z=2$

Answer 1

$x-y+z=4$ ... (1)

$2x+y-3z=0$ ...(2)

$x+y+z=2$ ...(3)

From (1), we can say that,

$z=4+y-x$

Using this value of $z$ in (2) and (3), we get

$2x+y-3(4+y-x)=0\Rightarrow 5x-2y=12$ ...(4)

$x+y+(4+y-x)=2\Rightarrow y=-1$ .... (5)

From (5), we get $y=-1$

Substituting in (4), we get

$x=\frac{10}{5}=2$

Using these values on $x$ and $y$, we get

$z=4+y-x=4-1-2=1$

$\therefore (x,y,z)\equiv (2,-1,1)$