Question

Solve the following systems of equations.
$2x+3y-z=6,x-y+7z=8,3x-y+2z=7$

Answer 1

Given equations

$2x+3y-z=6$ $\to \left(1\right)$

$x-y+7z=8$ $\to \left(2\right)$

$3x-y+2z=7$ $\to \left(3\right)$

$\left(1\right)+3\times \left(2\right)$

$2x+3y-z+3x-3y+21z=6+24$

$5x+20z=30$

$x+4z=6$ $\to \left(4\right)$

$\left(3\right)-\left(2\right)$

$2x-5z=-1$ $\to \left(5\right)$

$2\times \left(4\right)-\left(5\right)$

$2x+8z-2x+5z=12=1$

$13z=13$

$z=1$

$x=6-4z$

$x=2$

$\Rightarrow$ $2\left(2\right)+3y-1=6$

$y=1$

$\therefore$ $x=2,y=1,z=1$ is solution of given equations.