Question

$x-2y+z=0$
$y-z=2$
$2x-3z=10$ Find the value of $x,y,z$

• A. $x=1;y=0;z=-2$
• B. $x=1;y=0;z=1$
• C. $x=1;y=1;z=2$
• D. $x=2;y=0;z=-2$

Answer 1

Answer: (D)

$x=2;y=0;z=-2$

Given, $x-2y+z=0\to \left(1\right)$

$y-z=2\to \left(2\right)$

$2x-3z=10\to \left(3\right)$

From equation $2$

$y=2+z$

Substitute this 'y' value in equation $\left(1\right)$

$x-2y+z=0\to 1$

$x-2(2+z)+z=0$

$x-4-2z+z=0$

$x-4-z=0$

$x-z=4\to \left(4\right)$

Now from equation $\left(3\right)$ and equation $\left(4\right)$

$2x-3z=10\to \left(3\right)$

$x-z=4\to \left(4\right)$

Multiply $\left(3\right)$ by $1$ and $\left(4\right)$ by $2$ and subtract both equations

$2x-3z=10\to 3$

$2x-2z=8\to 4$

$-+-$

$-z=2$

$z=-2$

Substitute 'z' value in equation $\left(3\right)$

$2x-3(-2)=10$

$x=2$

Substitute 'z' value in equation $\left(2\right)$

We get

$y-(-2)=2$

$y=0$

$\therefore$By solving the given three equations we get

$x=2;y=0;z=-2$