Question

Solve the following equations:
$x-y+z=0$
$x+2y-z=0$
$x+y-3z=1$

• A. $x=\frac{1}{8},y=\frac{-2}{4}andz=\frac{-3}{8}$
• B. $x=\frac{1}{8},y=\frac{-1}{4}andz=\frac{-3}{8}$
• C. $x=\frac{1}{8},y=\frac{-1}{5}andz=\frac{-3}{8}$
• D. $x=\frac{1}{7},y=\frac{-1}{4}andz=\frac{-3}{8}$

Answer 1

The correct option is B $x=\frac{1}{8},y=\frac{-1}{4}andz=\frac{-3}{8}$

We have
$x-y+z=0$........(1)
$x+2y-z=0$.........(2)
$x+y-3z=1$..........(3)

Adding equations (1) and (2), we get

$2x+y=0$............(4)

Multiplying equation (2) by '-3', we get

$-3x-6y+3z=0$..........(5)

Adding equations (3) and (5), we get

$-2x-5y=1$.........(6)

Adding equations (4) and (6), we get

$-4y=1$

$\Rightarrow y=\frac{-1}{4}$

Putting $y=\frac{-1}{4}$ in equation (4), we get

$2x-\frac{1}{4}=0$

$\Rightarrow 2x=\frac{1}{4}$
$\Rightarrow x=\frac{1}{8}$

Putting $x=\frac{1}{8}andy=\frac{-1}{4}$ in equation (1), we get

$\frac{1}{8}+\frac{1}{4}+z=0$

$\Rightarrow z=\frac{-1}{8}-\frac{1}{4}$

$\Rightarrow z=\frac{-3}{8}$

Thus, we have
$x=\frac{1}{8},y=\frac{-1}{4}andz=\frac{-3}{8}$