Question

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

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

Answer 1

The correct option is B $x=-\frac{1}{2},y=\frac{1}{2}andz=\frac{1}{2}$

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

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

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

Adding equations (1) and (3), we get

$y-z=0$............(5)

Adding equations (4) and (5), we get

$2y=1$

$\Rightarrow y=\frac{1}{2}$

Putting $y=\frac{1}{2}$ in equation 5, we get

$\frac{1}{2}-z=0$

$\Rightarrow z=\frac{1}{2}$

Putting $y=\frac{1}{2}andz=\frac{1}{2}$ in equation (1), we get

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

$\Rightarrow x=-\frac{1}{2}$

Thus, we have

$x=-\frac{1}{2},y=\frac{1}{2}andz=\frac{1}{2}$