Question

Solve the following equations:
$x^2+xy+xz=18$,
$y^2+yz+yz+12=0$
$z^2+zx+zy=30$.

• A. $x=\pm 2,y=\mp 2,z=\pm 4$
• B. $x=\pm 3,y=\mp 2,z=\pm 5$
• C. $x=\pm 4,y=\pm 5,z=\pm 5$
• D. None of the above

Answer 1

The correct option is B $x=\pm 3,y=\mp 2,z=\pm 5$

$x^2+xy+yz=18$

$\Rightarrow x(x+y+z)=18$ ........(i)

$y^2+yz+yx+12=0$

$\Rightarrow y(x+y+z)=-12$ .......(ii)

$z^2+zx+zy=30$

$\Rightarrow z(x+y+z)=30$ ........(iii)

Dividing (i) by (ii)

$\Rightarrow \frac{x}{y}=-\frac{18}{12}=-\frac{3}{2}$

$y=-\frac{2}{3}x$ .......(a)

Dividing (i) by(iii)

$\frac{x}{z}=\frac{18}{30}=\frac{3}{5}$

$z=\frac{5}{3}x$ ...........(b)

Substituting (a) and (b) in (i)

$x(x-\frac{2}{3}x+\frac{5}{3}x)=18$

$\Rightarrow \frac{6x^2}{3}=18$

$\Rightarrow x^2=9\Rightarrow x=\pm 3$

Substituting $x$ in (a), we get

$\Rightarrow y=-\frac{2}{3}(\pm 3)=\mp 2$

Substituting $x$ in (b), we get

$\Rightarrow z=\frac{5}{3}(\pm 3)=\pm 5$

So, $x=\pm 3,y=\mp 2,z=\pm 5$