Question

Solve the following equations:
$9x+y-8z=0$,
$4x-8y+7z=0$,
$yz+zx+xy=47$.

• A. $x=\pm 3;y=\pm 5;z=\pm 4$
• B. $x=\pm 2;y=\pm 4;z=\pm 1$
• C. $x=x=\pm 4;y=3;z=\pm 5$
• D. None of these

Answer 1

The correct option is A $x=\pm 3;y=\pm 5;z=\pm 4$

Let $9x+y-8z=0$ .......(i)

$4x-8y+7z=0$ .......(ii)

$yz+zx+xy=47$ .......(iii)

Solving (i) and (ii) by cross multiplication

$\frac{x}{7-64}=\frac{y}{-32-63}=\frac{z}{-72-4}=k\Rightarrow \frac{x}{-57}=\frac{y}{-95}=\frac{z}{-76}=k\Rightarrow x=-57k,y=-95k,z=-76k$ .........(iv)

Substituting $x,y$ and $z$ in (iii), we get

$(-95k)(-76k)+(-76k)(-57k)+(-57k)(-95k)=47\Rightarrow 16967k^2=47\Rightarrow k^2=\frac{47}{16967}=\frac{1}{361}\Rightarrow k=\pm \frac{1}{19}$

Substituting $k$ in (iv), we get

$\Rightarrow x=\pm 3,y=\pm 5,z=\pm 4$

which is the required solution.