Question

Solve the following equation :
$7yz+3zx=4xy,21yz-3zx=4xy,x+2y+3z=19.$

Answer 1

$4xy-7yz-3zx=\mathrm{0.....}\left(i\right)4xy-21yz+3zx=\mathrm{0....}\left(ii\right)$

Adding $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow \frac{x}{z}=\frac{7}{2}\Rightarrow \frac{x}{7}=\frac{z}{2}....\left(iii\right)$

subtracting $\left(ii\right)$ from $\left(i\right)$

$\Rightarrow 14yz=6zx\Rightarrow \frac{x}{7}=\frac{y}{3}....\left(iv\right)$

From $\left(iii\right)$ and $\left(iv\right)$

$\Rightarrow \frac{x}{7}=\frac{y}{3}=\frac{z}{2}=k\Rightarrow x=7k,y=3k,z=2k$

Third equation is $x+2y+3z=19$

substituing $x,y$ and $z$

$\Rightarrow 7k+6k+6k=19\Rightarrow k=1\Rightarrow x=7,y=3,z=2$