Question

Solve for $x$ and $y$ if $3(2x+y)=7xy$ and $3(x+3y)=11x$.

Answer 1

$3\left(2x+y\right)=7xy$

$\Rightarrow$ $6x+3y=7xy$

$\Rightarrow$$\frac{6}{y}+\frac{3}{x}=7$

$\Rightarrow$$\frac{3}{x}+\frac{6}{y}=7$ ...... $\left(1\right)$

$\Rightarrow$$3\left(x+3y\right)=11x$

$\Rightarrow$$3x+9y=11$

$\Rightarrow$$8x=9y$

$\Rightarrow$$x=\frac{9y}{8}$

Putting $x=\frac{9y}{8}$ in $\left(1\right)$, we get

$\frac{3\times 8}{9y}+\frac{6}{y}=7$

$\frac{1}{y}\left(\frac{8}{3}+6\right)=7$

$7y=\frac{26}{3}$

$y=\frac{26}{21}$

From $\left(1\right)$, we get

$x=\frac{9y}{8}=\frac{9}{8}\times \frac{26}{21}$

So, $x=\frac{39}{28}$ and $y=\frac{26}{21}$