Question

Find the value of $m$ for equarion $x+y=\frac{18}{m}$ if $x$ and $y$ satisfying following equations

$x+4y=27xy$

$x+2y=21xy$.Find the value of $m$

Answer 1

$x+4y=27xy$

$=>\frac{x+4y}{27}=xyx+2y=21xy$

$=>x+2y=21\left(\frac{x+4y}{27}\right)$

$=>x+2y=7\left(\frac{x+4y}{9}\right)$

$=>x+2y=\frac{7x+28y}{9}$

$=>9x+18y=7x+28y$

$=>2x=10y$

$=>x=5yx+4y=27xy$

$=>5y+4y=27\times 5y\times y$

$=>9y=27\times 5y^2$

$=>y=\frac{9}{27\times 5}=\frac{1}{15}x+4y=27xy$

$=>x+\frac{4}{15}=27x\times \frac{1}{15}$

$=>x+\frac{4}{15}=\frac{9x}{5}$

$=>15x+4=27x$

$=>-12x=-4$

$=>x=\frac{1}{3}x+y=\frac{1}{m}$

$=>\frac{1}{3}+\frac{1}{15}=\frac{1}{m}$

$=>\frac{1}{m}=\frac{18}{45}$

$=>m=45$