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Solve the following system of equations:
$\frac{1}{(2x)}\ +\ \frac{1}{(3y)}\ =\ 2$
$\frac{1}{(3x)}\ +\ \frac{1}{(2y)}\ =\ \frac{13}{6}$
Given:
The given system of equations is:
$\frac{1}{(2x)}\ +\ \frac{1}{(3y)}\ =\ 2$
$\frac{1}{(3x)}\ +\ \frac{1}{(2y)}\ =\ \frac{13}{6}$
To do:
We have to solve the given system of equations.
Solution:
Let $\frac{1}{x}=u$ and $\frac{1}{y}=v$
This implies,
The given system of equations can be written as,
$\frac{1}{(2x)}\ +\ \frac{1}{(3y)}\ =\ 2$
$\frac{u}{2}+\frac{v}{3}=2$
$\frac{3u+2v}{6}=2$
$3u+2v=6(2)$
$3u+2v=12$-----(i)
$\frac{1}{(3x)}\ +\ \frac{1}{(2y)}\ =\ \frac{13}{6}$
$\frac{u}{3}+\frac{v}{2}=\frac{13}{6}$
$\frac{2u+3v}{6}=\frac{13}{6}$
$2u+3v=13$
$2u=13-3v$
$u=\frac{13-3v}{2}$
Substitute $u=\frac{13-3v}{2}$ in equation (i), we get,
$3(\frac{13-3v}{2})+2v=12$
Multiplying both sides by $2$, we get,
$2(\frac{3(13-3v)}{2})+2(2v)=2(12)$
$39-9v+4v=24$
$-5v=24-39$
$-5v=-15$
$v=\frac{-15}{-5}$
$v=3$
This implies,
$u=\frac{13-3(3)}{2}$
$u=\frac{13-9}{2}$
$u=\frac{4}{2}$
$u=2$
$x=\frac{1}{u}=\frac{1}{2}$
$y=\frac{1}{v}=\frac{1}{3}$
Therefore, the solution of the given system of equations is $x=\frac{1}{2}$ and $y=\frac{1}{3}$.