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Solve the following system of equations:
$\frac{4}{x}\ +\ 5y\ =\ 7$
$\frac{3}{x}\ +\ 4y\ =\ 5$
Given:
The given system of equations is:
$\frac{4}{x}\ +\ 5y\ =\ 7$
$\frac{3}{x}\ +\ 4y\ =\ 5$
To do:
We have to solve the given system of equations.
Solution:
The given system of equations can be written as,
$\frac{4}{x}+5y=7$
Let $\frac{1}{x}=k$,
$\Rightarrow 4k+5y=7$---(i)
$\frac{3}{x}+4y=5$
$\Rightarrow 3k+4y=5$
$\Rightarrow 3k=5-4y$
$\Rightarrow k=\frac{5-4y}{3}$----(ii)
Substitute $k=\frac{5-4y}{3}$ in equation (i), we get,
$4(\frac{5-4y}{3})+5y=7$
$\frac{4(5-4y)}{3}+5y=7$
Multiplying by $3$ on both sides, we get,
$3(\frac{20-16y}{3})+3(5y)=3(7)$
$20-16y+15y=21$
$-y=21-20$
$-y=1$
$y=-1$
Substituting the value of $y=-1$ in equation (ii), we get,
$k=\frac{5-4(-1)}{3}$
$k=\frac{5+4}{3}$
$k=\frac{9}{3}$
$k=3$
This implies,
$x=\frac{1}{k}=\frac{1}{3}$
Therefore, the solution of the given system of equations is $x=\frac{1}{3}$ and $y=-1$.