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 $y=-1$.

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Updated on: 10-Oct-2022

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