Solve the following system of equations:

$\frac{4}{x}\ +\ 3y\ =\ 8$
$\frac{6}{x}\ −\ 4y\ =\ −5$

Given:

The given system of equations is:

$\frac{4}{x}\ +\ 3y\ =\ 8$

$\frac{6}{x}\ −\ 4y\ =\ −5$

To do:

We have to solve the given system of equations.

Solution:

The given system of equations can be written as,

$\frac{4}{x}+3y=8$

Let $\frac{1}{x}=k$,

$\Rightarrow 4k+3y=8$---(i)

$\frac{6}{x}-4y=-5$

$\Rightarrow 6k-4y=-5$

$\Rightarrow 6k=4y-5$

$\Rightarrow k=\frac{4y-5}{6}$----(ii)

Substitute $k=\frac{4y-5}{6}$ in equation (i), we get,

$4(\frac{4y-5}{6})+3y=8$

$\frac{2(4y-5)}{3}+3y=8$ 

Multiplying by $3$ on both sides, we get,

$3(\frac{8y-10}{3})+3(3y)=3(8)$

$8y-10+9y=24$

$17y=24+10$

$17y=34$

$y=\frac{34}{17}$

$y=2$

Substituting the value of $y=2$ in equation (ii), we get,

$k=\frac{4(2)-5}{6}$

$k=\frac{8-5}{6}$

$k=\frac{3}{6}$

$k=\frac{1}{2}$

This implies,

$x=\frac{1}{k}=\frac{1}{\frac{1}{2}}$

$x=2$

Therefore, the solution of the given system of equations is $x=2$ and $y=2$.

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Updated on: 10-Oct-2022

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