Solve the following system of equations:
$\frac{4}{x}\ +\ 3y\ =\ 14$
$\frac{3}{x}\ –\ 4y\ =\ 23$

Given:

The given system of equations is:

$\frac{4}{x}\ +\ 3y\ =\ 14$

$\frac{3}{x}\ –\ 4y\ =\ 23$

To do:

We have to solve the given system of equations.

Solution:

The given system of equations can be written as,

$\frac{4}{x}+3y=14$

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

$\Rightarrow 4k+3y=14$---(i)

$\frac{3}{x}-4y=23$

$\Rightarrow 3k-4y=23$

$\Rightarrow 3k=4y+23$

$\Rightarrow k=\frac{4y+23}{3}$----(ii)

Substitute $k=\frac{4y+23}{3}$ in equation (i), we get,

$4(\frac{4y+23}{3})+3y=14$

$\frac{4(4y+23)}{3}+3y=14$ 

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

$3(\frac{16y+92}{3})+3(3y)=3(14)$

$16y+92+9y=42$

$25y=42-92$

$25y=-50$

$y=\frac{-50}{25}$

$y=-2$

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

$k=\frac{4(-2)+23}{3}$

$k=\frac{-8+23}{3}$

$k=\frac{15}{3}$

$k=5$

This implies,

$x=\frac{1}{k}=\frac{1}{5}$

Therefore, the solution of the given system of equations is $x=\frac{1}{5}$ and $y=-2$.

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Updated on: 10-Oct-2022

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