# How do you solve the linear equation system (4/x)-(3/y)=1 & (6/x)-(3/y)=-4?

May 4, 2016

$\left(x , y\right) = \left(- \frac{2}{5} , - \frac{3}{11}\right)$

#### Explanation:

$4 x - 3 y = 1$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow 4 y - 3 x = x y$ (after multiplying everything by $x y$)

$\textcolor{w h i t e}{\text{XXX}} \rightarrow 4 y - x y = 3 x$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow y \left(4 - x\right) = 3 x$

[1]$\textcolor{w h i t e}{\text{XXX}} \rightarrow y = \frac{3 x}{4 - x}$

$\frac{6}{x} - \frac{3}{y} = - 4$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow 6 y - 3 x = - 4 x y$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow 6 y + 4 x y = 3 x$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow y \left(6 + 4 x\right) = 3 x$

[2]$\textcolor{w h i t e}{\text{XXX}} \rightarrow y = \frac{3 x}{6 + 4 x}$

Combining [1] and [2]
$\textcolor{w h i t e}{\text{XXX}} \frac{3 x}{4 - x} = y = \frac{3 x}{6 + 4 x}$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow 4 - x = 6 + 4 x$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow - 5 x = 2$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow x = - \frac{2}{5}$

Substituting $\left(- \frac{2}{5}\right)$ for $x$ in [1]
$\textcolor{w h i t e}{\text{XXX}} y = \frac{3 \cdot \left(- \frac{2}{5}\right)}{4 - \left(- \frac{2}{5}\right)}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{- \frac{6}{5}}{\frac{20 + 2}{5}}$

$\textcolor{w h i t e}{\text{XXXX}} = - \frac{6}{22} = - \frac{3}{11}$

(Substitute these values back into the original equations to verify).