# How do you solve 5/x + 3/y = 4 and 25/x - 2/y = 3?

Aug 23, 2015

$\left\{\begin{matrix}x = 5 \\ y = 1\end{matrix}\right.$

#### Explanation:

Right from the start, you know that $x$ and $y$ cannot be equal to zero, since they represent denominators.

Now, start by rewriting your equations to get rid of the fractions. The common denominator will be $x \cdot y$, so multiply each term accordingly. For the first equation you will get

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

$5 y + 3 x = 4 x y$

For the second equation you will get

$\frac{25 y}{x y} - \frac{2 x}{x y} = \frac{3 x y}{x y}$

$25 y - 2 x = 3 x y$

Multiply the first equation by $- 5$ to get

$5 y + 3 x = 4 x y | \cdot \left(- 5\right)$

$- 25 y - 15 y = - 20 x y$

$\left\{\begin{matrix}- 25 y - 15 x = - 20 x y \\ 25 y - 2 x = 3 x y\end{matrix}\right.$

Notice that if you were to add these two equations together, the $y$-terms on the left sides of the equations would cancel out,

You will be left with

$- \textcolor{red}{\cancel{\textcolor{b l a c k}{25 y}}} - 15 x + \textcolor{red}{\cancel{\textcolor{b l a c k}{25 y}}} - 2 x = - 20 x y + 3 x y$

$- 17 x = - 17 x y$

You can further simplify this to get the value of $y$

$\textcolor{red}{\cancel{\textcolor{b l a c k}{- 17 x}}} = \textcolor{red}{\cancel{\textcolor{b l a c k}{- 17 x}}} \cdot y \implies y = \textcolor{g r e e n}{1}$

Take this value of $y$ into one of the two equations and find the value of $x$

$5 \cdot 1 + 3 x = 4 \cdot x \cdot 1$

$4 x - 3 x = 5 \implies x = \textcolor{g r e e n}{5}$

Since you have $x \ne 0$ and $y \ne 0$, the two solutions to your system of equations will be

$\left\{\begin{matrix}x = 5 \\ y = 1\end{matrix}\right.$

You can do a quick check to make sure that the calculations are correct

$\frac{5}{5} + \frac{3}{1} = 4$

$1 + 3 = 4 \text{ } \textcolor{g r e e n}{\sqrt{}}$

and

$\frac{25}{5} - \frac{2}{1} = 3$

$5 - 2 = 3 \text{ } \textcolor{g r e e n}{\sqrt{}}$