# How to solve for x and y if the equations are: (5/(x+y))-(2/(x-y))+1=0 and (15/(x+y))+(7/(x-y))-10=0 ?

Nov 12, 2017

$x = 3$ and $y = 2$

#### Explanation:

I'll use Gabriel Cramer's method to solve the system of equations.

First of all, let $x + y = p$ and $x - y = q$

Then the first equation is equal to

$\frac{5}{p} - \frac{2}{q} + 1 = 0$

or, $\frac{5}{p} - \frac{2}{q} = - 1$

LET ${a}_{1} = 5$, ${b}_{1} = - 2$, ${c}_{1} = - 1$

Also the second equation is equal to

$\frac{15}{p} + \frac{7}{q} - 10 = 0$

or, $\frac{15}{p} + \frac{7}{q} = 10$

LET ${a}_{2} = 15$, ${b}_{2} = 7$, ${c}_{2} = 10$

Now we have,

$\Delta = | \left({a}_{1} , {b}_{1}\right) , \left({a}_{2} , {b}_{2}\right) | = | \left(5 , - 2\right) , \left(15 , 7\right) | = 7 \cdot 5 - \left(- 2\right) \ast 15$

$= 35 + 30 = 65$

Also,

${\Delta}_{1} = | \left({c}_{1} , {b}_{1}\right) , \left({c}_{2} , {b}_{2}\right) | = | \left(- 1 , - 2\right) , \left(10 , 7\right) | = - 1 \cdot 7 - \left(- 2\right) \cdot 10$

$= 20 - 7 = 13$

Similarly, ${\Delta}_{2} = | \left({a}_{1} , {c}_{1}\right) , \left({a}_{2} , {c}_{2}\right) | = | \left(5 , - 1\right) , \left(15 , 10\right) | = 10 \cdot 50 - \left(- 1\right) \cdot 15$

$= 50 + 15 = 65$

According to Cramer's rule

$\frac{1}{p} = {\Delta}_{1} / \Delta$

or, $p = \frac{\Delta}{\Delta} _ 1 = \frac{65}{13} = 5$

Similarly , $\frac{1}{q} = {\Delta}_{2} / \Delta$

or $q = \frac{\Delta}{\Delta} _ 2 = \frac{65}{65} = 1$

Putting $p = x + y$ and $q = x - y$ we have

$x + y = 5$....equation(1)

and $x - y = 1$.....equation(2)

From equation(1) we have,

$x = 5 - y$....equation...(3)

Putting the value of $x$ from e$q n \left(3\right)$ to $e q n \left(2\right)$ we get

$5 - y - y = 1$

$2 y = 4$

$y = 2$

Putting the value of $y$ in $e q n \left(3\right)$ we get
$x = 5 - y = 5 - 2 = 3$

So the value $x$ is $3$ and $y$ is $2$.