# How to Solve bu using elimination method ?

## $\frac{5}{x + y} + \frac{2}{x - y} = 3$ $\frac{20}{x + y} - \frac{3}{x - y} = 1$

Jun 4, 2018

$x = - \frac{11}{2} , y = - \frac{9}{2}$

#### Explanation:

Substituting
$a = \frac{1}{x + y} , b = \frac{1}{x - y}$
so we have the System
$5 a + 2 b = 3$
$20 a - 3 b = 1$
Multiplying the first equation by $- 4$ and adding to the second we get
$b = - 1$
so
$a = - \frac{1}{10}$
Then we have
$1 = - x + y$
$- 10 = x + y$
from here we get
$x = - \frac{11}{2}$
$y = - \frac{9}{2}$

Jun 4, 2018

$x = 3 \mathmr{and} y = 2$

#### Explanation:

$\frac{5}{x + y} + \frac{2}{x - y} = 3$

$\frac{20}{x + y} - \frac{3}{x - y} = 1$

Let;

$\frac{1}{x + y} = a \mathmr{and} \frac{1}{x - y} = b$

Therefore;

$\frac{5}{x + y} + \frac{2}{x - y} = 3$

$5 \left(\frac{1}{x + y}\right) + 2 \left(\frac{1}{x - y}\right) = 3$

$5 a + 2 b = 3 - - - e q n 1$

Similarly..

$\frac{20}{x + y} - \frac{3}{x - y} = 1$

$20 \left(\frac{1}{x + y}\right) - 3 \left(\frac{1}{x - y}\right) = 1$

$20 a - 3 b = 1 - - - e q n 2$

Using Elimination Method!

$5 a + 2 b = 3 - - - e q n 1$

$20 a - 3 b = 1 - - - e q n 2$

Multiply $e q n 1$ by $3$ and $e q n 2$ by $2$

$3 \left(5 a + 2 b = 3\right)$

$2 \left(20 a - 3 b = 1\right)$

$15 a + 6 b = 9 - - - e q n 3$

$40 a - 6 b = 2 - - - e q n 4$

Adding both $e q n 3 \mathmr{and} e q n 4$ together..

$\left(15 a + 40 a\right) + \left(6 b + \left(- 6 b\right)\right) = 9 + 2$

$55 a + 6 b - 6 b = 11$

$55 a = 11$

$a = \frac{11}{55}$

$a = \frac{1}{5}$

Substituting the value of $a$ into $e q n 1$

$5 a + 2 b = 3 - - - e q n 1$

$5 \left(\frac{1}{5}\right) + 2 b = 3$

$\cancel{5} \left(\frac{1}{\cancel{5}}\right) + 2 b = 3$

$1 + 2 b = 3$

$2 b = 3 - 1$

$2 b = 2$

$b = \frac{2}{2}$

$b = 1$

But;

$\frac{1}{x + y} = a \mathmr{and} \frac{1}{x - y} = b$

$a = \frac{1}{x + y}$

$\frac{1}{5} = \frac{1}{x + y}$

Cross multiplying;

$1 \left(x + y\right) = 1 \left(5\right)$

$x + y = 5 - - - e q n 5$

Similarly..

$b = \frac{1}{x - y}$

$1 = \frac{1}{x - y}$

$\frac{1}{1} = \frac{1}{x - y}$

Cross multiplying;

$1 \left(x - y\right) = 1 \left(1\right)$

$x - y = 1 - - - e q n 6$

Solving simultaenously again..

$x + y = 5 - - - e q n 5$

$x - y = 1 - - - e q n 6$

Using Elimination Method!

Adding $e q n 5 \mathmr{and} e q n 6$ together;

$\left(x + x\right) + \left(y + \left(- y\right)\right) = 5 + 1$

$2 x + y - y = 6$

$2 x = 6$

$x = \frac{6}{2}$

$x = 3$

Substituting the value of $x$ into $e q n 6$

$x + y = 5 - - - e q n 5$

$3 + y = 5$

Collecting like terms;

$y = 5 - 3$

$y = 2$

Hence;

$x = 3 \mathmr{and} y = 2$

Jun 4, 2018

$x = 3 , y = 2$

#### Explanation:

$\frac{5}{x + y} + \frac{2}{x - y} = 3$
$\frac{20}{x + y} - \frac{3}{x - y} = 1$

Let x+y be a and x-y be b

$\therefore \frac{5}{a} + \frac{2}{b} = 3$----------(1)

$\therefore \frac{20}{a} - \frac{3}{b} = 1 - - - - - - - - - - - \left(2\right)$

$\therefore \left(1\right) \times 4$

$\therefore \frac{20}{a} + \frac{8}{b} = 12$------(3)

$\therefore \left(2\right) - \left(3\right)$

$\therefore - \frac{11}{b} = - \frac{11}{1}$

$\therefore \frac{b}{-} 11 = \frac{1}{-} 11$

multiply both sides by-11

$\therefore b = - \frac{11}{-} 11$

$\therefore b = 1$

substitute b=1 in (1)

$\therefore \frac{5}{a} + 2 = 3$

$\therefore \frac{5}{a} = 3 - 2$

$\therefore \frac{5}{a} = 1$

$\therefore \frac{a}{5} = 1$

multiply both sides by 5

$\therefore a = 5$

$\therefore a = x + y = 5$-----(4)

$\therefore b = x - y = 1$------(5)

$\therefore \left(4\right) + \left(5\right)$

$\therefore 2 x = 6$

$\therefore x = \frac{6}{2}$

$\therefore x = 3$

$\therefore 3 + y = 5$

$\therefore y = 5 - 3$ in (4)

$\therefore y = 2$