Question

solve the equation
(10/x + y )+ 2/x- y = 4
+15/x + y)+ 5/x - y = -2
find x and y ​

Answer 1

Given Equations :-

$:\implies\sf{ \dfrac{10}{x + y} + \dfrac{2}{x - y} = 4}$

$:\implies\sf{\dfrac{15}{x + y} + \dfrac{5}{x - y} = -2}$

Now, let's consider

$\sf{ u = \dfrac{1}{x + y}}$...(1)

$\sf{v = \dfrac{1}{x - y}}$...(2)

Then,

We get :

10u + 2v = 4 ⠀⠀...(3)

15u + 5v = -3 ⠀ ...(4)

It reduced in a pair of linear equations now -

For solving further

By using Equating coefficient Method :

We shall Multiply both side of equation (3) by 5 and equation (4) by 2

New Equations will be

$\tt\longrightarrow{50u + 10v = 20}$..(5)

$\tt\longrightarrow{30u - 10v = -4}$..(6)

By Using Elimination Method

Adding Equation (5) and (6) , we get :

→ (50u + 30u) + (10v - 10v) = 20 - 4

→ 80u = 16

$:\implies\sf{ u = \dfrac{16}{80}}$

$:\implies\bf{ u = \dfrac{1}{5}}$

Putting up the value of " u " in the Equation (3)

$:\implies\sf{ 10 \times \dfrac{1}{5} + 2v =4 }$

→ 2 + 2v = 4

→ 2v = 4 - 2

→ 2v = 2

$:\implies\sf{ v = \dfrac{2}{2}}$

v = 1

Now, we know

$:\implies\bf{ u = \dfrac{1}{5}}$

v = 1

Substituting the values in equation (1) and (2) we get :

$\sf{\dfrac{1}{x + y} = \dfrac{1}{5}}$

$\sf{\dfrac{1}{x - y} = 1}$

By cross Multiplication :

→ x + y = 5⠀⠀..(7)

→ x - y = 1 ⠀⠀...(8)

Again using Elimination Method

Adding equations (7) and (8) , we get

→ (x + x) + (y - y) = 5 + 1

→ 2x = 6

$:\implies\sf{ x = \dfrac{6}{2}}$

$:\implies\bf{ x = 3}$

Hence,

$\boxed{\therefore{\sf{x = 3}}}$

Putting the value of " x " in Equation (7)

$\implies$3 + y = 5

$\implies$y = 5 - 3

$\implies$y = 2

$\boxed{\therefore{\sf{y = 2}}}$