Question

Solve for x and y :15/x-y + 30/x+y =11 and 9/x-y +25/x+y =8

Answer 1

Let $\dfrac{1}{x+y}=a \:\:\:and\:\:\: \dfrac{1}{x-y}=b$

Given equations :$\dfrac{15}{x-y}+\dfrac{30}{x+y}=11$

                            $\dfrac{9}{x-y}+\dfrac{25}{x+y}=8$

Now, after substituting the assumed values of x- y and x + y, equations are

15b + 30a = 11 and 9b + 25a = 8

15b + 30a = 11       ...( i )

9b + 25a = 8        ...( ii )

Multiply by 9 on ( i ) and by 15 on ( ii ) : -

⇒[ 15b + 30a = 11 ] × 9

⇒[ 9b + 25b = 8   ] × 15

⇒ 135b + 270a = 99

⇒ 135b + 375a = 120

Now, Subtracting any equation from other.

135b + 270a = 99

135b + 375a = 120

-         -            -

______________

         105b = 21

______________

$b=\dfrac{21}{105}$

b = $\dfrac{1}{5}$

$\dfrac{1}{x-y} = \dfrac{1}{5}$

x - y = 5

x = 5 + y             ...( iii )

Substituting the value of b in ( i )

15b + 30a = 11    

$[ 15 \times \dfrac{1}{5}] + 30a = 11$

3 + 30a = 11

30a = 11 - 3

30a = 8

a = $\dfrac{8}{30}$

a = $\dfrac{4}{15}$

$\dfrac{1}{x+y}=\dfrac{4}{15}$

x + y = $\dfrac{15}{4}$

Substituting the value of x from ( iii )

5 + y + y = $\dfrac{15}{4}$

( 5 + 2y ) 4 = 15

20 + 8y = 15

8y = -5

y = - 8 / 5

From ( iii ), x + y = 5

                 x - 8 / 5 = 5

                 x = 5 + 8 / 5

                 x = 33 / 5