Solve the equation

6÷x+y=7÷x-y+3

1÷2(x+y)=1÷3(x-y)

Open in App

Solution

Let 1/x + y = a , 1/x - y = b

6a = 7b + 3 ---- (i)

a / 2 = b / 3 a = 2b / 3

Putting value of a in (i)
6*2b/3 = 7b + 3

4b = 7b + 3
-3b = 3
b = -1

a = -2/ 3

x + y = - 3 / 2 ------- (iii)
x - y = - 1 ----- (iv)

(iii) + (iv)
x + y + x - y = -3/2 - 1

2x = -5/2 x = - 5 /4

y = - 1 /4

Ans: x = - 5 / 4, y = - 1 / 4

Suggest Corrections

Similar questions

Solve the following systems of equations:

$\frac{6}{x + y} = \frac{7}{x - y} + 3$

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

Solve for x and y, using elimination method

6/(x+y)= 7/(x-y) + 3

1/2(x+y) = 1/3(x-y)

6/x+y =7/x-y +3

1/2(x+y) =1/3(x-y)

\frac{6}{x + y} = \frac{7}{x - y} + 3, \frac{1}{2 (x + y)} = \frac{1}{3 (x - y)}

x + y \neq 0 a n d x - y \neq 0

\frac{6}{x + y} = \frac{7}{x - y} + 3

\frac{1}{2 (x + y)} = \frac{1}{3 (x - y)},

x + y \neq 0

x - y \neq 0

Join BYJU'S Learning Program