​ Solve the following systems of equations: 6/(x+y) - 7/(x - y) +3 1/{2(x+y)} = 1/(3{x - y}) where, x + y ≠ 0 and x - y ≠ 0

Question

Solve the following systems of equations:

\(\frac{6}{x+y}\) - \(\frac{7}{x-y}\)+3

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

where, x + y ≠ 0 and x - y ≠ 0

Answer 1

Given:

The following systems of equations:

 \(\frac{6}{x+y}\) - \(\frac{7}{x-y}\)+3 ...... (1)

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

To find:

The value of x and y.

Solution:

Take \(\frac{1}{x+y}\) = u and \(\frac{1}{x-y}\) = v

The equations become:

6u = 7v+36u - 7v = 3 ...... (3)

And, 

⇒ \(\frac{1}{2}\)u = \(\frac{1}{3}\)v

3u = 2v

3u - 2v = 0 ...... (4)

To find the value of u and v,

Multiply eq. 4 with 2 and subtract it from eq. 3

⇒ 6u - 7v - 2(3u 2v) = 3 - 0

⇒ 6u -7v - 6u + 4v = 3

⇒ - 3v = 3

⇒ v = - 1

Put the value of v in eq. 4 to get,

⇒ 3u - 2( -1) = 0

⇒ 3u+2 = 0

⇒ 3u = - 2

u = \(\frac{-2}{3}\)

Now, 

⇒ x - y = - 1 .... (6)

Multiply eq. 6 with 2 and add to eq. 5

⇒ 2(x - y)+(- 2x - 2y) = - 2+3

⇒ - 2y - 2y - 2x - 2y = 1

⇒ - 4y = 1

⇒ y = \(\frac{-1}{4}\)

Now put the value of y in eq. 6 to get

Hence the values are

x = \(\frac{-5}{4}\), y = \(\frac{-1}{4}\)