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Question
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
Solution
Let `1/(x + y) = u and 1/(x - y) = v` Then, the given system of equation becomes
6u = 7v + 3
=> 6u - 7v = 3 .....(i)
And `u/2 = v/3`
=> 3u = 2v
=> 3u - 2v = 0 ......(ii)
Multiplying equation ii by 2, and equation (i) by 1, we get
6u - 7v = 3 ....(iii)
6u - 4v = 0 ......(iv)
Subtracting equation (iv) from equation (iii), we get
-7 + 4v = 3
=> -3v = 3
=> v = -1
Puttting v = -1 in equation (ii), we get
`3u - 2xx (-1) = 0`
=> 3u + 2 = 0
=> 3u = -2
`=> u = (-2)/3`
Now `u = (-2)/3`
=> `1/(x + 2) = (-2)/3`
`=> x + y = (-3)/2` ...(v)
And v= -1
`=> 1/(x - y) = -1`
=> x - y = -1 ...(vi)
Adding equation (v) and equation (vi), we get
`2x = (-3)/2 - 1`
`=> 2x = (-3-2)/2`
`=> 2x = (-5)/2`
`=> x = (-5)/4`
Putting x = (-5)/4 in equation (vi), we get
`(-5)/4 - y = -1`
`=> (-5)/4 + 1 = y`
`=> (-5+4)/4 = y`
`=> (-1)/4 = y`
`=> y = (-1)/4`
Hence, solution of the system of equation is `x = (-5)/4, y = (-1)/4`
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