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प्रश्न
Solve the following systems of equations:
`5/(x + 1) - 2/(y -1) = 1/2`
`10/(x + 1) + 2/(y - 1) = 5/2` where `x != -1 and y != 1`
उत्तर
Let `1/(x + 1) = u and 1/(y - 1) = v`
Then, the given system of equations becomes
`=> 5u - 2v = 1/2` .....(i)
`=> 10u + 2y = 5/2 ....(2)`
Adding equation (i) equation (ii), we get
`5u + 10u = 1/2 + 5/2`
`=> 15u = (1 + 5)/2`
`=> 15u = 6/2 = 3`
`=> u = 3/15 = 1/5`
Putting `u = 1/5` in equation (i) we get
`5 xx 1/5 - 2v = 1/2`
`=> 1 - 2v = 1/2`
`=> -2v = 1/2 - 1`
`=> -2v = (1 - 2)/2`
`=> -2v = (-1)/2`
`=> v = (-1)/(-4) = 1/4`
Now `u = 1/(x + 1)`
`=> 1/(x + 1) = 1/5`
=> x + 1 = 5
=> x = 5 -1 =4
And `v = 1/(y - 1)`
`=> 1/(y -1) = 1/4 `
`=> y - 1 = 4 `
=> y = 4 + 1 = 5
Hence, solution of the give system of equation is x = 4, y = 5
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