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प्रश्न
Solve the following systems of equations:
`2x - 3/y = 9`
`3x + 7/y = 2, y != 0`
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उत्तर
The given systems of equation is
`2x - 3/y = 9` .....(i)
`3x + 7/y = 2, y != 0 ` ...(ii)
Taking `1/y = u` the given equations becomes
2x - 3u = 9 ....(iii)
3x + 7u = 2 ....(iv)
From (iii), we get
2x = 9 + 3u
`=> x = (9 + 3u)/2`
Substituting `x = (9 + 3u)/2` in iv we get
`3((9 + 3u)/2) + 7u = 2`
`=> (27 + 9u + 14u)/2 = 2`
`=> 27 + 23u = 2xx2`
`=> 23u = 4 - 27`
`=> u = (-23)/23 = -1`
hence `y = 1/u = 1/(-1) = -1`
Putting u = -1 in `x = (9 + 3u)/2` we get
`x = (9 + 3(-1))/2 = (9-3)/2 = 6/2 = 3`
=> x = 3
Hence, solution of the given system of equation is x = 3, y = -1
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