1.

Solve for ` x and y` : ` (1)/(2x) - (1)/(y) = - 1, (1)/(x) + (1)/(2y)= 8 ( x ne 0 , y ne 0 )`

Answer» Putting ` (1)/(x) = u nd (1)/(y) = v `, the given equations become.
` (u)/(2) - v = - 1 rArr u - 2v = - 2 " " `… (i)
` u + (u)/(2) = 8 rArr 2u + v = 16 " " `… (ii)
Multiplying (ii) by 2 and adding the result with (i), we get
` u + 4u = - 2 + 32 `
` rArr 5u = 30 `
` rArr u = ( 30 )/( 5) = 6 `
Putting `u= 6 ` in (i), we get
` 6- 2 = - 2 rArr 2v = 8 rArr v = 4`.
Now, `u = 6 rArr (1)/(x) = 6 rArr 6x = 1 rArr x = (1)/(6)`
And, ` v = 4 rArr (1)/(y) =4 rArr 4y = 1 rArr y = (1)/(4)`
Hence, ` x = (1)/(6) and y = (1)/(4)`.


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