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\({7x -2y \over xy} = 5\) \({{8x + 7y} \over xy} =15 \)

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Consider, \({7x -2y \over xy} = 5\)

\({7x \over xy } - {2y \over xy } = 5 \)

\({-2 \over x} +{ 7 \over y} =5\)

\(-2({1 \over x}) + 7 ({1 \over y}) = 5 \) ...(I)

Now consider

\({{8x + 7y} \over xy }=15\)

∴ \({8x \over xy} + {7y \over xy}=15\)

∴ \({8 \over y} + {7 \over x} = 15\)

∴ \({7 \over x} + {8 \over y} = 15\)

∴ \(7 ({1 \over x}) + 8({1 \over y})= 15 \) ...(II)

Replace \(1 \over x \) by a and \(1 \over y\) by b in equations (I) and (II), we get

–2a + 7b = 5 ...(III)

7a + 8b = 15 ...(IV)

Multiplying eq. (III) by 7 and eq. (IV) by 2

–14a + 49b = 35 ...(V)

14a + 16b = 30 ...(VI)

Adding eq. (V) and (VI)

–14a + 49 b = 35

__+ 14 a + 16 b = 30 __

65b = 65

b =1

Place b = 1 in eq. (III)

–2a + 7(1) = 5

∴ –2a + 7 = 5

∴ –2a = 5 – 7

∴ –2a = –2

∴ a =1

Resubstituting the values of a and b

\({1 \over x} = 1\) and \({1 \over y}=1\)

∴ x = 1 ∴ y = 1

∴ (x, y) = (1, 1) is the required solutions.