Question

Find value of x and y full explanation​

Answer 1

Solution :

$\sf \dfrac{2}{2x+y} - \dfrac{1}{x-2y} + \dfrac{5}{9} = 0$

$\sf and \: \dfrac{9}{2x+y} - \dfrac{6}{x-2y} + 4 = 0$

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$\sf Let \: \dfrac{1}{2x+y} = a \: and \: \dfrac{1}{x-2y} = b$

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$\sf \implies 2a - b + \dfrac{5}{9} = 0 \: \: \: -(1)$

$\sf and \: 9a - 6b + 4 = 0 \: \: \: -(2)$

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$\sf From \: (1) : 2a - b + \dfrac{5}{9} = 0$

$\sf 2a - b = - \dfrac{5}{9}$

$\sf - b = - \dfrac{5}{9} - 2a$

$\sf - b = - (\dfrac{5}{9} + 2a)$

$\sf b = \dfrac{5}{9} + 2a$

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Put the above value of b in (2) : $\sf 9a - 6( \dfrac{5}{9} + 2a) + 4 = 0$

$\sf 9a - \dfrac{30}{9} - 12a + 4 = 0$

$\sf 9a - \dfrac{\cancel{30}^{10}}{\cancel{9}_{3}} - 12a + 4 = 0$

$\sf - 3a - \dfrac{10}{3} + 4 = 0$

$\sf \dfrac{-9a - 10 + 12}{3} = 0$

$\sf -9a - 10 + 12 = 0 × 3$

$\sf -9a + 2 = 0$

$\sf -9a = -2$

$\sf 9a = 2$

$\boxed{\sf a = \dfrac{2}{9}}$

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Now, put value of a in (1) : $\sf 2(\dfrac{2}{9}) - b + \dfrac{5}{9} = 0$

$\sf \dfrac{4}{9} - b + \dfrac{5}{9} = 0$

$\sf \dfrac{4 - 9b + 5}{9} = 0$

$\sf 4 - 9b + 5 = 0 × 9$

$\sf 9 - 9b = 0$

$\sf - 9b = - 9$

$\sf 9b = 9$

$\sf b = \dfrac{9}{9}$

$\sf b = \cancel{\dfrac{9}{9}}$

$\boxed{\sf b = 1}$

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$\sf Now, \: we \: have \: value \: of \: a = \dfrac{2}{9}$

$\sf and \: a = \dfrac{1}{2x+y}$

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$\sf \implies \dfrac{2}{9} = \dfrac{1}{2x+y}$

$\sf \implies \dfrac{2(2x+y)}{9} = 1$

$\sf \implies \dfrac{4x+2y}{9} = 1$

$\sf \implies 4x + 2y = 1 \times 9$

$\sf \implies 4x + 2y = 9 \: \: \: -(3)$

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$\sf Similarly, \: we \: have \: value \: of \: b = 1$

$\sf and \: b = \dfrac{1}{x-2y}$

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$\sf \implies 1 = \dfrac{1}{x-2y}$

$\sf \implies x-2y = 1 \: \: \: -(4)$

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$\sf From (4) : x-2y = 1$

$\sf x = 1 + 2y$

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Now, put the above value of x in (3) : $\sf 4(1+2y) + 2y = 9$

$\sf 4 + 8y + 2y = 9$

$\sf 4 + 10y = 9$

$\sf 10y = 9-4$

$\sf 10y = 5$

$\sf y = \dfrac{5}{10}$

$\sf y = \dfrac{\cancel{5}^{1}}{\cancel{10}_{2}}$

$\boxed{\sf y = \dfrac{1}{2}}$

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Put value of y in (4) : $\sf x-2(\dfrac{1}{2}) = 1$

$\sf x - \cancel{2} \times \dfrac{1}{\cancel{2}} = 1$

$\sf x - 1 = 1$

$\sf x = 1 + 1$

$\boxed{\sf x = 2}$

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$\sf \color{fuchsia} \therefore \: value \: of \: x = 2 \: and \: y = \dfrac{1}{2}$