Question

Solve the following systems of equations : 2/ x + 3/ y = 9/ x.y : 4/ x + 9/ y = 21/ x.y ​

Answer 1

AsnwEr:

Here We have,

$\;\;\;\;\;\;\;\star\;{\boxed{\sf{ \dfrac{2}{x} + \dfrac{3}{y} = \dfrac{9}{xy}}}}$

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$:\implies\sf \dfrac{2y + 3x}{xy} = \dfrac{9}{xy}$

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$:\implies\sf 2x + 3y = 9$⠀⠀⠀⠀⠀⠀⠀eq.(1)

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And,

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$\;\;\;\;\;\;\;\star\;{\boxed{\sf{ \dfrac{4}{x} + \dfrac{9}{y} = \dfrac{21}{xy}}}}$

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$:\implies\sf \dfrac{4y + 9x}{xy} = \dfrac{21}{xy}$

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$:\implies\sf 4y + 9x = 21$⠀⠀⠀⠀⠀⠀⠀eq.(2)

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★ From eq(1) -

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$:\implies\sf 2x + 3y = 9$

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$:\implies\sf 3x = 9 - 2y$

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$:\implies\sf \red{x = \dfrac{9 - 2y}{3}}$⠀⠀⠀⠀⠀⠀⠀eq.(3)

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★ Now, Substitute value of x from eq(3) in eq(2) -

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$:\implies\sf 4y + \cancel{9} \bigg( \dfrac{9 - 2y}{ \cancel{3}} \bigg) = 21$

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$:\implies\sf 4y + 3(9 - 2y) = 21$

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$:\implies\sf 4y + 27 - 6y = 21$

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$:\implies\sf - 2y = 21 - 27$

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$:\implies\sf - 2y = - 6$

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$:\implies\sf y = \cancel{ \dfrac{-6}{-2}}$

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$:\implies{\underline{\boxed{\sf{\pink{y = 3}}}}}\;\bigstar$

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★ Now, Put value of y in eq(3) -

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$:\implies\sf x = \dfrac{9 - 2 \times 3}{3}$

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$:\implies\sf x = \dfrac{9 - 6}{3}$

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$:\implies\sf x = \cancel{ \dfrac{3}{3}}$

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$:\implies{\underline{\boxed{\sf{\purple{y = 1}}}}}\;\bigstar$

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$\therefore$ Hence, Solution of the system of equation is x = 3 , y = 1.