Question

2/x+3/y=20.....(i)
7/x+2/y=53.....(ii)​

Answer 1

Question:

$\dfrac{2}{x} + \dfrac{3}{y} = 20$[Equation..1]

$\dfrac{7}{x} + \dfrac{2}{y} = 53$[Equation..2]

To Find:

$\underline{\underline{The\:value\:of\:x\:and\:y.}}$

Concept Used:

$\underline{\underline{By\:Elimination\:method}}$

Solution:

Solving Equation 1....

☞$\dfrac{2}{x} + \dfrac{3}{y} = 20$[Equation..1]

$\Rightarrow \dfrac{2}{x} + \dfrac{3}{y} = 20$

LCM of denominators is xy.

$\Rightarrow \dfrac{2y + 3x}{xy} = 20$

$\therefore 2y + 3x = 20xy$ [Equation..i]

Solving Equation 2....

☞$\dfrac{7}{x} + \dfrac{2}{y} = 53$[Equation..2]

$\Rightarrow \dfrac{7}{x} + \dfrac{2}{y} = 53$

LCM of denominators is xy.

$\Rightarrow \dfrac{7y + 2x}{xy} = 53$

$\therefore 7y + 2x = 53xy$ [Equation..ii]

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Putting Equation (i) and (ii), we get:

$2y + 3x = 20xy$

$7y + 2x = 53xy$

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$2y + 3x = 20xy$ × (2)

$7y + 2x = 53xy$ × (3)

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$4y + 6x = 40xy$

$21y + 6x = 159xy$

_______________[By Subtracting]

$-17y = -119$

$y = \dfrac{119}{17}$

$y = 7$

${\boxed{\therefore y = 7}}$

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Putting the value of y in Equation (1) :

$\dfrac{2}{x} + \dfrac{3}{y} = 20$

$\Rightarrow \dfrac{2}{x} + \dfrac{3}{7} = 20$

LCM of denominators is 7x.

$\Rightarrow \dfrac{14 + 3x}{7x} = 20$

$\Rightarrow 14 + 3x = 140x$

$\Rightarrow 14 = 140x - 3x$

$\Rightarrow 14 = 137x$

$\Rightarrow \dfrac{14}{137} = x$

${\boxed{\therefore x = \dfrac{14}{137}}}$

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$\underline{\underline{x = \dfrac{14}{137}\: y = 7}}$