Question

solve the following pair of equation by reducing them to a pair of linear equation ​

Answer 1

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Answer 2

$\huge \mathfrak \red{AnswEr}$

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• $= > x = 4$
• $= > y = 5$

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❥︎$\huge\underline\bold\green{SoluTion}$

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$= > \frac{5}{x - 1} + \frac{1}{y - 2} = 2.......i)$

$= > \frac{6}{x - 1} - \frac{3}{y - 2} = 1.......ii)$

here,

1/x-1 = m & 1/y-2 = n

put in equation i) & ii)

$= > 5m + n = 2......iii)$

$= > 6m - 3n = 1.......iv)$

now,

multiplying equation iii) by 3

$= > 15m + 3n = 6.......v)$

hence,

adding equation iv) and v)

6m - 3n = 1

+

15m + 3n = 6

21m. = 7

$= > m = \frac{7}{27}$

$= > m = \frac{1}{3}$

therefore,

puting value of m in equation iii)

$= > 5m + n = 2$

$= > n = 5m$

$= > n = 2 - \frac{5}{3}$

$= > n = \frac{6 - 5}{3}$

$= > n = \frac{1}{3}$

but,

$m = \frac{1}{x - 1}$

&

$n = \frac{1}{y - 2}$

therefore,

$= > m = \frac{1}{x - 1}$

$= > \frac{1}{3} = \frac{1}{x - 1}$

$= > 3 = x - 1$

$= > x = 3 + 1$

$= > x = 4$

similarly,

$= > n = \frac{1}{y - 2}$

$= > \frac{1}{3} = \frac{1}{y - 2}$

$= > 3 = y - 2$

$= > y = 3 + 2$

$= > y = 5$

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cross-check -

$= > \frac{ 5}{x - 1} + \frac{1}{y - 2} = 2$

$= > \frac{5}{4 - 1} + \frac{1}{5 - 2 } = 2$

$= > \frac{5}{3} + \frac{1}{3} = 2$

$= > \frac{6}{3} = 2$

$= > 2 = 2$

hence proved ans is correct ✔︎✔︎✔︎

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