Question

2/y+3/x=7/xy,1/y+9/x=11/xy by substitution method​

Answer 1

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$\huge\star\underbrace{\mathtt\red{⫷❥ᴀ᭄} \mathtt\green{n~ }\mathtt\blue{ §} \mathtt\purple{₩}\mathtt\orange{Σ} \mathtt\pink{R⫸}}\star\:$

$\: \star \underline{ \boxed { \bf \red{by \: substitution \: method}}}$

$\: \bullet \bf{ \frac{2}{y} + \frac{3}{x} = \frac{7}{xy} }$

$\to \bf{2x + 3y = 7......(1)}$

$\bullet \bf{ \frac{1}{y} + \frac{9}{x} = \frac{11}{xy} }$

$\to \bf{x + 9y = 11...(2)}$

$\implies\sf{ \: multiply \: in \: equation \: (1)by \: 3}$

$\: \bf{ \: we \: get \: 6x + 9y = 21.....(3)}$

$\: \clubsuit \boxed { \bf \pink{ \: subtracting \: both \: equation \: (3) \: and \: (1)}}$

$\: \bf{6x + 9y = 21}$

$\bf{x + 9y = 11}$

$\: \to \bf \red{5x = 10}$

$\: \to \bf \red{x = 2}$

$\: \underline{ \boxed { \bf{ \: substitute \: in \: equation \: (2)}}}$

$\to \bf{2 + 9y = 11}$

$\to \bf{9y = 9}$

$\: \to \bf \red{ y = 1}$

$\: \: \implies \boxed{ \bf \purple{ \: answer \: will \: be \: x = 2 \: and \: y = 1}}$