Question

$\frac{4}{x} + \frac{5}{y} = 7 ; \frac{3}{x} + \frac{4}{y} = 5$



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Answer 1

GIVEN :-

$\\ \bullet \: \: \: \: \: \sf \: \frac{4}{x} + \frac{5}{y} = 7 \\ \\ \bullet \: \: \: \: \sf \: \frac{3}{x} + \frac{4}{y} = 5$

TO FIND :-

• x and y.

SOLUTION :-

Let us consider ,

$\\ \sf \: \frac{1}{x} = p \: \: \: \: \: \: \: and \: \: \: \: \: \: \: \frac{1}{y} = q$

So, the equation becomes ,

$\\ \bullet \: \: \: \: \: \sf \: 4p + 5q = 7 \: \: \: \: - - (1) \\ \\ \bullet \: \: \: \: \: \sf \: 3p + 4q = 5 \: \: - - (2)$

Multiplying equation (1) by 3 ,

Multiplying equation (2) by 4 ,

We get..

$\\ (1) \: \: \: \: \: \sf \: 3(4 p + 5q) = 3(7) \\ \implies \sf 12p + 15q = 21 \: \: \: \: \: - - (3) \\ \\ (2) \: \: \: \: \: \: \sf 4(3p + 4q) = 4(5) \\ \implies \sf 12p + 16q = 20 \: \: \: \: - - (4)$

Subtracting equation (4) by equation (3) ,

→ 12p + 15q - (12p + 16q) = 21 - 20

→ 12p + 15q - 12p - 16q = 1

→ -q = 1

→ q = -1

Putting q = -1 in equation (1) ,

→ 4p + 5q = 7

→ 4p + 5(-1) = 7

→ 4p - 5 = 7

→ 4p = 7 + 5

→ 4p = 12

→ p = 3

Hence , q = -1 and p = 3.

$\\ \bullet \sf \: p = \frac{1}{x} \\ \\ \implies\sf \: 3 = \frac{1}{x} \\ \\ \implies \boxed{\sf \: x = \frac{1}{3} } \\ \\ \\ \bullet \sf \: q = \frac{1}{y} \\ \\ \implies \sf \: - 1 = \frac{1}{y} \\ \\ \implies \boxed{\sf \: y = - 1}$

Hence , x = 1/3 and y = -1.

Answer 2

$\huge \mathtt \pink{to \: solve} \pink \to$

$\frac{4}{x} + \frac{5}{y} = 7 ; \frac{3}{x} + \frac{4}{y} = 5</p><p></p><p>$

$\huge \mathtt \pink{solution} \pink \to$

$\displaystyle {4} \left( \frac{1}{x} \right) </p><p></p><p> + \displaystyle{5} \left( \frac{1}{y} \right) = 7. . . (i)$

$\displaystyle{3} \left( \frac{1}{x} \right) + 4 \displaystyle \left( \frac{1}{y} \right) = 5...(ii)</p><p></p><p>$

$\sf{replacing} \displaystyle \left( \frac{1}{x} \right) \sf \: by \: m \: and \: \displaystyle \left( \frac{1}{y} \right) \sf \: \\ \sf by \: n \: in \: equation \\ \sf (i)and(ii) \: we \: get \:$

$\text{4m + 5n = 7...(iii)}$

$\text{3m + 4n = 5...(iv)}$

$\mathtt{on \: solving \: these \: equation \: we \: get}$

$\text{m = 3, \: n = - 1}$

$\mathtt{now,m = \frac{1}{x}}$

$\therefore \boxed {3 = \frac{1}{x}}$

$\therefore\boxed{x = \frac{1}{3}}$

$[/tex]</p><p></p><p></p><p>[tex] \sf{n = \frac{1}{y}}$

$\therefore \boxed{ - 1 = \frac{1}{y}}$

$\therefore \boxed{y = - 1}$

$\therefore \mathtt{solution \: of \: given \: simultaneous \: equation \: is \: } \\ \\ \\ \\ \boxed{(x,y) = ( \frac{1}{3}, - 1)}$