Question

✓2 times the distance between (0,5)and(-5,0)is..​

Answer 1

Step-by-step explanation:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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