Question

what is the value of (1/x-1/y) if 6/x-3/y=10 and 2/x-5/y=6 ?​

Answer 1

Given Equation

$\tt \to \: \dfrac{6}{x} - \dfrac{3}{y} = 10 \: \: \: \: \: \: \: \: \: \: \: \: (i)$

$\tt \to \: \dfrac{2}{x} - \dfrac{5}{y} = 6 \: \: \: \: \: \: \: \: \: \: \: \: (ii)$

To find

$\tt \to \: \bigg( \dfrac{1}{x} - \dfrac{1}{y} \bigg)$

Now Let

$\tt \to \: \dfrac{1}{x} = u$

$\tt \to \: \dfrac{1}{y} = v$

We get

$\tt \to \: 6u - 3v = 10 \: \: \: \: \: \: \: \: \: \: \: \: \: (i)$

$\tt \to \: 2u - 5v = 6 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (ii)$

Using Substitution Method so take (i)eq

$\tt \to \: 6u - 3v = 10 \: \: \: \: \: \: \: \: \: \: \: \: \: (i)$

$\tt \to \: 6u = 10 + 3v$

$\tt \to \: u = \dfrac{10 + 3v}{6} \: \: \: \: \: \: \: \: \: \: (iii)$

Now Put the value of u on (ii)eq

$\tt \to \: 2u - 5v = 6 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (ii)$

$\tt \to2 \bigg( \dfrac{10 + 3v}{6} \bigg) - 5v = 6$

$\tt \to\dfrac{10 + 3v}{3} - 5v = 6$

$\tt \to10 + 3v - 15v = 18$

$\tt \to \: - 12v = 18 - 10$

$\tt \to \: - 12v = 8$

$\tt \to \: v = \dfrac{ - 8}{12} = \dfrac{ - 2}{3}$

$\tt \to \: v = \dfrac{ - 2}{3}$

Now Put the value of v in (iii) eq

$\tt \to \: u = \dfrac{10 + 3v}{6} \: \: \: \: \: \: \: \: \: \: (iii)$

$\tt \to \: u = \dfrac{10 + 3 \times \dfrac{ - 2}{3} }{6}$

$\tt \to \: u = \dfrac{10 - 2 }{6} = \dfrac{8}{6} = \dfrac{4}{3}$

$\tt \to \: u = \dfrac{4}{3}$

We get

$\tt \to \: \dfrac{1}{x} = \dfrac{4}{3}$

$\tt \to \: \dfrac{1}{y} = \dfrac{ - 2}{3}$

Now we have to find

$\tt \to \: \bigg( \dfrac{1}{x} - \dfrac{1}{y} \bigg)$

Put the value

$\tt \to \dfrac{4}{3} + \dfrac{2}{3}$

$\tt \to \: \dfrac{6}{3} = 2$

Answer is 2