Question

solve the following simultaneous equation​

Answer 1

Solution

$\bf \: we \: have$

$\tt \to \: \dfrac{10}{x + y} + \dfrac{2}{x - y} = 4$

$\tt \to \dfrac{15}{x + y} - \dfrac{5}{x - y} = - 2$

$\bf \: let$

$\tt \to \: \dfrac{1}{x + y} = u \: \: \: and \: \: \: \dfrac{1}{x - y} = v$

$\bf \: we \: get$

$\tt \to \: 10u + 2v = 4 \: \: \: \: \: \: (i) \\ \tt \to \: 15u - 5v = - 2 \: \: \: \: \: \: (ii)$

$\bf \: Now \: using \: substitution \: method$

$\tt \to \: 2(5u + v) = 4 \: \: \: \: \: \: (i)$

$\tt \to \: 5u + v = 2$

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

$\bf \: Now \: put \: the \: value \: of \: v \: on \: (ii)eq$

$\tt \to \: 15u \: - 5v \: = - 2 \: \: \: \: \: (ii)$

$\tt \to \: 15u - 5(2 - 5u) = - 2$

$\tt \to \: 15u - 10 + 25u = - 2$

$\tt \to \: 40u = 10 - 2$

$\tt \to40u = 8$

$\tt \to \: u = \dfrac{8}{40} = \dfrac{1}{5}$

$\bf \: Now \: put \: the \: value \: of \: u \: on \: (iii)eq$

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

$\tt \to \: v = 2 - 5 \times \dfrac{1}{5}$

$\tt \to \:v = 2 - 1 = 1$

$\bf Now \:put \: the \: value \: of \: u \: and \: v \: on \:$

$\tt \to \: \dfrac{1}{x + y} = \dfrac{1}{5} \: \: \: and \: \: \: \dfrac{1}{x - y} = 1$

$\tt \to \: x + y = 5 \: \: \: \: \: (i)\\ \tt \to \: x - y = 1 \: \: \: \: \: (ii)$

$\bf \: Add \: both \: the \: equation$

$\tt \to \: x + y + x - y = 5 + 1$

$\tt \to \: 2x = 6$

$\tt \to \: x = 3$

$\bf \: Now \: put \: the \: value \: of \: x \: on \: (i)eq$

$\tt \to \: x + y = 5$

$\tt \to3 + y = 5$

$\tt \to \: y = 2$

$\bf \: Answer$

$\tt \: x = 3 \: \: and \: y \: = 2$