Question

10/x+y + 2/x-y =4 ,15/x+y - 5/x-y= -2 by special elimination metgod

Answer 1

$let \: \frac{1}{x + y} = a \\ \frac{1}{x - y} = b \\ place \: these \: assumptions \: in \\ \: \: main \: equation \\ 10a + 2b = 4 \\ 15a - 5b = - 2 \\ multiply \: eq1 \: by \: 5 \: and \: 2 \: by \: 2 \\ 50a + 10b = 20 \\ 30a - 10b = - 4 \\ add \: both \: equation \\ 80a = 16 \\ a = \frac{16}{80} = \frac{1}{5} \\ 10 \times \frac{1}{5} + 2b = 4 \\ 2b = 4 - 2 = 2 \\ b = 1 \\ now \: put \: the \: values \: of \: a \: and \: b \\ \frac{1}{x + y} = \frac{1}{5} \\ \frac{1}{x - y} = 1 \\ on \: cross \: multiplying \: both \: we \: get \\ x + y = 5 \\ x - y = 1 \\ \: add \: both \: equations \\ 2x = 6 \\ x = \frac{6}{2} = 3 \\ 3 + y = 5 \\ y = 5 - 3 = 2 \\ so \: x = 3 \: \: \: y = 2 \: is \: the \: solution \: of \: these \: equations$