Question

Solve for x&y : (3/x+y)+(11/x-y)=10. (4/x+y)+(5/x-y)=13

Answer 1

Answer:

$x=\frac{1363}{93} \ and \ y=-\frac{1334}{93}$

Step-by-step explanation:

$\frac{3}{x+y}+\frac{11}{x-y}=10\\\frac{4}{x+y}+\frac{5}{x-y}=13\\let \ \frac{1}{x+y}=a \ and \ \frac{1}{x-y}=b \ we \ get,\\3a+11b=10 \ ...(A)\times 4\\4a+5b=13 \ ...(B)\times 3\\\\ \ we \ get\\\therefore \ 12a+44b-(12a+15b)=40-39\\\\\therefore \ 29b=1 \implies b=\frac{1}{29}\\put \ the \ value \ of \ b \ in (A),$

$\therefore \ 3a+\frac{11}{29}=10 \implies a=\frac{93}{29}\\now,\\a=\frac{1}{x+y}=\frac{93}{29}\implies x+y=\frac{29}{93}\\b=\frac{1}{x-y}=\frac{1}{29}\implies x-y=29\\solving \ these \ two \ equations \ we \ get\\x=\frac{1363}{93} \ and \ y=-\frac{1334}{93}$

Answer 2

Answer:

x=1363/93. y=-1334/93

HOPE IT HELPS