Question

$\frac{6}{x+y}=\frac{7}{x-y}+3$

$\frac{1}{2(x+y)}=\frac{1}{3(x-y)}$

• A. $x=0,y=0$
• B. $x=\frac{5}{9},y=\frac{-1}{9}$
• C. $x=\frac{-5}{4},y=\frac{-1}{4}$
• D. None of these

Answer 1

Answer: (C)

$x=\frac{-5}{4},y=\frac{-1}{4}$

$\frac{6}{x+y}=\frac{7}{x-y}+3=>\frac{6}{x+y}=\frac{7+3(x-y)}{x-y}=>\frac{6}{x+y}=\frac{7+3x-3y}{x-y}=>6(x-y)=(7+3x-3y)(x+y)=>6x-6y=7x+7y+3x^2+3xy-3xy-3y^2=>3x^2-3y^2+x+13y=0\frac{1}{2(x+y)}=\frac{1}{3(x-y)}=>3x-3y=2x+2y=>3x-2x=2y+3y=>x=5y3\times (5y{)}^2-3y^2+5y+13y=0=>75y^2-3y^2+18y=0=>72y^2+18y=0=>18y(4y+1)=0=>4y=-1=>y=\frac{-1}{4}x=5\times \left(\frac{-1}{4}\right)=\frac{-5}{4}$