Question

3x-y/3x+y=1/3 then find x:y and also check your answer

Answer 1

$\frac{3x-y}{3x+y}= \frac{1}{3}\\ \\ divide\ both\ numerator\ and\ denominator\ by\ y\\ \\ \frac{(3x-y)/y}{(3x+y)/y}= \frac{1}{3}\\ \\ \frac{3(x/y) -(y/y)}{3(x/y)+(y/y)}= \frac{1}{3}\\ \\ \frac{3(x/y) -1}{3(x/y)+1}= \frac{1}{3}\\ \\ let\ \frac{x}{y} =t\\ \\ \frac{3t-1}{3t+1}= \frac{1}{3}\\ \\ 3(3t-1)=3t+1\\ \\9t-3=3t+1\\ \\6t=4\\ \\t= \frac{4}{6}\\ \\ \frac{x}{y} = \frac{4}{6}= \frac{2}{3}\\ \\x:y=\boxed{2:3}$