Question

$x-y=\frac{x+y}{7}=\frac{xy}{4}$ than xy=?

Answer 1

$\underset{}{x-y=\frac{x+y}{7}}=\underset{}{\frac{xy}{4}}\therefore xy=?\begin{array}{cc}x-y=\frac{x}{7}+\frac{y}{7}& \frac{4y}{3}+y=\frac{7xy}{4}\\ \Rightarrow x-\frac{x}{7}=y+\frac{y}{7}& \Rightarrow \frac{7y}{3}=\frac{7xy}{4}\\ \Rightarrow \frac{6x}{7}=\frac{8y}{7}& \Rightarrow \frac{y}{3}-\frac{xy}{4}=0\\ \Rightarrow 3x=4y& \Rightarrow y(\frac{1}{3}-\frac{x}{y})=0\\ \Rightarrow x=\frac{4y}{3}& \Rightarrow \underset{\underset{y=0}{\downarrow }}{x=0}orx=\frac{4}{3}\Rightarrow y=\frac{3}{4}x=\frac{3}{4}\times \frac{4}{3}=1\end{array}\therefore xy=0or\frac{4}{3}\times 1=\frac{4}{3}$