Question

If $x^2+y^2-xy=3\mathrm{and}y-x=1,\mathrm{then}\frac{xy}{x^2+y^2}=\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ {x}^2+y^2-xy=3...\left(1\right) \\ y-x=1...\left(2\right) \\ \mathrm{Now}, \\ y-x=1 \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow {\left(y-x\right)}^2=1^2 \\ \Rightarrow x^2+y^2-2xy=1\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{\left(a-b\right)}^2=a^2+b^2-2ab\right) \\ \Rightarrow x^2+y^2-xy-xy=1 \\ \Rightarrow 3-xy=1\left(\mathrm{From}\left(1\right)\right) \\ \Rightarrow xy=2...\left(3\right) \\ \mathrm{Thus}, \\ \frac{xy}{x^2+y^2}=\frac{2}{3+xy}\left(\mathrm{From}\left(1\right)\mathrm{and}\left(3\right)\right) \\ =\frac{2}{3+2} \\ =\frac{2}{5} \end{gathered}$$

Hence, if $x^2+y^2-xy=3\mathrm{and}y-x=1,\mathrm{then}\frac{xy}{x^2+y^2}=\underline{\frac{2}{5}}.$