Question Number 4973 by hung260677 last updated on 28/Mar/16
Answered by Rasheed Soomro last updated on 30/Mar/16
$$\frac{\mathrm{3}{x}^{\mathrm{2}} +{xy}−{y}^{\mathrm{2}} }{{x}+{y}}=\mathrm{9}\:\wedge\:\frac{\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} {y}}{{x}+{y}}=\mathrm{20} \\ $$$$\Rightarrow\:{x}+{y}=\frac{\mathrm{3}{x}^{\mathrm{2}} +{xy}−{y}^{\mathrm{2}} }{\mathrm{9}}\:\wedge\:{x}+{y}=\frac{\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} {y}}{\mathrm{20}} \\ $$$$\Rightarrow\:\frac{\mathrm{3}{x}^{\mathrm{2}} +{xy}−{y}^{\mathrm{2}} }{\mathrm{9}}=\frac{\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} {y}}{\mathrm{20}} \\ $$$$\Rightarrow\:\mathrm{60}{x}^{\mathrm{2}} +\mathrm{20}{xy}−\mathrm{20}{y}^{\mathrm{2}} =\mathrm{18}{x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} {y} \\ $$$$\Rightarrow\mathrm{18}{x}^{\mathrm{3}} −\mathrm{60}{x}^{\mathrm{2}} −\mathrm{9}{x}^{\mathrm{2}} {y}−\mathrm{20}{xy}+\mathrm{20}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$${Any}\:{way}\:{of}\:{factorization}? \\ $$$${Continue} \\ $$