Question Number 11748 by tawa last updated on 30/Mar/17
$$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4y}^{\mathrm{2}} \:+\:\mathrm{xy}\:=\:\mathrm{6}\:\:\:\:\:…………..\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\mathrm{3x}^{\mathrm{2}} \:+\:\mathrm{8y}^{\mathrm{2}} \:=\:\mathrm{14}\:\:\:\:\:\:\:\:\:\:\:…………..\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$
Answered by mrW1 last updated on 30/Mar/17
$$\left({ii}\right)−\left({i}\right)×\mathrm{2}: \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{xy}=\mathrm{2} \\ $$$$\Rightarrow{y}=\frac{{x}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}{x}} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{8}×\frac{\left({x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} }=\mathrm{14} \\ $$$$\mathrm{3}{x}^{\mathrm{4}} +\mathrm{2}\left({x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}\right)=\mathrm{14}{x}^{\mathrm{2}} \\ $$$$\mathrm{5}{x}^{\mathrm{4}} −\mathrm{22}{x}^{\mathrm{2}} +\mathrm{8}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{22}\pm\sqrt{\mathrm{22}^{\mathrm{2}} −\mathrm{4}×\mathrm{5}×\mathrm{8}}}{\mathrm{10}}=\frac{\mathrm{22}\pm\mathrm{18}}{\mathrm{10}}=\mathrm{4},\frac{\mathrm{2}}{\mathrm{5}} \\ $$$${x}_{\mathrm{1},\mathrm{2}} =\pm\mathrm{2} \\ $$$${x}_{\mathrm{3},\mathrm{4}} =\pm\sqrt{\frac{\mathrm{2}}{\mathrm{5}}} \\ $$$${y}_{\mathrm{1},\mathrm{2}} =\frac{{x}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}{x}}=\pm\mathrm{2}×\frac{\mathrm{4}−\mathrm{2}}{\mathrm{2}×\mathrm{4}}=\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${y}_{\mathrm{3},\mathrm{4}} =\frac{{x}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}{x}}=\pm\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}×\frac{\frac{\mathrm{2}}{\mathrm{5}}−\mathrm{2}}{\mathrm{2}×\frac{\mathrm{2}}{\mathrm{5}}}=\mp\mathrm{2}\sqrt{\frac{\mathrm{2}}{\mathrm{5}}} \\ $$
Commented by tawa last updated on 30/Mar/17
$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$