Question Number 168400 by leicianocosta last updated on 10/Apr/22
Commented by cortano1 last updated on 10/Apr/22
![{ ((5(√7) πx+2πy = ((2π(√7))/7))),((x −2πy = 10π^2 )) :} x(5(√7) π+1)=((2π(√7))/7)+10π^2 x=((2π(√7)+70π^2 )/(35π(√7)+7)) ; [π=((22)/7)]](https://www.tinkutara.com/question/Q168401.png)
$$\:\:\:\:\begin{cases}{\mathrm{5}\sqrt{\mathrm{7}}\:\pi{x}+\mathrm{2}\pi{y}\:=\:\frac{\mathrm{2}\pi\sqrt{\mathrm{7}}}{\mathrm{7}}}\\{{x}\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}\pi{y}\:=\:\mathrm{10}\pi^{\mathrm{2}} }\end{cases} \\ $$$$\:\:\:\:{x}\left(\mathrm{5}\sqrt{\mathrm{7}}\:\pi+\mathrm{1}\right)=\frac{\mathrm{2}\pi\sqrt{\mathrm{7}}}{\mathrm{7}}+\mathrm{10}\pi^{\mathrm{2}} \\ $$$$\:\:\:{x}=\frac{\mathrm{2}\pi\sqrt{\mathrm{7}}+\mathrm{70}\pi^{\mathrm{2}} }{\mathrm{35}\pi\sqrt{\mathrm{7}}+\mathrm{7}}\:;\:\left[\pi=\frac{\mathrm{22}}{\mathrm{7}}\right] \\ $$