\(\left\{{}\begin{matrix}\frac{2y-5x}{3}+5=\frac{y+27}{4}-2x\\\frac{x+1}{3}+y=\frac{6y-5x}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\frac{2y-5x}{3}+5+2x=\frac{y+27}{4}\\\frac{x+1}{3}+y=\frac{6y-5x}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\frac{2y+x+15}{3}=\frac{y+27}{4}\\\frac{x+3y+1}{3}=\frac{6y-5x}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8y+4x+60=3y+81\\7x+21y+7=18y-15x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x+5y=21\\22x+3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+5y=21\\66x+9y=-21\end{matrix}\right.\Leftrightarrow70x+14y=0\Leftrightarrow5x+y=0\Leftrightarrow20x+4y=0;4x+5y=21\Leftrightarrow20x+25y=105\Leftrightarrow\left(20x+25y\right)-\left(20x+4y\right)=105\Leftrightarrow21y=105\Leftrightarrow y=5.\text{Thay vào ta được:}4x+25=21\Leftrightarrow4x=-4\Leftrightarrow x=-1\)
\(\text{Thử lại ta thấy thỏa mãn: Vậy: x=-1;y=5}\)
\(\left\{{}\begin{matrix}\frac{2}{3}y-\frac{5}{3}x-\frac{1}{4}y+2x=\frac{27}{4}-5\\\frac{1}{3}x+\frac{5}{7}x+y-\frac{6}{7}y=-\frac{1}{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{3}x+\frac{5}{12}y=\frac{7}{4}\\\frac{22}{21}x+\frac{1}{7}y=-\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=5\end{matrix}\right.\)
