-Ta có: $\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{12}=\frac{2x+3y-1}{6x}=\frac{(2x+3y-1)-(2x+3y-1)}{12-6x}=\frac{0}{12-6x}=0 $
-Vì:
+$\frac{2x+1}{5}=0\Rightarrow 2x+1=0\Rightarrow x=\frac{-1}{2}$
+$\frac{3y-2}{7}=0\Rightarrow 3y-2=0\Rightarrow y=\frac{2}{3}$