Ta có : \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{\left(x-2\right)+\left(y-3\right)}{3+4}=\frac{x-2+y-3}{7}=\frac{x+y-5}{7}=\frac{0}{7}=0\)
Nên : \(\frac{x-2}{3}=0\Rightarrow x-2=0\Rightarrow x=2\)
\(\frac{y-3}{4}=0\Rightarrow y-3=0\Rightarrow y=3\)
Vậy x = 2 ; y = 3