$\frac{x-2}{3}=\frac{4y-1}{7}=\frac{y-3}{5}$
Bài giải
Ta có: $\frac{x-2}{3}=\frac{4y-1}{7}=\frac{y-3}{5}\Rightarrow \frac{x-2}{3}=\frac{4y-1}{7}=\frac{4y-12}{20}$
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
$\frac{x-2}{3}=\frac{4y-1}{7}=\frac{4y-12}{20}=\frac{\left( 4y-1 \right)-\left( 4y-12 \right)}{7-20}=\frac{-11}{13}$
$\Rightarrow \left\{ \begin{align}
& \frac{x-2}{3}=\frac{-11}{13} \\
& \frac{4y-1}{7}=\frac{-11}{13} \\
\end{align} \right.$\[\Rightarrow \left\{ \begin{align}
& x-2=\frac{-33}{13} \\
& 4y-1=\frac{-77}{13} \\
\end{align} \right.\]$\Rightarrow \left\{ \begin{align}
& x=\frac{-33}{13}+2 \\
& 4y=\frac{-77}{13}+1 \\
\end{align} \right.$$\Rightarrow \left\{ \begin{align}
& x=\frac{-7}{13} \\
& 4y=\frac{-64}{13} \\
\end{align} \right.$$\Rightarrow \left\{ \begin{align}
& x=\frac{-7}{13} \\
& y=\frac{-16}{13} \\
\end{align} \right.$
Vậy $\left( x;y \right)=\left( \frac{-7}{13};\frac{-16}{13} \right)$