\(\dfrac{x}{y}=\dfrac{3}{5}\)\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{2x}{10}=\dfrac{3y}{9}\)
Theo t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{2x}{10}=\dfrac{3y}{9}=\dfrac{2x-3y}{10-9}=\dfrac{54}{1}=54\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=54\\\dfrac{y}{3}=54\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=270\\y=162\end{matrix}\right.\)
\(\dfrac{x}{y}=\dfrac{3}{5}\&2x-3y=54\)
\(\dfrac{x}{y}=\dfrac{3}{5}\Leftrightarrow5x=3y\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{5}\)
\(\Leftrightarrow\dfrac{2x}{2.3}=\dfrac{3y}{3.5}\Leftrightarrow\dfrac{2x}{6}=\dfrac{3y}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{2x}{6}=\dfrac{3y}{15}=\dfrac{2x-3y}{6-15}=\dfrac{54}{-9}=-6\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=-6\\\dfrac{y}{5}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-18\\y=-30\end{matrix}\right.\)
\(\dfrac{x}{y}=\dfrac{3}{5}\) và 2x-3y=54
Ta có: \(\dfrac{x}{y}=\dfrac{3}{5}\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{5}\)
+) \(\dfrac{x}{3}=\dfrac{2x}{6}\)(điều 1)
+) \(\dfrac{y}{5}=\dfrac{3y}{15}\)(điều 2 )
\(\Rightarrow\)\(\dfrac{2x}{6}=\dfrac{3y}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{6}=\dfrac{3y}{15}=\dfrac{2x-3y}{6-15}=\dfrac{54}{-9}=-6\)
\(\Rightarrow x=-18;y=-30\)
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