a) Ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5k\\y=7k\end{matrix}\right.\)
Do \(x-y=-10\Leftrightarrow5k-7k=-10\)
\(\Leftrightarrow\left(5-7\right)k=-10\)
\(\Leftrightarrow\left(-2\right)k=-10\)
\(\Leftrightarrow k=\left(-10\right):\left(-2\right)=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=5k=5.5=25\\y=7k=7.5=35\end{matrix}\right.\)
b) Xét \(3x=4y\Leftrightarrow\dfrac{x}{4}=\dfrac{y}{3}=m\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4m\\y=3m\end{matrix}\right.\)
Do \(y+x=14\Leftrightarrow3m+4m=14\)
\(\Leftrightarrow\left(3+4\right)m=14\)
\(\Leftrightarrow7m=14\)
\(\Leftrightarrow m=14:7=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=4m=4.2=8\\y=3m=3.2=6\end{matrix}\right.\)
c) Ta có \(\dfrac{4}{x}=\dfrac{2}{y}=n\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{n}\\y=\dfrac{2}{n}\end{matrix}\right.\)
Do \(2x-y=12\Leftrightarrow2.\dfrac{4}{n}-\dfrac{2}{n}=12\)
\(\Leftrightarrow\dfrac{8}{n}-\dfrac{2}{n}=12\)
\(\Leftrightarrow\dfrac{6}{n}=12\)
\(\Leftrightarrow n=\dfrac{6}{12}=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{n}=\dfrac{4}{\dfrac{1}{2}}=8\\y=\dfrac{2}{n}=\dfrac{2}{\dfrac{1}{2}}=4\end{matrix}\right.\)
