Question

Tìm \(x,y\) biết

\(\dfrac{2x}{3}=\dfrac{-y}{5}\) và\(x+5y=3\)

Answer 1

ta có : \(x+5y=3\Leftrightarrow x=3-5y\)

ta có : \(\dfrac{2x}{3}=\dfrac{-y}{5}\Leftrightarrow10x=-3y\Leftrightarrow10x+3y=0\)

\(\Leftrightarrow10\left(3-5y\right)+3y=0\Leftrightarrow30-50y+3y=0\)

\(\Leftrightarrow47y=30\Leftrightarrow y=\dfrac{30}{47}\)

\(\Rightarrow x+5.\dfrac{30}{47}=3\Leftrightarrow x=\dfrac{-9}{47}\)

vậy \(y=\dfrac{30}{47};x=\dfrac{-9}{47}\)

Answer 2

Từ \(\dfrac{2x}{3}=\dfrac{-y}{5}\Rightarrow2.5x=-3y\)

\(\Rightarrow10x=-3y\Rightarrow\dfrac{x}{-3}=\dfrac{y}{10}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{-3}=\dfrac{y}{10}=\dfrac{5y}{50}=\dfrac{x+5y}{-3+50}=\dfrac{3}{47}\)

\(\Rightarrow\left\{{}\begin{matrix}x=10.\dfrac{3}{47}=\dfrac{30}{47}\\y=-3.\dfrac{3}{47}=-\dfrac{9}{47}\end{matrix}\right.\)