a) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{-15}{5}=-3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).3=-9\end{matrix}\right.\)
b) \(3x=7y\Rightarrow\dfrac{x}{7}=\dfrac{y}{3}=\dfrac{x-y}{7-3}=\dfrac{-16}{4}=-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\left(-4\right).7=-28\\y=\left(-4\right).3=-12\end{matrix}\right.\)
c) Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=k\Rightarrow x=3k;y=4k\)
\(\Leftrightarrow xy=3k.4k=12k^2=12\)
\(\Rightarrow k=\left\{-1;1\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=3;y_1=4\\x_2=-3;y_2=-4\end{matrix}\right.\)
a, \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{-15}{5}=-3\)
<=> \(\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).3=-9\end{matrix}\right.\)
@Phạm Hải Minh
b, 3x = 7y => \(\dfrac{x}{7}=\dfrac{y}{3}=\dfrac{x-y}{7-3}=\dfrac{-16}{4}=-4\)
<=> \(\left\{{}\begin{matrix}x=\left(-4\right).7=-28\\y=\left(-4\right).3=-12\end{matrix}\right.\)
@Phạm Hải Minh
c, Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=k\Rightarrow x=3k;y=4k\)
<=> xy = 3k.4k = 12k2 = 12
=> k = {-1; 1}
=> \(\left\{{}\begin{matrix}x_1=3;y_1=4\\x_2=-3;y_2=-4\end{matrix}\right.\)
@Phạm Hải Minh