a) \(\left\{{}\begin{matrix}3x=4y\\x+y=-56\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}3x-4y=0\\x+y=-56\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}3x-4y=0\\4x+4y=-224\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}7x=-224\\x+y=-56\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{-224}{7}=-32\\x+y=-56\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-32\\-32+y=-56\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-32\\y=-24\end{matrix}\right.\) vậy \(x=-32;y=-24\)
b/ Theo đề ta có: \(\dfrac{x}{4}=\dfrac{x}{5}\Rightarrow\dfrac{x}{12}=\dfrac{x}{15}\)
\(\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{12}=\dfrac{y}{8}\)
\(\Rightarrow\dfrac{x}{12}=\dfrac{y}{8}=\dfrac{z}{15}\) và \(x+y-z=10\)
Áp dụng t/c của dãy tỉ số = nhau ta có:
\(\dfrac{x}{12}=\dfrac{y}{8}=\dfrac{z}{15}=\dfrac{x+y-z}{12+8-15}=\dfrac{10}{5}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot12=24\\y=2\cdot8=16\\z=2\cdot15=30\end{matrix}\right.\)
Vậy.............
a) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}3x=4y\\x+y=-56\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x-4y=0\\x+y=-56\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x-4y=0\\4x+4y=-224\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}7x=-224\\x+y=-56\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\dfrac{-224}{7}=-32\\x+y=-56\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-32\\-32+y=-56\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-32\\y=-24\end{matrix}\right.\) vậy x= -32; y= -24
