Question

\(\left\{{}\begin{matrix}2x+\dfrac{12}{y+3}=7\\3x-\dfrac{4}{y+3}=5\end{matrix}\right.\)

Answer 1

ĐKXĐ: y<>-3

\(\left\{{}\begin{matrix}2x+\dfrac{12}{y+3}=7\\3x-\dfrac{4}{y+3}=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+\dfrac{12}{y+3}=7\\9x-\dfrac{12}{y+3}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=22\\3x-\dfrac{4}{y+3}=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2\\\dfrac{4}{y+3}=3x-5=3\cdot2-5=6-5=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2\\y+3=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\left(nhận\right)\end{matrix}\right.\)