Question

Giải hệ pt: \(\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{2}\\\dfrac{x+4}{y+8}=\dfrac{4}{9}\end{matrix}\right.\)

Answer 1

ĐKXĐ: \(y\ne-8\)

\(\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{2}\\\dfrac{x+4}{y+8}=\dfrac{4}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=3y\\9\left(x+4\right)=4\left(y+8\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-3y=0\\9x+36=4y+32\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-3y=0\\9x-4y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}18x-27y=0\\18x-8y=-8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}18x-27y-18x+8y=0+8\\2x=3y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-19y=8\\x=\dfrac{3}{2}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{8}{19}\\x=\dfrac{3}{2}\cdot\dfrac{-8}{19}=\dfrac{-12}{19}\end{matrix}\right.\)