Question

Tìm x, y ∈ Q biết: \(\dfrac{1-x}{3}=\dfrac{2y-1}{8}\) và 2x + y = 6

Answer 1

\(\dfrac{1-x}{3}=\dfrac{2y-1}{8}\)

=>8(1-x)=3(2y-1)

=>8-8x=6y-3

=>-8x-6y=-11

=>8x+6y=11

mà 2x+y=6

nên ta có hệ phương trình:

\(\left\{{}\begin{matrix}8x+6y=11\\2x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x+6y=11\\8x+4y=24\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2y=-13\\2x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{13}{2}\\2x=6-y=6+\dfrac{13}{2}=\dfrac{25}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{25}{4}\\y=-\dfrac{13}{2}\end{matrix}\right.\)