`2x-3y=1`
`<=>3y=2x-1`
`<=>x=(2x-1)/3` thay vào pt dưới ta có:
`x^2-x*(2x-1)/3=24`
`<=>x^2-2x^2/3+x/3=24`
`<=>x^2/3+x/3=24`
`<=>x^2+x-72=0`
`<=>(x-8)(x+9)=0`
`<=>[(x=8),(x=-9):}`
`<=>[({(x=8),(y=5):}),({(x=-9),(y=-19/3):}):}`
\(\left\{{}\begin{matrix}2x-3y=1\\x^2-xy=24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=3y+1\\x^2-xy=24\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1,5y+0,5\\\left(1,5y+0,5\right)^2-y\left(1,5y+0,5\right)-24=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1,5y+0,5\\2,25y^2+1,5y+0,25-1,5y^2-0,5y-24=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1,5y+0,5\\0,75y^2+y-23,75=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1,5y+0,5\\\left[{}\begin{matrix}y=5\\y=-\dfrac{19}{3}\end{matrix}\right.\end{matrix}\right.\)
Khi y=5 thì \(x=1,5\cdot5+0,5=7,5+0,5=8\)
Khi y=-19/3 thì \(x=\dfrac{3}{2}\cdot\dfrac{-19}{3}+\dfrac{1}{2}=-\dfrac{19}{2}+\dfrac{1}{2}=-\dfrac{18}{2}=-9\)
