Sửa lại đề \(T=2x+5y\)
\(\left\{{}\begin{matrix}2x+y=4\\6x-5y=-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x+5y=20\\6x-5y=-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}16x=8\\2x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=4-2x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=3\end{matrix}\right.\)
\(\Rightarrow T=2x+5y=2.\dfrac{1}{2}+5.3=16\)
có hệ pt :
\(\left\{{}\begin{matrix}2x+y=4\\6x-5y=-12\end{matrix}\right.\)
\(\left\{{}\begin{matrix}y=4-2x\\6x-5y=-12\end{matrix}\right.\)
\(\left\{{}\begin{matrix}y=4-2x\\6x-5\left(4-2x\right)=-12\end{matrix}\right.\)
\(\left\{{}\begin{matrix}y=4-2x\\6x-20+10x=-12\end{matrix}\right.\)
\(\left\{{}\begin{matrix}y=4-2x\\16x=8\end{matrix}\right.\)
\(\left\{{}\begin{matrix}y=4-2\cdot\dfrac{1}{2}=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Thay `x = 1/2 ; y = 3` vào `T` ta có :(T sửa đề `T=2x + 5y`)
`T = 2*1/2 + 5*3`
`T = 1 + 15`
`T = 16`
Vậy hệ PT có nghiện `(x,y)` là `(1/2 , 3)` thì `T = 16`
