a. Ta có:
\(\left\{{}\begin{matrix}a+b=2\\3a+b=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a=6\\a+b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=2-3=-1\end{matrix}\right.\)
`=>y=3x-1`
b. Ta có:
\(\left\{{}\begin{matrix}2a+b=1\\4a+b=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a=-3\\2a+b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{3}{2}\\-3+b=1\end{matrix}\right. \Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{3}{2}\\b=1+3=4\end{matrix}\right.\)
`=>y=-3/2x+4`
c. Ta có:
\(\left\{{}\begin{matrix}0x+b=1\\4a+b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=1\\4a=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=1\\a=-\dfrac{1}{4}\end{matrix}\right.\)
`=>y=-1/4x+1`
d. Ta có:
\(\left\{{}\begin{matrix}a+b=1\\2a+b=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-3\\a+b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-3\\b=1-a=1+3=4\end{matrix}\right.\)
`=>y=-3x+4`
