Đáp án:
2) b) \(m > - 1\)
Giải thích các bước giải:
\(\begin{array}{l}
1)Do:\left( d \right)//\left( {d'} \right)\\
\to \left\{ \begin{array}{l}
2a - 1 = 3\\
2 \ne 8
\end{array} \right.\\
\to 2a = 4\\
\to a = 2\\
2)\left\{ \begin{array}{l}
x = 2y + m\\
2\left( {2y + m} \right) + y = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 2y + m\\
5y = 3 - 2m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{3 - 2m}}{5}\\
x = 2.\dfrac{{3 - 2m}}{5} + m = \dfrac{{6 - 4m + 5m}}{5}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{3 - 2m}}{5}\\
x = \dfrac{{m + 6}}{5}
\end{array} \right.\\
a)Thay:m = - 1\\
\to \left\{ \begin{array}{l}
x = 1\\
y = 1
\end{array} \right.\\
b)\dfrac{{m + 6}}{5} > \dfrac{{3 - 2m}}{5}\\
\to 3m > - 3\\
\to m > - 1
\end{array}\)
