Question

10．二元一次方程组$\left\{\begin{array}{l}x-y=3\\ 2x+y=6\end{array}\right.$的解为$\left\{\begin{array}{l}x=3\\ y=3\end{array}\right.$．

Answer 1

分析 先用加减消元法求出x的值，再用代入消元法求出y的值即可．

解答 解：$\left\{\begin{array}{l}x-y=3①\\ 2x+y=6②\end{array}\right.$，①+②得，3x=9，解得x=3，把x=3代入①得，3-y=3，解得y=0，
故方程组的解为$\left\{\begin{array}{l}x=3\\ y=3\end{array}\right.$．
故答案为：$\left\{\begin{array}{l}x=3\\ y=3\end{array}\right.$．

点评 本题考查的是解二元一次方程组，熟知解二元一次方程组的加减消元法和代入消元法是解答此题的关键．