Question

13．对于方程组$\left\{\begin{array}{l}x+y=4…（1）\\ x-y=2…（2）\end{array}\right.$，在$\left\{\begin{array}{l}x=-1\\ y=5\end{array}\right.$，$\left\{\begin{array}{l}x=3\\ y=1\end{array}\right.$，$\left\{\begin{array}{l}x=6\\ y=4\end{array}\right.$这3对数值中，$\left\{\begin{array}{l}x=-1\\ y=5\end{array}\right.$，$\left\{\begin{array}{l}x=3\\ y=1\end{array}\right.$是方程①的解．$\left\{\begin{array}{l}x=3\\ y=1\end{array}\right.$，$\left\{\begin{array}{l}x=6\\ y=4\end{array}\right.$是方程②的解．

Answer 1

分析 把各组数值代入两方程进行检验即可．

解答 解：对于$\left\{\begin{array}{l}x=-1\\ y=5\end{array}\right.$，代入①得，-1+5=4成立，代入②得，-1-5=-6≠2，故此对数值是方程①的解；
对于$\left\{\begin{array}{l}x=3\\ y=1\end{array}\right.$，代入①得，3+1=4成立，代入②得，3-1=2成立，故此对数值是方程①②的解；
对于$\left\{\begin{array}{l}x=6\\ y=4\end{array}\right.$，代入①得，6+4=10≠4，代入②得，6-4=2，故此对数值是方程②的解．
故答案为：$\left\{\begin{array}{l}x=-1\\ y=5\end{array}\right.$，$\left\{\begin{array}{l}x=3\\ y=1\end{array}\right.$；$\left\{\begin{array}{l}x=3\\ y=1\end{array}\right.$，$\left\{\begin{array}{l}x=6\\ y=4\end{array}\right.$．

点评 本题考查的是二元一次方程组的解，熟知一般地，二元一次方程组的两个方程的公共解，叫做二元一次方程组的解是解答此题的关键．