Question

6．若方程组$\left\{\begin{array}{l}2a-3b=-1\\ 3a+5b=27\end{array}\right.$的解集是$\left\{\begin{array}{l}a=4\\ b=3\end{array}\right.$，则方程组$\left\{\begin{array}{l}2（x+2）-3（y-1）=-1\\ 3（x+2）+5（y-1）=27\end{array}\right.$的解是$\left\{\begin{array}{l}{x=2}\\{y=4}\end{array}\right.$．

Answer 1

分析 仿照已知方程组的解求出x与y的值，即可确定出所求方程组的解．

解答 解：根据方程组$\left\{\begin{array}{l}2a-3b=-1\\ 3a+5b=27\end{array}\right.$的解集是$\left\{\begin{array}{l}a=4\\ b=3\end{array}\right.$，
方程组$\left\{\begin{array}{l}2（x+2）-3（y-1）=-1\\ 3（x+2）+5（y-1）=27\end{array}\right.$中，可得$\left\{\begin{array}{l}{x+2=4}\\{y-1=3}\end{array}\right.$，
解得：$\left\{\begin{array}{l}x=2\\ y=4\end{array}\right.$，
故答案为：$\left\{\begin{array}{l}{x=2}\\{y=4}\end{array}\right.$

点评 此题考查了二元一次方程组的解，方程组的解即为能使方程组中两方程都成立的未知数的值．