Question

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

Answer 1

分析 由方程组$\left\{\begin{array}{l}{2a-3b=m}\\{3a+5b=n}\end{array}\right.$和方程组$\left\{\begin{array}{l}{2（x-1）-3（y+2）=m}\\{3（x-1）+5（y+2）=n}\end{array}\right.$可知，a=x-1，b=y+2，由$\left\{\begin{array}{l}{a=3}\\{b=-1}\end{array}\right.$，可以求得x、y的值，本题得以解决．

解答 解：∵方程组$\left\{\begin{array}{l}{2a-3b=m}\\{3a+5b=n}\end{array}\right.$和方程组$\left\{\begin{array}{l}{2（x-1）-3（y+2）=m}\\{3（x-1）+5（y+2）=n}\end{array}\right.$，
∴x-1=a，y+2=b，
∵方程组$\left\{\begin{array}{l}{2a-3b=m}\\{3a+5b=n}\end{array}\right.$的解是$\left\{\begin{array}{l}{a=3}\\{b=-1}\end{array}\right.$，
∴$\left\{\begin{array}{l}{x-1=3}\\{y+2=-1}\end{array}\right.$，
解得$\left\{\begin{array}{l}{x=4}\\{y=-3}\end{array}\right.$．
故答案为：$\left\{\begin{array}{l}{x=4}\\{y=-3}\end{array}\right.$．

点评 本题考查二元一次方程组的解，解题的关键是运用整体的数学思想解答问题．