Question

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

Answer 1

分析 由方程組$\left\{\begin{array}{l}{2a-3b=m}\\{3a+5b=n}\end{array}\right.$和方程組$\left\{\begin{array}{l}{2（x-1）-3（y+2）=m}\\{3（x-1）+5（y+2）=n}\end{array}\right.$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，本題得以解決．

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

點評 本題考查二元一次方程組的解，解題的關(guān)鍵是運用整體的數(shù)學(xué)思想解答問題．