Question

1．若方程组$\left\{\begin{array}{l}{2a-3b=13}\\{3a+5b=30.9}\end{array}\right.$的解为$\left\{\begin{array}{l}{a=8.3}\\{b=3.2}\end{array}\right.$，则方程组$\left\{\begin{array}{l}{2（x+2）-3（y-1）=13}\\{3（x+2）+5（y-1）=30.9}\end{array}\right.$的解为$\left\{\begin{array}{l}{x=6.3}\\{y=4.2}\end{array}\right.$．

Answer 1

分析 仿照已知方程组的解法求出所求方程组的解即可．

解答 解：∵方程组$\left\{\begin{array}{l}{2a-3b=13}\\{3a+5b=30.9}\end{array}\right.$的解为$\left\{\begin{array}{l}{a=8.3}\\{b=3.2}\end{array}\right.$，
∴由方程组$\left\{\begin{array}{l}{2（x+2）-3（y-1）=13}\\{3（x+2）+5（y-1）=30.9}\end{array}\right.$可得：$\left\{\begin{array}{l}{x+2=8.3}\\{y-1=3.2}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{x=6.3}\\{y=4.2}\end{array}\right.$，
故答案为：$\left\{\begin{array}{l}{x=6.3}\\{y=4.2}\end{array}\right.$

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．