Question

9．已知方程组$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=2}\\{y=-3}\end{array}\right.$，那么方程组$\left\{\begin{array}{l}{{a}_{1}（x-1）+{b}_{1}（y+2）={c}_{1}}\\{{a}_{2}（x-1）+{b}_{2}（y+2）={c}_{2}}\end{array}\right.$的解为（　　）

• A． $\left\{\begin{array}{l}{x=2}\\{y=-3}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=3}\\{y=-5}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=1}\\{y=-5}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=3}\\{y=-1}\end{array}\right.$

Answer 1

分析 根据对应相等，由方程组$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=2}\\{y=-3}\end{array}\right.$，可以得到方程组$\left\{\begin{array}{l}{{a}_{1}（x-1）+{b}_{1}（y+2）={c}_{1}}\\{{a}_{2}（x-1）+{b}_{2}（y+2）={c}_{2}}\end{array}\right.$的解，本题得以解决．

解答 解：∵方程组$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=2}\\{y=-3}\end{array}\right.$，
∴在方程组$\left\{\begin{array}{l}{{a}_{1}（x-1）+{b}_{1}（y+2）={c}_{1}}\\{{a}_{2}（x-1）+{b}_{2}（y+2）={c}_{2}}\end{array}\right.$中，$\left\{\begin{array}{l}{x-1=2}\\{y+2=-3}\end{array}\right.$，
解得，$\left\{\begin{array}{l}{x=3}\\{y=-5}\end{array}\right.$，
故选B．

点评 本题考查二元一次方程组的解，解答本题的关键是明确解二元一次方程组的方法，找准对应关系．