Question

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

• 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 根據(jù)對應(yīng)相等，由方程組$\left\{\begin{array}{l}{{a}_{1}x+_{1}y={c}_{1}}\\{{a}_{2}x+_{2}y={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=2}\\{y=-3}\end{array}\right.$，可以得到方程組$\left\{\begin{array}{l}{{a}_{1}（x-1）+_{1}（y+2）={c}_{1}}\\{{a}_{2}（x-1）+_{2}（y+2）={c}_{2}}\end{array}\right.$的解，本題得以解決．

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

點評 本題考查二元一次方程組的解，解答本題的關(guān)鍵是明確解二元一次方程組的方法，找準對應(yīng)關(guān)系．