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$\left\{\begin{array}{l}{x=3}\\{y=4}\end{array}\right.$，老师让同学们解方程组$\left\{\begin{array}{l}{3{a}_{1}x+4{b}_{1y}=5{c}_{1}}\\{3{a}_{2}x+4{b}_{2}y=5{c}_{2}}\end{array}\right.$，小聪先觉得这道题好像条件不够，后将方程组中的两个方程同除以5，整理得$\left\{\begin{array}{l}{{a}_{1}•\frac{3}{5}x+{b}_{1}•\frac{4}{5}y={c}_{1}}\\{{a}_{2}•\frac{3}{5}x+{b}_{2}\frac{4}{5}y={c}_{2}}\end{array}\right.$，运用换元思想，得$\left\{\begin{array}{l}{\frac{3}{5}x=3}\\{\frac{4}{5}y=4}\end{array}\right.$，所以方程组$\left\{\begin{array}{l}{3{a}_{1}x+4{b}_{1}y=5{c}_{1}}\\{3{a}_{2}x+4{b}_{2}y=5{c}_{2}}\end{array}\right.$的解为$\left\{\begin{array}{l}{x=5}\\{y=5}\end{array}\right.$，即得出方程组$\left\{\begin{array}{l}{{a}_{1}x-{b}_{1}y=m}\\{{a}_{2}x-{b}_{2}y=n}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=8}\\{y=10}\end{array}\right.$，请你求出方程组$\left\{\begin{array}{l}{{a}_{1}（x-2）-{b}_{1}（y+1）=m}\\{{a}_{2}（x-2）-{b}_{2}（y+1）=n}\end{array}\right.$的解．

Answer 1

分析 根据示例，运用换元思想，即可列出简易方程组，很容易求出方程组的解．

解答 解：∵$\left\{\begin{array}{l}{{a}_{1}x-{b}_{1}y=m}\\{{a}_{2}x-{b}_{2}y=n}\end{array}\right.$，
$\left\{\begin{array}{l}{{a}_{1}（x-2）-{b}_{1}（y+1）=m}\\{{a}_{2}（x-2）-{b}_{2}（y+1）=n}\end{array}\right.$，
又∵$\left\{\begin{array}{l}{{a}_{1}x-{b}_{1}y=m}\\{{a}_{2}x-{b}_{2}y=n}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=8}\\{y=10}\end{array}\right.$，
∴$\left\{\begin{array}{l}{x-2=8}\\{y+1=10}\end{array}\right.$，
解得$\left\{\begin{array}{l}{x=10}\\{y=9}\end{array}\right.$．

点评 考查了二元一次方程组的解，本题给出了一些材料，考查了同学们的阅读分析能力，需要同学们有一定的逻辑分析能力．