Question

9．已知$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=3}\\{y=4}\end{array}\right.$，则方程组$\left\{\begin{array}{l}{3{a}_{1}（x-1）+4{b}_{1}（y+3）={c}_{1}}\\{3{a}_{2}（x-1）+4{b}_{2}（y+3）={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=2}\\{y=-2}\end{array}\right.$．

Answer 1

分析 根据二元一次方程组的解，即可解答．

解答 解：将$\left\{\begin{array}{l}{x=3}\\{y=4}\end{array}\right.$代入$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$得：$\left\{\begin{array}{l}{3{a}_{1}+4{b}_{1}={c}_{1}}\\{3{a}_{2}+4{b}_{2}={c}_{2}}\end{array}\right.$，
将$\left\{\begin{array}{l}{3{a}_{1}+4{b}_{1}={c}_{1}}\\{3{a}_{2}+4{b}_{2}={c}_{2}}\end{array}\right.$代入方程组$\left\{\begin{array}{l}{3{a}_{1}（x-1）+4{b}_{1}（y+3）={c}_{1}}\\{3{a}_{2}（x-1）+4{b}_{2}（y+3）={c}_{2}}\end{array}\right.$得：
$\left\{\begin{array}{l}{x-1=1}\\{y+3=1}\end{array}\right.$
解得：$\left\{\begin{array}{l}{x=2}\\{y=-2}\end{array}\right.$，
故答案为：$\left\{\begin{array}{l}{x=2}\\{y=-2}\end{array}\right.$．

点评 本题考查了二元一次方程组的解，解决本题的关键是解二元一次方程组．