Question

11．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$\left\{\begin{array}{l}{x=5}\\{y=-2}\end{array}\right.$，则方程组$\left\{\begin{array}{l}{{a}_{1}（x-3）+{b}_{1}（y+1）={c}_{1}}\\{{a}_{2}（x-3）+{b}_{2}（y+1）={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=8}\\{y=-3}\end{array}\right.$．

Answer 1

分析 根据已知方程组的解求出x与y的值，即为所求方程组的解．

解答 解：根据题意得：x-3=5，y+1=-2，
解得：x=8，y=-3，
则方程组$\left\{\begin{array}{l}{{a}_{1}（x-3）+{b}_{1}（y+1）={c}_{1}}\\{{a}_{2}（x-3）+{b}_{2}（y+1）={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=8}\\{y=-3}\end{array}\right.$．
故答案为：$\left\{\begin{array}{l}{x=8}\\{y=-3}\end{array}\right.$．

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．