Question

5．若$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=3}\\{y=4}\end{array}\right.$，则$\left\{\begin{array}{l}{3{a}_{1}（x-1）+{b}_{1}（y+3）={c}_{1}}\\{3{a}_{2}（x-1）+{b}_{2}（y+3）={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$．

Answer 1

分析 根据已知得出方程组$\left\{\begin{array}{l}{x-1=3}\\{y+3=4}\end{array}\right.$，求出方程组的解即可．

解答 解：∵$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=3}\\{y=4}\end{array}\right.$，
∴$\left\{\begin{array}{l}{x-1=3}\\{y+3=4}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$，
即$\left\{\begin{array}{l}{3{a}_{1}（x-1）+{b}_{1}（y+3）={c}_{1}}\\{3{a}_{2}（x-1）+{b}_{2}（y+3）={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$，
故答案为：$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$．

点评 本题考查了二元一次方程组的解和解二元一次方程组，能根据题意得出关于x、y的方程组是解此题的关键．