Question

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

Answer 1

分析 根据二元一次方程组的解的定义得到3x=3，2（y-1）=4，于是得到结论．

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

点评 本题考查了二元一次方程组的解，熟练掌握二元一次方程组的解的定义是解题的关键．