Question

12．若方程组$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解为$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$，那么方程组$\left\{\begin{array}{l}{2{a}_{1}x+3{b}_{1}y=5{c}_{1}}\\{2{a}_{2}x+3{b}_{2}y=5{c}_{2}}\end{array}\right.$的解为$\left\{\begin{array}{l}{x=5}\\{y=5}\end{array}\right.$．

Answer 1

分析 根据方程组$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解为$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$，代入变形后即可建立与方程组$\left\{\begin{array}{l}{2{a}_{1}x+3{b}_{1}y=5{c}_{1}}\\{2{a}_{2}x+3{b}_{2}y=5{c}_{2}}\end{array}\right.$的关系，从而可以解答本题．

解答 解：∵方程组$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解为$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$，
∴$\left\{\begin{array}{l}{2{a}_{1}+3{b}_{1}={c}_{1}}\\{2{a}_{2}+3{b}_{2}={c}_{2}}\end{array}\right.$
∴$\left\{\begin{array}{l}{2{a}_{1}×5+3{b}_{1}×5={c}_{1}×5}\\{2{a}_{2}×5+3{b}_{2}×5={c}_{2×5}}\end{array}\right.$
即$\left\{\begin{array}{l}{2{a}_{1}×5+3{b}_{1}×5=5{c}_{1}}\\{2{a}_{2}×5+3{b}_{2}×5=5{c}_{2}}\end{array}\right.$，
∴方程组$\left\{\begin{array}{l}{2{a}_{1}x+3{b}_{1}y=5{c}_{1}}\\{2{a}_{2}x+3{b}_{2}y=5{c}_{2}}\end{array}\right.$的解为：$\left\{\begin{array}{l}{x=5}\\{y=5}\end{array}\right.$，
故答案为：$\left\{\begin{array}{l}{x=5}\\{y=5}\end{array}\right.$．

点评 本题考查二元一次方程的解，解题的关键是利用数学中转化的思想建立已知条件与所求问题之间的关系．