Question

17．若方程组$\left\{\begin{array}{l}{a_1}x+{b_1}y={c_1}\\{a_2}x+{b_2}y={c_2}\end{array}\right.$的解是$\left\{\begin{array}{l}x=3\\ y=4\end{array}\right.$，则方程组$\left\{\begin{array}{l}\frac{1}{2}{a_1}x+\frac{1}{3}{b_1}y={c_1}\\ \frac{1}{2}{a_2}x+\frac{1}{3}{b_2}y={c_2}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=6}\\{y=12}\end{array}\right.$．

Answer 1

分析 根据题意列出x与y的方程，分别求出方程的解即可得到方程组的解．

解答 解：方程组$\left\{\begin{array}{l}\frac{1}{2}{a_1}x+\frac{1}{3}{b_1}y={c_1}\\ \frac{1}{2}{a_2}x+\frac{1}{3}{b_2}y={c_2}\end{array}\right.$可变形为$\left\{\begin{array}{l}{{a}_{1}•\frac{1}{2}x+{b}_{1}•\frac{1}{3}y={c}_{1}}\\{{a}_{2}•\frac{1}{2}x+{b}_{2}•\frac{1}{3}y={c}_{2}}\end{array}\right.$，
∵$\left\{\begin{array}{l}{a_1}x+{b_1}y={c_1}\\{a_2}x+{b_2}y={c_2}\end{array}\right.$的解是$\left\{\begin{array}{l}x=3\\ y=4\end{array}\right.$，
∴$\frac{1}{2}$x=3，$\frac{1}{3}$y=4，
∴x=6，y=12，
∴方程组$\left\{\begin{array}{l}\frac{1}{2}{a_1}x+\frac{1}{3}{b_1}y={c_1}\\ \frac{1}{2}{a_2}x+\frac{1}{3}{b_2}y={c_2}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=6}\\{y=12}\end{array}\right.$，
故答案为：$\left\{\begin{array}{l}{x=6}\\{y=12}\end{array}\right.$．

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成立的未知数的值．弄清题意是解本题的关键．