Question

15．方程組$\left\{\begin{array}{l}x-5y=2\\ 2x+3y=9\end{array}\right.$的解為$\left\{\begin{array}{l}{x=\frac{41}{13}}\\{y=\frac{5}{13}}\end{array}\right.$．

Answer 1

分析 方程組利用代入消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{x-5y=2①}\\{2x+3y=9②}\end{array}\right.$，
由①變形得：x=5y+2③，
把③代入②得：10y+4+3y=9，即y=$\frac{5}{13}$，
把y=$\frac{5}{13}$代入③得：x=$\frac{41}{13}$，
則方程組的解為$\left\{\begin{array}{l}{x=\frac{41}{13}}\\{y=\frac{5}{13}}\end{array}\right.$．
故答案為：$\left\{\begin{array}{l}{x=\frac{41}{13}}\\{y=\frac{5}{13}}\end{array}\right.$

點(diǎn)評 此題考查了解二元一次方程組，利用了消元的思想，消元的方法有：代入消元法與加減消元法．