Question

16．方程組$\left\{\begin{array}{l}2x-y=5\\ 3x-y=7\end{array}\right.$的解是（　�。�

• A． $\left\{\begin{array}{l}x=3\\ y=1\end{array}\right.$
• B． $\left\{\begin{array}{l}x=1\\ y=-4\end{array}\right.$
• C． $\left\{\begin{array}{l}x=2\\ y=-1\end{array}\right.$
• D． $\left\{\begin{array}{l}x=\frac{12}{5}\\ y=-\frac{1}{5}\end{array}\right.$

Answer 1

分析 方程組利用加減消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{2x-y=5①}\\{3x-y=7②}\end{array}\right.$，
②-①得：x=2，
把x=2代入①得：y=-1，
則方程組的解為$\left\{\begin{array}{l}{x=2}\\{y=-1}\end{array}\right.$，
故選C

點評 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．