题目内容

16.方程组$\left\{\begin{array}{l}2x-y=5\\ 3x-y=7\end{array}\right.$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.$\left\{\begin{array}{l}x=\frac{12}{5}\\ y=-\frac{1}{5}\end{array}\right.$

分析 方程组利用加减消元法求出解即可.

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,
则方程组的解为$\left\{\begin{array}{l}{x=2}\\{y=-1}\end{array}\right.$,
故选C

点评 此题考查了解二元一次方程组,利用了消元的思想,消元的方法有:代入消元法与加减消元法.

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