题目内容
20.方程组$\left\{\begin{array}{l}{2x-y=5}\\{3x+4y=2}\end{array}\right.$的解是( )| A. | $\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$ | B. | $\left\{\begin{array}{l}{x=-1}\\{y=2}\end{array}\right.$ | C. | $\left\{\begin{array}{l}{x=2}\\{y=-1}\end{array}\right.$ | D. | $\left\{\begin{array}{l}{x=-2}\\{y=1}\end{array}\right.$ |
分析 方程组利用加减消元法求出解即可.
解答 解:$\left\{\begin{array}{l}{2x-y=5①}\\{3x+4y=2②}\end{array}\right.$,
①×4+②得:11x=22,即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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