题目内容
13.方程组$\left\{\begin{array}{l}{x+y=10}\\{2x+y=16}\end{array}\right.$的解是( )| A. | $\left\{\begin{array}{l}{x=6}\\{y=4}\end{array}\right.$ | B. | $\left\{\begin{array}{l}{x=4}\\{y=8}\end{array}\right.$ | C. | $\left\{\begin{array}{l}{x=4}\\{y=6}\end{array}\right.$ | D. | $\left\{\begin{array}{l}{x=9}\\{y=-1}\end{array}\right.$ |
分析 方程组利用代入消元法求出解即可.
解答 解:$\left\{\begin{array}{l}{x+y=10①}\\{2x+y=16②}\end{array}\right.$,
②-①得:x=6,
把x=6代入①得:y=4,
则方程组的解为$\left\{\begin{array}{l}{x=6}\\{y=4}\end{array}\right.$,
故选A
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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