题目内容
18.方程组$\left\{\begin{array}{l}{x-y=0}\\{x+y=2}\end{array}\right.$的解为( )| A. | $\left\{\begin{array}{l}{x=1}\\{y=-1}\end{array}\right.$ | B. | $\left\{\begin{array}{l}{x=-1}\\{y=1}\end{array}\right.$ | C. | $\left\{\begin{array}{l}{x=2}\\{y=0}\end{array}\right.$ | D. | $\left\{\begin{array}{l}{x=1}\\{y=1}\end{array}\right.$ |
分析 方程组利用加减消元法求出解即可.
解答 解:$\left\{\begin{array}{l}{x-y=0①}\\{x+y=2②}\end{array}\right.$,
①+②得:2x=2,即x=1,
①-②得:-2y=-2,即y=1,
则方程组的解为$\left\{\begin{array}{l}{x=1}\\{y=1}\end{array}\right.$,
故选D
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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