题目内容
1．二元一次方程组$\left\{\begin{array}{l}{x+y=4}\\{x-y=2}\end{array}\right.$的解是（ ）| A． | $\left\{\begin{array}{l}{x=1}\\{y=3}\end{array}\right.$ | B． | $\left\{\begin{array}{l}{x=3}\\{y=1}\end{array}\right.$ | C． | $\left\{\begin{array}{l}{x=1}\\{y=-2}\end{array}\right.$ | 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：$\left\{\begin{array}{l}{x+y=4①}\\{x-y=2②}\end{array}\right.$，
①+②得：2x=6，即x=3，
把x=3代入①得：y=1，
则方程组的解为$\left\{\begin{array}{l}{x=3}\\{y=1}\end{array}\right.$，
故选B
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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