题目内容
17.解方程:$\left\{\begin{array}{l}{2x-y=7}\\{x+2y=6}\end{array}\right.$.分析 解决本题关键是寻找式子间的关系,寻找方法消元.可想法把y的系数化为相反数,然后用加法化去,达到消元的目的.
解答 解:$\left\{\begin{array}{l}{2x-y=7①}\\{x+2y=6②}\end{array}\right.$,
①×2+②得5x=20,
解得x=4,
把x=4代入①得8-y=7,
解得y=1.
故原方程组的解为$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$.
点评 本题考查二元一次方程组的解法,有加减法和代入法两种,一般选用加减法解二元一次方程组较简单.
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