题目内容
17.解下列方程组(1)$\left\{\begin{array}{l}2x-y=5\\ x+4y=2\end{array}\right.$
(2)$\left\{\begin{array}{l}3x+2y=7\\ 6x-2y=11\end{array}\right.$.
分析 (1)方程组利用加减消元法求出解即可;
(2)方程组利用加减消元法求出解即可.
解答 解:(1)$\left\{\begin{array}{l}{2x-y=5①}\\{x+4y=2②}\end{array}\right.$,
①×4+②得:9x=22,即x=$\frac{22}{9}$,
把x=$\frac{22}{9}$代入②得:y=-$\frac{1}{9}$,
则方程组的解为$\left\{\begin{array}{l}{x=\frac{22}{9}}\\{y=-\frac{1}{9}}\end{array}\right.$;
(2)$\left\{\begin{array}{l}{3x+2y=7①}\\{6x-2y=11②}\end{array}\right.$,
①+②得:9x=18,即x=2,
把x=2代入①得:y=$\frac{1}{2}$,
则方程组的解为$\left\{\begin{array}{l}{x=2}\\{y=\frac{1}{2}}\end{array}\right.$.
点评 此题考查了解二元一次方程组,熟练掌握运算法则是解本题的关键.
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