题目内容

5.解方程组
(1)$\left\{\begin{array}{l}{3x-2y=9}\\{x+2y=-1}\end{array}\right.$
(2)$\left\{\begin{array}{l}{6x-3y=-3}\\{5x-9y=-35}\end{array}\right.$.

分析 (1)应用加减消元法,求出方程组$\left\{\begin{array}{l}{3x-2y=9}\\{x+2y=-1}\end{array}\right.$的解是多少即可.
(2)应用加减消元法,求出方程组$\left\{\begin{array}{l}{6x-3y=-3}\\{5x-9y=-35}\end{array}\right.$的解是多少即可.

解答 解:(1)$\left\{\begin{array}{l}{3x-2y=9…①}\\{x+2y=-1…②}\end{array}\right.$
①+②,可得4x=8,
解得x=2,
把x=2代入②,可得
2+2y=-1,
解得y=-1.5,
∴方程组的解为$\left\{\begin{array}{l}{x=2}\\{y=-1.5}\end{array}\right.$.

(2)$\left\{\begin{array}{l}{6x-3y=-3…①}\\{5x-9y=-35…②}\end{array}\right.$
①×3-②,可得13x=26,
解得x=2,
把x=2代入①,可得
6×2-3y=-3,
解得y=5,
∴方程组的解为$\left\{\begin{array}{l}{x=2}\\{y=5}\end{array}\right.$.

点评 此题主要考查了二元一次方程组的解法,要熟练掌握,解答此类问题的关键是注意代入消元法和加减消元法的应用.

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