题目内容
5.解方程组:(1)$\left\{\begin{array}{l}{2x+5y=8}\\{3x+2y=5}\end{array}\right.$;
(2)$\left\{\begin{array}{l}{2x+3y=6}\\{3x-2y=-2}\end{array}\right.$.
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)$\left\{\begin{array}{l}{2x+5y=8①}\\{3x+2y=5②}\end{array}\right.$,
①×3-②×2得:11y=14,即y=$\frac{14}{11}$,
把y=$\frac{14}{11}$代入①得:x=$\frac{9}{11}$,
则方程组的解为$\left\{\begin{array}{l}{x=\frac{9}{11}}\\{y=\frac{14}{11}}\end{array}\right.$;
(2)$\left\{\begin{array}{l}{2x+3y=6①}\\{3x-2y=-2②}\end{array}\right.$,
①×2+②×3得:13x=6,即x=$\frac{6}{13}$,
把x=$\frac{6}{13}$代入①得:y=$\frac{22}{13}$,
则方程组的解为$\left\{\begin{array}{l}{x=\frac{6}{13}}\\{y=\frac{22}{13}}\end{array}\right.$.
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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