题目内容

6.解下列方程组
(1)$\left\{\begin{array}{l}{m-\frac{n}{2}=2}\\{2m+3n=12}\end{array}\right.$               
(2)$\left\{\begin{array}{l}{2x+3y=15}\\{\frac{x+1}{7}=\frac{y+4}{5}}\end{array}\right.$.

分析 (1)方程组整理后,利用加减消元法求出解即可;
(2)方程组整理后,利用加减消元法求出解即可.

解答 解:(1)方程组整理得:$\left\{\begin{array}{l}{2m-n=4①}\\{2m+3n=12②}\end{array}\right.$,
②-①得:4n=8,即n=2,
把n=2代入①得:m=3,
则方程组的解为$\left\{\begin{array}{l}{m=3}\\{n=2}\end{array}\right.$;
(2)方程组整理得:$\left\{\begin{array}{l}{2x+3y=15①}\\{5x-7y=23②}\end{array}\right.$,
①×7+②×3得:29x=174,即x=6,
把x=6代入①得:y=1,
则方程组的解为$\left\{\begin{array}{l}{x=6}\\{y=1}\end{array}\right.$.

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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