题目内容

16.解方程组$\left\{\begin{array}{l}{5(x+y)-3(x-y)=2}\\{2(x+y)+4(x-y)=6}\end{array}\right.$若设(x+y)=A,(x-y)=B,则原方程组可变形为$\left\{\begin{array}{l}{5A-3B=2}\\{2A+4B=6}\end{array}\right.$,解方程组得$\left\{\begin{array}{l}{A=1}\\{B=1}\end{array}\right.$,所以$\left\{\begin{array}{l}{x+y=1}\\{x-y=1}\end{array}\right.$解方程组得$\left\{\begin{array}{l}{x=1}\\{y=0}\end{array}\right.$,我们把某个式子看成一个整体,用一个字母去代替它,这种解方程组的方法叫换元法,请用这种方法解方程组$\left\{\begin{array}{l}{\frac{x+y}{2}+\frac{x-y}{3}=6}\\{2(x+y)-3x+3y=24}\end{array}\right.$.

分析 设x+y=A,x-y=B,方程变形后,利用代入消元法求出A与B的值,进而确定出x与y的值即可.

解答 解:设x+y=A,x-y=B,
方程组变形得:$\left\{\begin{array}{l}{\frac{A}{2}+\frac{B}{3}=6}\\{2A-3B=24}\end{array}\right.$,
整理得:$\left\{\begin{array}{l}{3A+2B=36①}\\{2A-3B=24②}\end{array}\right.$,
①×3+②×2得:13A=156,即A=12,
把A=12代入②得:B=0,
∴$\left\{\begin{array}{l}{x+y=12}\\{x-y=0}\end{array}\right.$,
解得:$\left\{\begin{array}{l}{x=6}\\{y=6}\end{array}\right.$.

点评 此题考查了解二元一次方程组,利用了消元的思想,消元的方法有:代入消元法与加减消元法.

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