Question

13．阅读填空：对于方程组$\left\{\begin{array}{l}{4（x+y）-3（x-y）=14}\\{\frac{x+y}{2}+\frac{x-y}{3}=6}\end{array}\right.$，不妨设$\frac{x+y}{2}=u，\frac{x-y}{3}=v$，则原方程组变为以u，v为未知数的方程组$\left\{\begin{array}{l}{8u-9v=14}\\{u+v=6}\end{array}\right.$，解得$\left\{\begin{array}{l}{u=4}\\{v=2}\end{array}\right.$，从而原方程组的解是$\left\{\begin{array}{l}{x=7}\\{y=1}\end{array}\right.$，这种解法称之为“换元法”

Answer 1

分析 根据设出的u与v，将方程组变形，求出解确定出u与v的值，进而求出x与y的值．

解答 解：阅读填空：对于方程组$\left\{\begin{array}{l}{4（x+y）-3（x-y）=14}\\{\frac{x+y}{2}+\frac{x-y}{3}=6}\end{array}\right.$，
不妨设$\frac{x+y}{2}$=u，$\frac{x-y}{3}$=v，则原方程组变为以u，v为未知数的方程组$\left\{\begin{array}{l}{8u-9v=14}\\{u+v=6}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{u=4}\\{v=2}\end{array}\right.$，从而原方程组的解是$\left\{\begin{array}{l}{x=7}\\{y=1}\end{array}\right.$，这种解法称之为“换元法”．
故答案为：$\left\{\begin{array}{l}{8u-9v=14}\\{u+v=6}\end{array}\right.$；$\left\{\begin{array}{l}{u=4}\\{v=2}\end{array}\right.$；$\left\{\begin{array}{l}{x=7}\\{y=1}\end{array}\right.$

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．