Question

13．閱讀填空：對于方程組$\left\{\begin{array}{l}{4（x+y）-3（x-y）=14}\\{\frac{x+y}{2}+\frac{x-y}{3}=6}\end{array}\right.$，不妨設(shè)$\frac{x+y}{2}=u，\frac{x-y}{3}=v$，則原方程組變?yōu)橐評，v為未知數(shù)的方程組$\left\{\begin{array}{l}{8u-9v=14}\\{u+v=6}\end{array}\right.$，解得$\left\{\begin{array}{l}{u=4}\\{v=2}\end{array}\right.$，從而原方程組的解是$\left\{\begin{array}{l}{x=7}\\{y=1}\end{array}\right.$，這種解法稱之為“換元法”

Answer 1

分析 根據(jù)設(shè)出的u與v，將方程組變形，求出解確定出u與v的值，進(jìn)而求出x與y的值．

解答 解：閱讀填空：對于方程組$\left\{\begin{array}{l}{4（x+y）-3（x-y）=14}\\{\frac{x+y}{2}+\frac{x-y}{3}=6}\end{array}\right.$，
不妨設(shè)$\frac{x+y}{2}$=u，$\frac{x-y}{3}$=v，則原方程組變?yōu)橐評，v為未知數(shù)的方程組$\left\{\begin{array}{l}{8u-9v=14}\\{u+v=6}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{u=4}\\{v=2}\end{array}\right.$，從而原方程組的解是$\left\{\begin{array}{l}{x=7}\\{y=1}\end{array}\right.$，這種解法稱之為“換元法”．
故答案為：$\left\{\begin{array}{l}{8u-9v=14}\\{u+v=6}\end{array}\right.$；$\left\{\begin{array}{l}{u=4}\\{v=2}\end{array}\right.$；$\left\{\begin{array}{l}{x=7}\\{y=1}\end{array}\right.$

點(diǎn)評 此題考查了解二元一次方程組，熟練掌握運(yùn)算法則是解本題的關(guān)鍵．