Question

15．方程组$\left\{\begin{array}{l}{\frac{4}{x}+\frac{1}{y}=9}\\{\frac{6}{x}-\frac{2}{y}=3}\end{array}\right.$的解为$\left\{\begin{array}{l}{x=\frac{2}{3}}\\{y=\frac{1}{3}}\end{array}\right.$．

Answer 1

分析 方程组整理后，利用加减消元法求出解即可．

解答 解：设$\frac{1}{x}$=a，$\frac{1}{y}$=b，方程组课化为得：$\left\{\begin{array}{l}{4a+b=9①}\\{6a-2b=3②}\end{array}\right.$，
①×2+②得：14a=21，即a=$\frac{3}{2}$，
把a=$\frac{3}{2}$代入①得：b=3，
∴$\left\{\begin{array}{l}{x=\frac{2}{3}}\\{y=\frac{1}{3}}\end{array}\right.$，
经检验$\left\{\begin{array}{l}{x=\frac{2}{3}}\\{y=\frac{1}{3}}\end{array}\right.$是原方程组的解．
故答案为：$\left\{\begin{array}{l}{x=\frac{2}{3}}\\{y=\frac{1}{3}}\end{array}\right.$．

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．