Question

15．方程组$\left\{\begin{array}{l}x+y=6\\ xy=8\end{array}\right.$的解是$\left\{\begin{array}{l}{{x}_{1}=2}\\{{y}_{1}=4}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=4}\\{{y}_{2}=2}\end{array}\right.$．

Answer 1

分析 由x+y=6可得y=6-x，代入xy=8，就可求出x，从而可求出对应y的值，就可解决问题．

解答 解：由x+y=6可得y=6-x，
代入xy=8可得
x（6-x）=8，
整理得
x²-6x+8=0，
解得x₁=2，x₂=4．
当x=2时，y=4；
当x=4时，y=2．
∴原方程组的解为$\left\{\begin{array}{l}{{x}_{1}=2}\\{{y}_{1}=4}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=4}\\{{y}_{2}=2}\end{array}\right.$．
故答案为：$\left\{\begin{array}{l}{{x}_{1}=2}\\{{y}_{1}=4}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=4}\\{{y}_{2}=2}\end{array}\right.$．

点评 本题主要考查了用代入消元法解方程组、解一元二次方程等知识，运用代入消元法是解决本题的关键．