Question

7．二元一次方程x+y=6的正整數(shù)解為$\left\{\begin{array}{l}{{x}_{1}=1}\\{{y}_{1}=5}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=2}\\{{y}_{2}=4}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{3}=3}\\{{y}_{3}=3}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{4}=4}\\{{y}_{4}=2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{5}=5}\\{{y}_{5}=1}\end{array}\right.$．

Answer 1

分析 根據(jù)二元一次方程的解的定義，可得出5組一元一次方程x+y=6的正整數(shù)解．

解答 解：當(dāng)x=1時(shí)，y=5；
當(dāng)x=2時(shí)，y=4；
當(dāng)x=3時(shí)，y=3；
當(dāng)x=4時(shí)，y=2；
當(dāng)x=5時(shí)，y=1；
∴方程x+y=6的正整數(shù)解為：$\left\{\begin{array}{l}{{x}_{1}=1}\\{{y}_{1}=5}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=2}\\{{y}_{2}=4}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{3}=3}\\{{y}_{3}=3}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{4}=4}\\{{y}_{4}=2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{5}=5}\\{{y}_{5}=1}\end{array}\right.$；
故答案為：$\left\{\begin{array}{l}{{x}_{1}=1}\\{{y}_{1}=5}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=2}\\{{y}_{2}=4}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{3}=3}\\{{y}_{3}=3}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{4}=4}\\{{y}_{4}=2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{5}=5}\\{{y}_{5}=1}\end{array}\right.$．

點(diǎn)評 本題考查了二元一次方程的整數(shù)解的情況，在求一個二元一次方程的整數(shù)解時(shí)，往往采用“給一個，求一個”的方法，即先給出其中一個未知數(shù)（一般是系數(shù)絕對值較大的）的值，再依次求出另一個的對應(yīng)值．