Question

9．x+3y=7的正整數(shù)解$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$，$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$．

Answer 1

分析 先求出y的值（把x當(dāng)作已知數(shù)），求出x的范圍，再范圍內(nèi)取符合條件的數(shù)即可．

解答 解：x+3y=7，
y=$\frac{7-x}{3}$，
即x＞0，$\frac{7-x}{3}$＞0，
解得：0＜x＜7，
∵x、y為正整數(shù)，
∴符合的解為：$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$，$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$，
故答案為：$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$，$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$．

點(diǎn)評(píng) 本題考查了解二元一次方程，二次一次方程的解的應(yīng)用，解此題的關(guān)鍵是求出x的范圍．