pt đã cho <=> y(3x+1)= 4-x (1)
<=> y =\(\dfrac{4-x}{3x+1}\) ĐKXĐ:\(x\ne\dfrac{-1}{3}\)
Do y nguyên => \(\dfrac{4-x}{3x+1}\) nguyên => (4-x) \(⋮\) (3x+1)
=> 3(x-4)\(⋮\)(3x+1)
=> (3x+1)-13 \(⋮\) (3x+1)
=> 13\(⋮\) (3x+1)
=> 3x+1 \(\inƯ\left(13\right)=\left\{1,-1,13,-13\right\}\)
\(\left[{}\begin{matrix}3x+1=1\\3x+1=-1\\3x+1=13\\3x+1=-13\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{-2}{3}\left(loại\right)\\x=4\left(tm\right)\\x=-\dfrac{14}{3}\left(loại\right)\end{matrix}\right.\)
Với x=0 thay vào (1) tìm dc y=4
Với x=4 thay vào (1) tìm dc y=0
Vậy (x,y) \(\in\left\{\left(4,0\right);\left(0,4\right)\right\}\)
