$\begin{array}{l}\dfrac{7a+3b}{19}\rm\ nguyên\\\Leftrightarrow 7a+3b\ \vdots\ 19\\\text{- Xét tổng : $\\\quad5(7a+3b)+4(-4a+b)\\=35a+15b-16a+4b\\=19a+19b\ \vdots\ 19\\\to5(7a+3b)+4(-4a+b)\ \vdots\ 19\\\text{mà $5(7a+3b)\ \vdots\ 19$ (vì $7a+3b\ \vdots\ 19$}\\\to 4(-4a+b)\ \vdots\ 19\\\text{mà $(4,19)=1$}\\\to -4a+b\ \vdots\ 19\\\to\dfrac{-4a+b}{19}\ \rm nguyên$}\\\text{- Vậy với $\dfrac{7a+3b}{19}$ nguyên thì $\dfrac{-4a+b}{19}$ cũng nguyên} \end{array}$