Do you understand division?

Notice that

$\dfrac{8a^2-14b-56a+2ab}{a-7} = \dfrac{(8a+2b)(a-7)}{a-7} = 8a+2b$

and

$\dfrac{16ab+6b+4b^2+24a}{2b+3} = \dfrac{(8a+2b)(2b+3)}{2b+3} = 8a+2b$

So the system is equivalent to

$\begin{cases} b^2 \mid 8a+2b \\ a+b \mid \dfrac{b^2}{2}-2a-b \\ 8a+2b \mid b^2 \end{cases}$

$b^2 \mid 8a+2b \implies 8a+2b \geq b^2$ (I)

$8a+2b \mid b^2 \implies 8a+2b \leq b^2$ (II)

For both (I) and (II) to be valid, we need $b^2 = 8a+2b$

Substituting $b^2$ in the second statement:

$a+b \mid \dfrac{8a+2b}{2}-2a-b \implies a+b \mid 4a+b-2a-b \implies a+b \mid 2a$

We have $\dfrac{2a}{a+b} < \dfrac{2a}{a} = 2 \implies \dfrac{2a}{a+b} = 1 \implies 2a = a+b \implies a = b$

Knowing that $a =b$ , the first statement becomes

$a^2 \mid 10a \implies a^2 \leq 10a$ (III)

and the third statement becomes

$10a \mid a^2 \implies a^2 \geq 10a$ (IV)

For both (III) and (IV) to be valid, we need $a^2 = 10a \implies a = 10 = b$

The only solution is $(10,10)$ , so the answer is $10+10 = \boxed{20}$