Known that: $3pq-5p+4q=22$, Find the value of $p + q$

Question

I have this question:

Known that: $$3pq-5p+4q=22$$ Find the value of $p + q$

I have solved 2 variables with 2 equations or more, but I have never encountered 1 equation with 2 variables. The answer is a positive integer. Can I have a hint or a guide?

Thanks!

Answer 1

The given equation can be written as $$(3p+4)(3q-5)=46$$ Since $p$ and $q$ are both integers (as what OP has mentioned in the comments). Therefore, we want factors of $46=ab$ such that \begin{align*} 3p+4 & =a \\ 3q-5 & =b \end{align*} Thus $$3(p+q)=a+b+1 \implies a+b+1 \equiv 0 \pmod{3}.$$ But the only possible values for $a,b \in \{\pm 1, \pm 2, \pm 23, \pm 46\}$ (of course with $ab=46)$. However the solutions that satisfy $a+b+1 \equiv 0 \pmod{3}$ are $(a,b) \in \{(1,46), (46,1),(-2,-23),(-23,-2)\}$.

Hence $p+q \in \{16,-8\}$.

Answer 2

Multiply with $3$, add a constant term and factorize to get $$ (3p+4)(3q-5)=3(3pq−5p+4q)-20=46 $$

If the quadratic equation $$0=z^2-az+46$$ has the solutions $z_1=(3p+4)$, $z_2=(3q−5)$ then by the Viete rules $$a=z_1+z_2=(3p+4)+(3q−5)=3(p+q)-1$$ which is independent of the order of the roots.

As $46=z_1·z_2=1·46=2·23=(-2)·(-23)=(-1)·(-46)$ only has that many integer factorizations, one quickly finds the only solutions \begin{align} 3(p+q)-1&=47:&p+q&=16;\\ 3(p+q)-1&=25:&not~&possible\\ 3(p+q)-1&=-25:&p+q&=-8\\ 3(p+q)-1&=-47:&not~&possible\\ \end{align}