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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!

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closed as unclear what you're asking by Dietrich Burde, астон вілла олоф мэллбэрг, Adam Hughes, S.C.B., user223391 Jan 30 '17 at 7:34

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    $\begingroup$ What is the nature of $p,q$ $\endgroup$ – lab bhattacharjee Jan 29 '17 at 12:27
  • $\begingroup$ Are $p$ & $q$ integers? $\endgroup$ – Anonymous_original Jan 29 '17 at 12:28
  • $\begingroup$ Yes, they are integers $\endgroup$ – Adola Jan 29 '17 at 12:30
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    $\begingroup$ @A---B When you spot an obviously errorneously used tag, you can just remove it (possibly but not necessarily also add a more appropriate one). Well spotted anyway! $\endgroup$ – Jyrki Lahtonen Jan 29 '17 at 12:47
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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 $p+q \in \{16,-8\}$.

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  • $\begingroup$ the rhs should be $46$, not $86$. In that case there are solutions, like $(14,2)$. $\endgroup$ – themaker Jan 29 '17 at 13:08
  • $\begingroup$ @themaker I have fixed the error. It should be $16$ and $-8$. $\endgroup$ – Anurag A Jan 29 '17 at 13:16
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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}

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  • $\begingroup$ if you multiply by $3$, the right side is $66$, not $46$. See my solution. $\endgroup$ – Anurag A Jan 29 '17 at 13:06
  • $\begingroup$ Yes, and $66-20=46$. You need to do both, multiply with 3 and factorize, to get that result, as you did (but somehow got a sign wrong). $\endgroup$ – LutzL Jan 29 '17 at 13:08
  • $\begingroup$ you are right my mistake. Thanks, $\endgroup$ – Anurag A Jan 29 '17 at 13:11

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