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Recently, I came across this system of equations,

\begin{align} \begin{cases} \dfrac{2x}{p(p+4)}=\dfrac{2x}{q(q-4)}=3 & (1) \\ \dfrac{x}{p+q} = 4 & (2) \end{cases} \end{align}

or

\begin{align} \begin{cases} \dfrac{2x}{p(p+4)}=\dfrac{2x}{q(q-4)}=3 & (1) \\ \dfrac{y}{p+4} = \dfrac{x-y}{q-4} = 4 & (2) \end{cases} \end{align}

It would be a great help if someone can help me solve these.

Thanks!

Leucippus
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  • Are $p$ and $q$ prime numbers? Please use MathJax. Here is a tutorial. – Dietrich Burde Nov 18 '22 at 19:23
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Nov 18 '22 at 19:28
  • @DietrichBurde No, p and q are not prime numbers. –  Nov 18 '22 at 19:29
  • Are p and q integers or real numbers ? – Jean Marie Nov 18 '22 at 19:53
  • @JeanMarie x, y, p and q, all of them are variables. In the context of the question, all of them would be positive integers. –  Nov 18 '22 at 20:08

1 Answers1

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Consider the first set of fractions: \begin{align} \frac{2 \, x}{p (p+4)} &= \frac{2 \, x}{q (q-4)} = 3 \\ \frac{x}{p+q} &= 4 \end{align} it would seem that the relations lead to an equation for $p$ and $q$. This follows from solving for $x$ in the second equation, $x = 4 \, (p+q)$, and using it in the first which gives $$ \frac{8 \, (p+q)}{p (p+4)} = 3 \hspace{10mm} \frac{8 \, (p+q)}{q (q-4)} = 3 \to \frac{8}{3} \, (p+q) = p (p+4) = q (q-4). $$ This gives $p^2 + 4 \, p = q^2 - 4 \, q$ and factors to $(p+q) (p - q + 4) = 0$ which gives $q = -p$ or $q = p + 4$.

The second set of equations follows a similar pattern. First solve for $y$ to obtain $y = 4 \, (p+4)$ and then solve for $x$ which is $x = 4 \, (p+q)$. Then find the relation between $p$ and $q$ from the first line (which is $p (p+4) = q (q-4)$).

Leucippus
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