How do you solve an equation like this? We have an equation:
$ \dfrac{p-1}{q} = \dfrac{q-1}{2p+1} = \dfrac {3}{5}$
How do you solve these types of equations? For example, if we have:
$\dfrac{1}{x} = \dfrac{3}{2} $, we use:
$1\times 2 = 3\times x$
$x = 1.5$
What is a similair approach to my equation?
 A: Essentially, you can solve this using the process you describe, but twice, to generate two equations in two variables, and then solving for each variable as a "system of equations".
We have:
$$ \dfrac{p-1}{q} = \dfrac{q-1}{2p+1} = \dfrac {3}{5}$$
Equation 1:  $\quad 5(p-1) = 3q \iff 5p - 3q = 5$
Equation 2:  $\quad 5(q-1) = 3(2p +1) \iff -6p +5q=8$
So your system of two equations in two unknowns becomes:
$$5p - 3q = 5\tag{1}$$
$$-6p +5q=8\tag{2}$$
Can you take it from there? 
You can express (1) as a function of p (isolate p), and then substitute the expression obtained for p, into p in (2), and then solve for q, then p,
or
You can use "row operations": multiply (1) by 5 (both sides), and (2) by 3 (both sides):
$$25p-15q=25\tag{1}$$
$$-18p+15q = 24\tag{2}$$
Now add the equations (q disappears), solve for p, then "plug" p into one of the original equations and solve for q:
$$7p = 49n\implies p = 7$$
Now...From (1), originally, above 
$$5p-3q=5 \implies 5(7) - 3q = 5\implies 35 - 3q = 5$$
$$\implies -3q=-30 \implies q = 10$$
A: You have two equation with two variables:
$$1) \quad \frac{p-1}{q} = \frac{3}{5}$$ and
$$2) \frac{q-1}{2p+1}=\frac{3}{5}  $$
$1)$ Implies that $5(p-1)= 3q$, which gives $q= \frac{5(p-1)}{3}$. Then you can insert $q= \frac{5(p-1)}{3}$ in $(2)$. 
Then find what is $p$, and then you can find what $q$ is. 
A: $$ \frac{p-1}{q} = \frac{q-1}{2p+1} = \frac {3}{5}$$ so 
$$\frac{p-1}{q}=\frac {3}{5}$$ and $$\frac {3}{5}=\frac{q-1}{2p+1}.$$
\begin{eqnarray}
5p-5&=3q&\\
6p+3&=&5q-5
\end{eqnarray}
$$5p-3q=5$$
$$6p-5q=-8$$
$$30p-18q=30$$
$$30p-25q=-40$$
$$-7q=-70$$
$$q=10$$
$$p=7.$$
A: From $$ \dfrac{p-1}{q} = \dfrac{q-1}{2p+1} = \dfrac {3}{5}$$ follow the system of linear equations with two unknowns
$$5(p-1)=3q$$
$$5(q-1)=3(2p+1)$$
or
$$5p-3q=5$$
$$6p-5q=-8$$
wich can be solved using for example Cramer rule
$\Delta=-25+18=-7$
$\Delta_p=-25-24=-49$
$\Delta_q=-40-30=-70$
$$p=\frac{\Delta_p}{\Delta}=\frac{-49}{-7}=7$$,$$ q=\frac{\Delta_q}{\Delta}=\frac{-70}{-7}=10$$
