If $\displaystyle p \cdot q \neq 0$

Then the RHS of the second equation is $\displaystyle 3$ times the RHS of the

first equation, so the same must be true of the LHS. So multiplying

the first equation by 3 and then equating coefficients of $\displaystyle x$ and

$\displaystyle y$, which gives:

$\displaystyle p+q=3(p-q)$,

$\displaystyle q-7=3(3p-5)$.

Which is a pair of linear equations which can be solved for $\displaystyle p$ and $\displaystyle q$

to give $\displaystyle p$ and $\displaystyle q$ which give an infinite number of solutions

to the original equations.