My daughter got the following optional question for 8th grade homework mixed into her simultaneous equations questions. All of the other questions were simple for me to solve using substitution or elimination which is what she is being taught at the moment but this one seems to me to be at a much higher grade.
A high school play sold 168 tickets to one showing. If tickets were bought in advance, they were half price. If they made $2130 from that one show, how many tickets were sold in advance?
I put together a few formulas but had to use guesswork for the last step which just felt wrong because math should be about processes, not guesswork.
a = advance
r = regular
v = advance price
$$a + r = 168$$ $$a(v) + r(2v) = 2130$$ $$a > 0$$ $$r > 0$$
Solving I get:
$$v(168-r) + 2rv = 2130$$ $$168v - rv + 2rv = 2130$$ $$168v + rv = 2130$$ $$v(168 + r) = 2130$$
At this point I guessed v at 10 and worked out r and a from that. This concerned me because I can't teach my daughter to make educated guesses. What is the process/trick involved with solving these sort of questions?
I'll slightly reform the situation, but the outcome is rather like yours. I'll then explain at the end where to go from there.
Let \begin{align} N_A&=\text{Number of Advanced tickets sold}\\ N_R&=\text{Number of Regular tickets sold} \end{align} Let \begin{align} P_R &=\text{Price of a Regular ticket}\\ P_A &= \text{Price of an Advanced ticket} \end{align} It is clear (as you note, but you've written the inverse relation) that $$P_A=\frac{P_R}{2}$$ Thus we do have two simultaneous equations, viz \begin{align} N_A+N_R &= 168 \qquad \tag{1}\\ N_A P_A + N_R P_R &= 2130 \qquad \tag{2} \end{align} We can re-write $(2)$ as \begin{align} N_A \frac{P_R}{2}+N_R P_R &=2130 \\ \implies N_A P_R + 2N_R P_R &= 4260 \qquad \tag{3} \end{align} We wish to find the value of $N_A$. Using $(1)$ we re-write $(3)$ as \begin{align} N_A P_R + 2(-N_A+168) P_R &= 4260 \\ \implies -N_A P_R + 336P_R &= 4260 \\ \implies N_A P_R &= -4260+336P_R \\ \implies N_A &= \frac{336P_R-4260}{P_R}\\ \implies N_A &= 336-\frac{4260}{P_R} \end{align} This is as far as we can go without knowing the price of a regular ticket. If your daughter's teacher did not give it, then as you state it becomes guesswork. all I can infer is that for the number of advanced tickets to be a positive number (clearly) then $P_R \geq 12.68$ but again, this was guesswork.