GIven data:
A merchant could sell one model of digital cameras at list price and receive $162 for all of them.
If he had three more cameras, he could sell each one for $9 less and still receive $162.
To Find the list price of each camera.
Let Price = P and Quantity = Q
PQ = 162
(P-9)(Q + 3) = 162 ...(1)
From the first equation,
Q=162/P
Then,(P-9)(162/P+3)=162
[tex]\begin{gathered} \mleft(P-9\mright)(\frac{162}{P}+3)=162 \\ 3P-\frac{1458}{P}+135=162 \\ \text{ X P on both sides} \\ P\cdot P-\frac{1458}{P}P+135P=162P \\ 3P^2-1458+135P=162P \\ P=27,\: P=-18 \\ \end{gathered}[/tex]price acnnot be negative so conside P = 27.
[tex]\begin{gathered} \mleft(27-9\mright)\mleft(Q+3\mright)=162 \\ \frac{\left(27-9\right)\left(Q+3\right)}{18}=\frac{162}{18} \\ Q+3=9 \\ Q+3-3=9-3 \\ Q=6 \end{gathered}[/tex]Hence P= 27 and Q=6.
