If x cameras are sold at price c each one, the total earning is $231, so we can write the following equation:
[tex]x\cdot c=231[/tex]
Then, if x+4 cameras are sold at price c-12 each one, the total earning is still $231, so we can write a second equation:
[tex](x+4)(c-12)=231[/tex]
Let's equate the right sides of each equation, since they have the same value:
[tex]\begin{gathered} x\cdot c=(x+4)(c-12)\\ \\ xc=xc-12x+4c-48\\ \\ -12x+4c-48=0\\ \\ -3x+c=12\\ \\ 3x=c-12\\ \\ x=\frac{c}{3}-4 \end{gathered}[/tex]
Now, let's use this value of x in the first equation and solve it for c:
[tex]\begin{gathered} (\frac{c}{3}-4)c=231\\ \\ \frac{c^2}{3}-4c=231\\ \\ c^2-12c=693\\ \\ c^2-12c-693=0\\ \\ c=\frac{12\pm\sqrt{144+4\cdot693}}{2}\\ \\ c_1=33\\ \\ c_2=-21 \end{gathered}[/tex]
Since a negative cost is not valid, the answer is 33.
