I'm stuck in a situation involving rental. Typically I would assign a set of constants based on their price but I'm confused on how to use it or if it applies here.
The problem is as follows:
A tv station in Kaohsiung hires the services of a Germany-based company which rents a set of equipment for tv-show production being a total of $19$ with these being tv cameras, video editing consoles and 3D chroma key graphics computers with rental cost of $80$, $50$ and $60$ $\textrm{USD}$ respectively for which obtains a profit of $\textrm{1240 USD}$ per week. If the company whose rents the equipment raised their rent prices in $\textrm{20 USD}$ for each tv camera and raised $\textrm{10 USD}$ for each chroma key computers and decreased in $\textrm{10 USD}$ the rent costs for each video editing consoles the profit would become into $\textrm{1420 USD}$ per week. How many video editing consoles does the company rents to the tv station?
What I tried to solve this particular problem was to build up a system of equations as follows:
I assigned the letters:
$$\begin{array}{c} \textrm{t, for tv cameras,}\\ \textrm{v, for the video editing consoles}\\ \textrm{c, for the chroma key}\end{array}$$ Short reference: Chroma key graphics computers, so they would represent the number of items respectively.
Since it is mentioned in the passage that what the company would obtain by a week is $1240\,USD$ I'm not worrying about the days as the latter passage of the problem also compares it with what will be obtained in a week being $1420\,USD$. So the convertion factors would be the dollars per number of objects. With that in mind I came up with this equation.
For the first part this would become into:
$t\times \frac{80\,\$}{\textrm{1 tv camera}}+ v\times \frac{50\,\$}{\textrm{1 video console}} + c \times \frac{60\,\$}{\textrm{1 chroma key computer}} = 1240\,\$$
In the second part it alludes that the price is riased for the tv cameras and chroma key computers and decreased for their chroma key computers, so I figured that it would be like this:
$t\times \frac{\left(80+20\right)\,\$}{\textrm{1 tv camera}}+ v\times \frac{\left(50-10\right)\,\$}{\textrm{1 video console}} + c \times \frac{\left(60+10\right)\,\$}{\textrm{1 chroma key computer}} = 1420\,\$$
Finally the last part would account for the total number of objects becoming into.
$t + v + c = 19$
Organizing all the three would be this system of $3 \times 3$ : (note that purposes of brevity and better displaying I ommited the wording and labels placed in the previous equations)
$$\begin{array}{c}80 t + 50 v + 60 c = 1240 \\ 100 t + 40 v + 70 c = 1420\\ t + v + c = 19\end{array}$$
Now there are many methods for solving this system but I preferred to use simple substitution of the first variable on the second and third rows.
By simplifying:
$$8t + 5v + 6c =124$$
$$8t = 124 -5v -6c$$
$$10t +4v + 7c= 142$$
Now I'm multiplying by $5$ to the $8t$ and by $4$ to $10t$ so it can be better replaced one onto the other.
$$5 \left( 8t = 124 -5v -6c \right)$$
$$4 \left( 10t +4v + 7c= 142 \right)$$
$$40 t = 620 - 25v - 30c$$
$$40 t + 16v + 28c = 568$$
Replacing the first in the second one becomes:
$$\left(620 - 25v - 30c\right) + 16v + 28c = 568$$
$$- 9 v - 2 c = -620 + 568 = -52$$
$$9v + 2 c = 52$$
Now all is left to multiply by $40$ to the third equation in the system so it can easily be substitued using the second expression:
$$40 \left(t+v+c=19\right)$$
$$40t + 40 v + 40 c = 760$$
Therefore:
$$\left(620 - 25v -30c\right) + 40v + 40c = 760$$
$$15v + 10c= 760 - 620 = 140$$
In short the remaining equations left are
$$\begin{array}{c}9v + 2 c = 52\\ 15v + 10 c = 140\end{array}$$
Simplifying the second one:
$$3v + 2 c = 28$$
So now all that is left is to multiply by $-1$ to any of both, which I used the second to obtain a quicker subtraction in the right side of the equation
$$-1 \left( 3v + 2 c = 28 \right)$$
$$-3 v - 2 c = -28$$
So all the two together becomes into:
$$\begin{array}{c}9v + 2 c = 52\\ -3v - 2 c = -28\end{array}$$
Summing them up:
$$6v= 24$$
With $$v = 4$$
Using any equation of the two I obtained $c$
$$c=\frac{52-9\times 4}{2}= \frac{52-36}{2}=\frac{16}{2}=8$$
And using the first equation:
$$t+v+c=19$$
$$t = 19 -\left(v+c\right)=19 - \left(4 + 8 \right) = 19 - 12 = 7$$
Therefore the number of tv cameras is $7$. Although what it was asked the number of video consoles which I obtained first being $4$.
In my book there appears these alternatives.
$$\begin{array}{c} 1.\,7\,\textrm{units} \\ 2.\,4\,\textrm{units} \\ 3.\,6\,\textrm{units} \\ 4.\,5\,\textrm{units} \\ 5.\,8\,\textrm{units} \end{array}$$
I would select the second option $4$. But was it right what I've just did?. My method seem to obtain an answer, but I don't know if it could had been simplified any further. Therefore I hope someone can check if my steps are consistent and help me with an easier method.