Question

Stuck on the algebra portion of 2C-13.

Answer 1

Scott I'm naturally pretty bad (slow) at word problems, however I always manage to solve them by running through the problem once using actual numbers, this may work well for you also.
First of all we want an equation that gives us an answer in revenue as that's what the problem is asking, also this must be a function of the price of a ticket.
\[revenue = r(p)\] where p is the price of the ticket.
This at least gives a way of thinking about the problem, and can be done for most word problems.
Now, lets run through what would happen if a ticket price was increased say $5 dollars.
According to the problem we would lose 2 people. So 100-2 = 98 people, now how much revenue do we have, 98x$205 dollars $20090 revenue, ok all good. If we keep doing this we see that $210 dollars rids us of 4 people 100-4 = 96, then 96 people x 210 dollars is $20160 revenue and so on.So lets see if we can build an equation out of a ticket price of $205. Ok we have p equals ticket cost so
\[p-200\] gives us the five dollar amounts over or under $200, so in our test case this is $205-200= $5, now we need a way to turn this into people, ie we need to subtract 2 people from the 100 seats, so $5 times some value x will gives us 2 people,
\[5x=2, x=\frac{2}{5}x\]
good we can use this to convert our $5 amounts we get from
\[p-200\]
to convert to people ie.
\[(p-200)\frac{2}{5}= people\]
for our example of $205 we have
\[($205-200)\frac{2}{5}=2\]
Now we subtract this from the 100 people
\[100 - 2; 100 - (205-200)\frac{2}{5} = 98\]
Ok, so we are nearly there, now all we have to do is multiply this number of people
by the ticket amount, which in our example is $205, and we have the revenue of the starting equation 98x205 - 20090. So our equation is now

\[(100-(205-200)\frac{2}{5})*205\]
so if we rewrite it using p for the amounts of $250
we get
\[(100-(p-200)\frac{2}{5})p\]
then just simplify it
\[(100p-(\frac{2}{5}p-80)p)\]
\[->(100p-(\frac{2}{5}p^2-80p))\]
\[->180p-\frac{2}{5}p^2\]
and we can use this to calculate any amount in five dollar increments.
The important thing to take away from this is that we used real numbers for the most part to derive the equation, if nothing else works for you this method may help.

Answer 2

A stray x is in the
\[5x=2\]
above
ie \[x=\frac{2}{5}\]
but im sure you picked that up anyway ;)