Write a linear equation that represents this scenario. 
Emma is planning her summer and would like to work enough to travel and buy a new laptop. She can earn $90$ dollars each day, after deductions, and she can work a maximum of $40$ days in July and August, combined. She expects each day of travel will cost her $150$ dollars and the laptop she hopes to buy costs $700$ dollars.
Write a linear equation that represents the number of days Emma can work and travel and still earn enough for her laptop.

This is what I've come up with: $90d = 150p + 700$,
(Standard form: $90d - 150p -700 = 0$)
where $d$ represents days worked, and $p$ represents travel days.
The question goes on to ask about how many days she will need to work she wants to travel, so I want to make sure i have to correct equation before I answer.
 A: $$\begin{array}{lrr}
\text{Dr: Cash}&\$90d\\
\quad\text{Cr: Salary income}&&\$90d\\
\hline
\text{Dr: Travel expense}& 150p\\
\quad\text{Cr: Cash}&&150p\\
\hline
\text{Dr: Laptop}&700\\
\quad\text{Cr: Cash}&&700
\end{array}$$
Consider only Emma's cash account over the summer period. If the days Emma works is just enough for her travel and her laptop:
$$90d = 150p + 700$$
A: You need to define your terms for people to have a good chance of understanding the equation you've come up with.  I infer that $d$ must be the number of days worked, so $90d$ is her money earned.  Similarly, $p$ is her days travelled so that $150p$ is her money spent on travel.  Is $c$ supposed to be money spent on the laptop?  In that case is $c=700$?
Try plugging in some numbers that should work and see if the equation is doing what it is supposed to.  For example, if she buys the laptop and travels 10 days she will have spent \$700 + 10(\$150) = \$2200.  She would have to work \$2200/\$90 = 24.444... days to pay for this.  If you plug in d = 24.444... and p = 10 does your equation come out right?
The comment from @RecklessReckoner is a good one.
