Alex E. answered 08/30/20
I’ve assisted hundreds of repeat students over the past ten years
QUESTION:
A concession stand sells hotdogs and hamburgers. A hamburger cost 75 cents more
than a hotdog. In one afternoon the stand sells 40 hotdogs and 50 hamburgers and collects
$226.50. Find the price of a hamburger and the price of a hotdog.
MY SOLUTION:
First what are you asked to find?
Ans...
the price of a hamburger and the price of a hotdog.
Label your unknowns as variables. I give somewhat meaningful name.
Let
b = burger price and d = dog price.
So there are two unknowns. My approach, form a system of relations (equations here) containing just as many relationships as unknowns. So here there are 2 unknowns so let’s look to form 2 relationships from the information given.
Note: A system of equations is just a set or collection of equations RELATED through an ‘and’ statement which here means a solution of system of equations MUST SATISFY All equations in the system, i.e. the solution is a solution to EVERY equation in the system.
First relationship found in problem....
A hamburger cost 75 cents more than a hotdog. This translates to...
b = 0.75 + d
so one equation of the system is
b = 0.75 + d
Second relationship found in problem....
the stand sells 40 hotdogs and 50 hamburgers and collects $226.50 ...translates to...
40 d + 50 b = 226.50
Note: really ‘and’ in the sentence here just tells us to what to associate, NOT ADD. The add part here really come from collects which means to sum together all previous values to form a result.
so a second equation to form the system of 2 equations is
40d + 50b = 226.50
Great now there is more than one way to solve as system (part of what I find to be the beauty in math); however, I’ll stick to the most direct one I see... substitution
Note: since this is a system and we are interested in it’s solution, our solution will work for all equations in the system. So when solving for the solution the d in the first equation is the exact same d as in the second equation and similarly with the b of the two equations.
Since b is solved for in our first equation substitute the b in the second equation for its value.
So now the second equation is rewritten as
40d + 50(0.75 + d) = 226.50.
Now we are left with a single variable equation that may be solved for d. After finding d, back substitution of it’s found value into any equation of the system will find b. Then you done.
Im going to go eat and leave you a chance to find the answer from here. I’ll come back and update it in a few hours with how I’d finish it out in detail.
**********************
Alright, picking up where I left off,
40d + 50(0.75 + d) = 226.50
⇒ 40d + 37.5 + 50d = 226.50
⇒ 90d + 37.5 = 226.50
- 37.5 - 37.50
⇒ 90d = 189.00
90 90
⇒ d = 2.1
Now with d=2.1 back substitute the value into any equation of the system, after all a solution to the system is a solution for ALL equations composing the system.
So I’ll choose the first one and if you want you can choose the other, totally your choice!
b = 0.75 + d, Substitute d for 2.1
= 0.75 + 2.1
= 2.85
Alright, since b = burger price, it sells for $2.85, and since d = dog price, it sells for $2.10.
BTW the steps I’ve outlined in this problem most likely can be used for other problems you’re currently working on as well. Hope this helps!