Applied Application Equations & Inequalities

[jlaw]

Hello! I'm grateful there is help for baby mathematicians like myself. Thank you in advance for sharing your time and talents with the needy.

Ironically, I know the answer to the following applied applications problem and can write an equation that proves to be true (Well... maybe. You be the judge.) but I can not figure out how to solve the equation without the answer already plugged into the equation or how to describe "how to solve it". It is a homework assignment in an intermediate algebra college course.

Problem: For a student club fund-raiser, the number of $2 raffle tickets printed was three times the number of $5 tickets. If all the tickets are sold, receipts from the $5 tickets will be $50 less than those from the $2 tickets. How many $5 tickets were printed?The answer is 50 tickets.

This is my best attempt at an equation for this problem:
x = 2(3y)
y = (x-50)/5

Now if 50 (the answer) is substituted for y in the first equation then x= 300, which I then substitute 300 for x in the second equation and the solution for y = 50 which is the answer but can you please share with me how in the world to solve the equations without first knowing the anwer or is an entirely new equation needed to solve this problem?

Other questionable attempts, to hunt down a workable formula,are listed below for your viewing entertainment
a. 2(3y) = 5x - 50
6y = 5x - 50

b. (3y)/2 * (x-50)/5

c. 5(50)+50 = x in which x=300
(x-50)/5 = y in which y = 50 (number of $5 tickets printed)

Grateful First Timer,
Julia

 

[Denis]

Well, you sure gave it your best shot, Julia!

Let x = number of $5 tickets
Then 3x = number of $2 tickets ; still with me?

sale from $5 tickets = 5x
sale from $2 tickets = 2(3x) = 6x ; ya'll ok with that?

Since difference is $50, then:
6x - 5x = 50
x = 50 (number of $5 tickets) ; kapishsky?

 

[jlaw]

Thank you Denis!!

Uh yah....it's a sure bet that I would have never got that on my own (...eh hem) expertise.

Kapish, the formula to solve the equation clearly provides the correct answer (50)...
Still, I have some questions, now why is the right hand of the equal sign 50 and not -50 (for $50 less than the $2 tickets). And why is the 50 on the otherside of the equal sign instead of being next to 5x as -50. And will I ever learn to figure these kinds of applications on my own?!? And Oh yah...What is the Quote button for on this site?

P.S. Ref: Werewolf
Hope all your vet shots are current....

 

[Denis]

Still, I have some questions, now why is the right hand of the equal sign 50 and not -50 (for $50 less than the $2 tickets). And why is the 50 on the otherside of the equal sign instead of being next to 5x as -50.
What is the Quote button for on this site?

First, on your "Quote button" question: I clicked on it on your last post and the full post appeared here;
I then deleted some of it, leaving what appears above; ok?

Not sure how to properly answer your other question; I'll try it like this:
if A is worth 2 more than B; then:
A = B + 2, OR A - B = 2, OR B = A - 2, OR B - A = -2

 

[jlaw]

Denis Thank You So Much for your help!
When other variables are changed the formula still works!!!
I will send in the homework assignment now.

Gratitudes,
Julia