Ok so Im not too good at algebra or math for that matter...but Im trying to help my friends kid solve this word problem... A group of 19 students want to see the show at the planetarium. Tickets cost 11 for members and 13 for non members. The membership costs $5 per year. Write an equation that can be used to determine n, the number of visits per year for which the cost of being a members is equal to the cost of a non member. Thanks in advance.
They generally do that in early algebra to see if you understand that you need to round up to the nearest whole number for that kind of problem.
Been in finance too long to remember that far back. Son, 3rd grade, came home with math homework one night last year and had a hard time understanding why multiplication is being introduced by use of arrays. Showed him the "old school" way and he picked it up in nothing flat.
That's actually the simplified form FYI @OP. As there are 19 students, each for which the membership costs 5, then we get: 19(11n + 5) = 19(13n) where n is the number of visits (and 11 and 13 are the cost per visit for member / non-member respectively). Indeed the 19's cancel out, which furthermore means the number of students is insignificant to the solution (OP mention this if you want to impress Solving for n gives 2.5, meaning only two or one visits a year and membership doesn't profit. 3 or more visits per year and it's better to become members. EDIT: if you solve (algebraically) for n, you get n = 5 / 2. So that's the simplest equation for the given assignment.
One visit for a member is actually 16 (11+5), two visits is 27 (2x11+5), three visits equals 38 (3x11+5). One visit for a non-member is 13, two visits is 26, three visits is 39. So, 3 (3x11+5=38) and 1/11 (of the 11 member admission charge = 1) visits for a member would equal the 39 the non-member pays.
the equation is 11n + 5=13n and the answer is 2.5 you are way off...its not like the member pays an extra 5 bucks each time they go, its a one time deal. You need to spread the 5 dollars into the amount of times the member goes to match the amount it would cost the same with a nonmember....
Of course, the 5 is a one-time fee, but it must be accounted for in the total cost, no? The answer should be 3, there is no ½ visit.
The problem is that it specifically says, "the number of visits per year for which the cost of being a members is equal to the cost of a non member". Equal means that the answer is 2.5, if it asked "what is the minimum number of times a student would have to go for it to be worth the membership cost?" then 3 is the answer.
Being that it's a math problem, the strict answer is 2.5 (although the practical answer is 3). Eitherway, any sensible professor/teacher would accept both as correct.