SOLUTION: On the first day, Cindy sells 6 adult tickets and 5 children's tickets for a total of 112.50. on the second day, she sells 8 adult tickets and 4 children's tickets for a total of

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: On the first day, Cindy sells 6 adult tickets and 5 children's tickets for a total of 112.50. on the second day, she sells 8 adult tickets and 4 children's tickets for a total of      Log On


   

Question 1179258: On
the first day, Cindy sells 6 adult tickets and 5 children's tickets for a total of 112.50. on the second day, she sells 8 adult tickets and 4 children's tickets for a total of 130.00
Found 4 solutions by MathLover1, ikleyn, josgarithmetic, greenestamps:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

let the price of adult tickets be x and the price of children's tickets y
if on the first day, Cindy sells 6 adult tickets and 5 children's tickets for a total of 112.50, we have
6x%2B5y=112.50.....solve for x
6x=112.50-5y
x=112.50%2F6-5y%2F6
x=18.75-5y%2F6...........eq.1
if on the second day, she sells 8 adult tickets and 4 children's tickets for a total of 130.00, we have
8x%2B4y=130.00.....solve for x
8x=130.00-4y
x=130.00%2F8-4y%2F8
x=16.25-0.5y...........eq.2
from eq.1 and eq.2 we have
18.75-5y%2F6=16.25-0.5y
18.75-16.25=5y%2F6-0.5y
2.5=2y%2F6
2.5%2A6=2y
%282.5%2A6%29%2F2=y
2.5%2A3=y
7.5=y
go to
x=16.25-0.5y...........eq.2, substitute y
x=16.25-0.5%2A7.5
x=12.5

the price of adult tickets is 12.5 and the price of children's tickets 7.5


Answer by ikleyn(52775) About Me  (Show Source):
You can put this solution on YOUR website!
.

Had you attach your questions to the post (as all civilized persons do),

I would show you more elegant solution without using fractions.


But since you neglected to request in a polite form, I will not do it . . .



Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
c, child ticket price
a, adult ticket price
system%285c%2B6a=112.5%2C4c%2B8a=130%29

%281%2F2%29%2AE2:
system%285c%2B6a=112.5%2C2c%2B4a=65%29

5%2AE2
2%2AE1:
system%2810c%2B12a=225%2C10c%2B20a=325%29

E2-E1:
8a=100
highlight%28a=12.5%29

and then
c=%28112.5-6%2A12.5%29%2F5
highlight%28c=7.5%29

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


The first tutor always solves systems of equations using substitution. That might be her preferred method; but when both equations are given in Ax+By=C form, I think some sort of elimination is much easier.

Below is an unusual method for solving the problem. I first present the solution informally using logical reasoning; then I show the corresponding formal algebraic solution. The method shown can make the solution of some similar problems easier than standard elimination; but for most problems a more straightforward solution using elimination will be fastest.

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"On the first day, Cindy sells 6 adult tickets and 5 children's tickets for a total of 112.50."
"On the second day, she sells 8 adult tickets and 4 children's tickets for a total of 130.00"

The difference between the two sales is 2 more adult tickets and 1 fewer children's tickets for $17.50 more.

Apply that difference four more times to make the number of children's tickets zero. Doing that gives us 8+4(2) = 16 adult tickets and 4-4(1) = 0 children's tickets for a total of $130+4($17.50)=$200.

So the cost of each adult ticket is $200/16 = $12.50.

Then use the purchase on the second day to determine the cost of each children's ticket:
8($12.50)+4x = $130
4x = $30
x = $30/4 = $7.50

ANSWER: Adult ticket price $12.50; Children's ticket price $7.50

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Algebraically, this unusual solution method looks like this:

system%286a%2B5c=112.5%2C8a%2B4c=130%29

Find the difference between the two equations:



Multiply the second equation by 4:

system%288a%2B4c=130%2C8a-4c=70%29

Add the two equations to eliminate c:

16a=200
a+=+200%2F16+=+12.50

Use that value for a to find c:

8%2812.50%29%2B4c+=+130
4c+=+30
c+=+7.50

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So why use this unusual method for solving the problem?

In a formal algebraic solution, finding the difference between the two original equations gives us an equation in which one of the coefficients is -1. That will make solving the system by elimination easier, because it will keep the coefficients smaller.

And if a formal algebraic solution is not required, this method of solving the problem makes an informal solution easier, for the same reason.