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
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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) (Show Source): You can put this solution on YOUR website!
let the price of adult tickets be and the price of children's tickets
if on the first day, Cindy sells adult tickets and children's tickets for a total of , we have
.....solve for
...........eq.1
if on the second day, she sells adult tickets and children's tickets for a total of , we have
.....solve for
...........eq.2
from eq.1 and eq.2 we have
go to
...........eq.2, substitute
the price of adult tickets is and the price of children's tickets
Answer by ikleyn(52776) (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) (Show Source): You can put this solution on YOUR website!
c, child ticket price
a, adult ticket price
:
:
:
and then
Answer by greenestamps(13198) (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:
Find the difference between the two equations:
Multiply the second equation by 4:
Add the two equations to eliminate c:
Use that value for a to find c:
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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.
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