Solving equations simultaneously
By keljad
@keljad (69)
Australia
October 12, 2006 11:28pm CST
2 Adults and 4 Children pay $76.00 entry while a group of 4 adults and 1 child pay $82.00 I need to solve to find the admissions cost of adults and children. can anyone help with an explination regards keljad
3 responses
@AJMSmith (112)
•
17 Feb 07
Assuming there are two types of ticket then
4 Adults and 8 Children cost $152 (twice the 2A, 4C for $76)
less
4 Adults and 1 Child costing $82
leaves
7 Children @ $70 = $10 per child
Which means that 2 Adults cost $36 and an Adult cost $18
Check : 2 * 18 + 4 * 10 = 36 + 40 = 76. 4 * 18 + 10 = 72 + 10 = 82
@RAMPersona (2033)
• Philippines
23 Oct 06
*Algebra
Solution:
Let a = adult entry, $
c = child entry, $
Given:
eq1. 2a+4c = 76
4a+c = 82
c = 82-4a
Substitute it to eq1.
2a+4(82-4a) = 76
2a+328-16a = 76
328-76 = 16a-2a
252 = 14a
* a = $18
And,
c = 82-4a
c = 82-4(18)
c = 82-72
* c = $10
@kiwidipa (2852)
• United States
23 Oct 06
well, how old are the children, is there a cost for them being say 10 or older? Or is 5 and under free? I don't think you could solve this unless you could get more information. Like what is it for? An amusement park? A zoo, aquarium. All have different price tags
