cost $12 for adults and 5$ for children .on the first day of the fair 312 tickets were sold for a total of 2204$.How many adults ticket and how many child tickets were sold

Question

cost $12 for adults and 5$ for children .on the first day of the fair 312 tickets were sold for a total of 2204$.How many adults ticket and how many child tickets were sold

Answer 1

Hi Falze R.

I noticed you have “pre-algebra” as your subject so if you have not covered solving systems of equations by Elimination just yet here is another way to do your problem.

For tickets

Let x = adults

Let y = children

Given that 312 tickets were sold,

x + y =312

We can put y in terms of x by solving the above for y as follows:

y = 312 - x

For the total sales we can substitute y = 312 - x and have everything in terms of one variable x

Total sales were $2204

Given that adult tickets are $12 we have 12x for adults.

Given that child tickets are $5 we have 5(312 - x)

So the total cost can be expressed as follows:

12x + 5(312 -x) = 2204

With the above you can just multiply through, combine like terms and solve for x to get x = 92, the same value obtained in the Elimination Raymond B used above.

Later on or maybe sooner depending on how your class moves along you will learn about the various methods for solving systems of equations like Graphing, Substitution, and Elimination.

If you have not started Systems of Equations specifically just yet I hope this helps.

Answer 2

2204=12A+5C

312 = A+B

multiply by 5

1560 =5A +5B

subtract from 1st equation

644 = 7A

A = 644/7 = 92 Adults

C=312-92 = 220 Children

check the answer

12(92) +5(220) = 1104+1100 = 2204