Admission prices

Question

Solve with substitution and then solve with elimination method show work pls I really need help

Admission prices at a local weekend fair were $6 for children and 22 for adults. The total money collected was 5382 and 473 people attended the fair how many children and how many adults attended the fair?

Answer 1

Let c be the number of children's tickets sold and a b the number of adult tickets sold.

Then (because there were a total of 473 people in attendance):

a + c = 473

Then to determine the amount of money made, we would take the number of children's tickets sold times $6 and the number of adult tickets sold times $22 or:

6c + 22a = 5382

To solve by substitution: take the first equation and subtract a from both sides to get:

c = 473 - a

Then, because a is the same as "473 - a", plug "473 - a" into the second equation wherever there is a c making:

6(473 - a) + 22a = 5382

Distribute and combine like terms to solve for a. Once you get a, plug it back into the equation: c = 473 - a to figure out c.

To solve by elimination, put the two equations over one another in order (a right above the other a, c right above the other c, number right above the other number like this:

a + c = 473

22a + 6c = 5382

Look for a way to adjust one equation so one of the variables can be eliminated if the two equations are added together. If the first equation were to be multiplied by -6 on both sides, there would be a -6c that would eliminate the +6c in the second equation. So multiply the first equation (both sides) by -6 to get this:

-6a - 6c = -2838

Put that over the second equation and add the two equations together:

-6a - 6c = -2838

22a + 6c = 5382

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16a = 2544

solve for a

Then plug the a back in (as before) into c = 473 - a to figure out c