SOLUTION: Please help me solve this equations The circus is coming. Cindy is selling tickets. she sold 8 adults tickets and 5 children's tickets for a total of $180.50 in sales. The second

Question 1135264: Please help me solve this equations
The circus is coming. Cindy is selling tickets. she sold 8 adults tickets and 5 children's tickets for a total of $180.50 in sales. The second day she received $261.00 for 12 adult tickets and 6 children tickets. What is the price of a child's ticket? What is the price of an adult ticket?
Write two algebraic equations that model this circus ticket scenario.
Show how you solved the problem by using elimination method, the substitution method, or the graphical method

Found 3 solutions by rothauserc, MathTherapy, greenestamps:
Answer by rothauserc(4718)   (Show Source): You can put this solution on YOUR website!
let a be the cost of an adult's ticket and c be the cost of a child's ticket, then
:
1) 8a +5c = 180.50
:
2) 12a +6c = 261
:
solve equation 1 for a
:
8a = 180.50 -5c
:
3) a = (180.50 -5c)/8
:
substitute for a in equation 2
:
12(180.50 -5c)/8 +6c = 261
:
multiply both sides of = by 8
:
12(180.50 -5c) +48c = 2088
:
2166 -60c +48c = 2088
:
-12c = -78
:
c = 6.50
:
Use equation 3 to find the value of a
:
a = (180.50 -5(6.50))/8 = 18.50
:
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cost of adult ticket is $18.50 and cost of child ticket is $6.50
:
check the answer by substituting in equations 1 and 2
:
equation 1
:
8(18.50) +5(6.50) = 180.50
:
148 +32.50 = 180.50
:
180.50 = 180.50
:
equation 2
:
12(18.50) +6(6.50) = 261
:
222 +39 = 261
:
261 = 261
:
answer checks
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:
Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!

Please help me solve this equations
The circus is coming. Cindy is selling tickets. she sold 8 adults tickets and 5 children's tickets for a total of $180.50 in sales. The second day she received $261.00 for 12 adult tickets and 6 children tickets. What is the price of a child's ticket? What is the price of an adult ticket?
Write two algebraic equations that model this circus ticket scenario.
Show how you solved the problem by using elimination method, the substitution method, or the graphical method

Never use substitution like the other person did unless asked to do so or if one of the variables has a coefficient of 1 on it
Let the cost of an adult's ticket sold, be A, and the cost of a child's ticket, C
We then get:
4A + C = 80.5 -------- Subtracting eq (i) from eq (ii)
C = 80.5 - 4A ------ eq (iii)
8A + 5(80.5 - 4A) = 180.5 ------- Substituting 80.5 - 4A for C in eq (i)
8A + 402.5 - 20A = 180.5
- 12A = - 222
A, or cost of an adult's ticket =
Substitute the value of A into any of the ORIGINAL equations to get the value of C, a child's ticket-cost.
Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!

If the two equations are both in the form Ax+By=C, then elimination is nearly always easier than substitution.

Multiply the first equation by 3 and the second by 2. That will make the coefficients of A both 24; then subtracting one equation from the other will eliminate A and let you solve for C.

Substitute the value you found for C back in one of the original equations to solve for A.

The tickets cost $6.50 for a child and $18.50 for an adult.
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