SOLUTION: I understand the concept but cannot arrive at the correct answers I think I am messing up the last equation. I keep getting the number of adult tickets is 74 but I know that is not

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Question 623771: I understand the concept but cannot arrive at the correct answers I think I am messing up the last equation. I keep getting the number of adult tickets is 74 but I know that is not right. A movie theater charges $7 for adults, $5 for children, and $4 for seniors over age 60. The theater sold 222 tickets and took in $1383. If twice as many adult tickets were sold as the total of children and senior tickets, how many tickets of each kind were sold.
Found 2 solutions by math-vortex, MathTherapy:
Answer by math-vortex(648)   (Show Source): You can put this solution on YOUR website!
Hi, there-- 

A movie theater charges $7 for adults, $5 for children, and $4 for seniors over age 60. The 
theater sold 222 tickets and took in $1383. If twice as many adult tickets were sold as the total 
of children and senior tickets, how many tickets of each kind were sold.

Let x be the number of adult tickets
Let y be the number of child tickets
let z be the number of senior tickets

Equations:
The theater sold 222 tickets
x + y + z = 222

The theater took in 1383. 7x is the income from adult tickets, 5y is the income from child tickets, 
and 4z is the income from senior tickets.
7x + 5y + 4z = 1383

Twice as many adult tickets were sold as the total of child and senior tickets.
[number of adult tickets] = [2] * [total of child tickets and senior tickets]
x = 2(y + z)

These are the three equations.

I solved the system and got:
x = 148 adult tickets
y = 51 child tickets
z = 23 senior tickets

Since you got 74 for the number of adult tickets, I'm thinking that you wrote your third equation as
2x = y + z. It's easy to mix up where to put the multiplier. I always ask myself which value is 
larger (e.g. adult tickets or child plus senior tickets) The wording of the problem indicates that 
the number of adult tickets is larger, so we need to multiply the child plus senior side of the 
equation by 2 to achieve equality.

Let me know if you want to see the system of equations solved out. I got the impression you
understood that part.

Ms.Figgy
math.in.the.vortex@gmail.com

Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!
I understand the concept but cannot arrive at the correct answers I think I am messing up the last equation. I keep getting the number of adult tickets is 74 but I know that is not right. A movie theater charges $7 for adults, $5 for children, and $4 for seniors over age 60. The theater sold 222 tickets and took in $1383. If twice as many adult tickets were sold as the total of children and senior tickets, how many tickets of each kind were sold.

Let amount of adults', children's, and seniors' tickets sold be A, C, and S, respectively

Then:
A + C + S = 222 ---------- eq (i)
7A + 5C + 4S = 1,383 ---- eq (ii)
A = 2(C + S) ---- A = 2C + 2S

2C + 2S + C + S = 222 ----- Substituting 2C + 2S for A in eq (i)
3C + 3S = 222
3(C + S) = 3(74)
C + S = 74 ------ eq (iii)

7(2C + 2S) + 5C + 4S = 1,383 ----- Substituting 2C + 2S for A in eq (ii)
14C + 14S + 5C + 4S = 1,383
19C + 18S = 1,383 ---- eq (iv)

C + S = 74 ------ eq (iii)
19C + 18S = 1,383 ---- eq (iv)
- 18C – 18S = - 1,332 ------- Multiplying eq (iii) by – 18 ---- eq (v)
C = 1,383 – 1,332 ------ Adding eqs (iv) and (v)

C, or amount of children’s tickets sold =

51 + S = 74 ----- Substituting 51 for C in eq (iii)
S = 74 – 51

S, or amount of seniors’ tickets sold =

Since A = 2C + 2S, then A = 2(51) + 2(23) ----- A = 102 + 46

A, or amount of adults’ tickets sold =

I'm sure you can do the check.

Send comments and “thank-yous” to “D” at MathMadEzy@aol.com
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