SOLUTION: Formulate a system of equations for the situation below and solve. A theater has a seating capacity of 750 and charges $4 for children, $6 for students, and $8 for adults. At a

Question 1201132: Formulate a system of equations for the situation below and solve.
A theater has a seating capacity of 750 and charges $4 for children, $6 for students, and $8 for adults. At a certain screening with full attendance, there were half as many adults as children and students combined. The receipts totaled $4600. How many children attended the show?
______children
Found 5 solutions by math_tutor2020, josgarithmetic, ikleyn, MathTherapy, greenestamps:
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Answer: 200 children

===================================================================================================================

Work Shown:

c = number of children
s = number of students
a = number of adults
These are nonnegative integers {0,1,2,3,...}

"A theater has a seating capacity of 750 ... At a certain screening with full attendance"
c+s+a = total number of people
c+s+a = total seating capacity = 750
c+s+a = 750
Let's refer to this as equation (1).

4c = revenue from the children only
6s = revenue from the students only
8a = revenue from the adults only
4c+6s+8a = total revenue = 4600 dollars
4c+6s+8a = 4600
2(2c+3s+4a) = 4600
2c+3s+4a = 4600/2
2c+3s+4a = 2300
Let's refer to this as equation (2).

"there were half as many adults as children and students combined"
c+s = number of children and students combined
0.5(c+s) = half the previous amount = number of adults
0.5(c+s) = a
Let's refer to this as equation (3).

System of equations:
c+s+a = 750
2c+3s+4a = 2300
0.5(c+s) = a

Let's multiply both sides of the third equation by 2, then get everything to one side.
0.5(c+s) = a
a = 0.5(c+s)
2*a = 2*0.5(c+s)
2a = c+s
0 = c+s-2a
c+s-2a = 0
This is the updated version of equation (3).

Updated system of equations:
c+s+a = 750
2c+3s+4a = 2300
c+s-2a = 0

Narrow your focus to equations (1) and (3)
c+s+a = 750
c+s-2a = 0
Both start with c+s
Let's subtract straight down to cancel out each copy of c+s
More specifically...
The c variables cancel.
The s variables cancel.
For the 'a' variables we have a - (-2a) = a+2a = 3a
For the right hand sides we have 750-0 = 750
So,
3a = 750
3a/3 = 750/3
a = 250 is the number of adults

Plug that value of 'a' into the first equation of the system.
Then isolate c.
c+s+a = 750
c+s+250 = 750
c+s = 750-250
c+s = 500
c = 500-s
We'll come back to this later.

Revisit the second equation.
Plug in a = 250 and do a bit of algebra.
2c+3s+4a = 2300
2c+3s+4*250 = 2300
2c+3s+1000 = 2300
2c+3s = 2300-1000
2c+3s = 1300

Then plug in c = 500-s
This will allow us to solve for s.
2c+3s = 1300
2(500-s)+3s = 1300
1000-2s+3s = 1300
1000+s = 1300
s = 1300-1000
s = 300 is the number of students

We'll use that value of s to determine c.
c = 500-s
c = 500-300
c = 200 is the number of children

-------------------------------------------
Summary

c = 200, s = 300, a = 250

200 children
300 students
250 adults

Check:
c+s+a = 750
200+300+250 = 750
500+250 = 750
750 = 750 ....... first equation works
and
2c+3s+4a = 2300
2*200+3*300+4*250 = 2300
400+900+1000 = 2300
1300+1000 = 2300
2300 = 2300 ....... second equation works
and
0.5(c+s) = a
0.5(200+300) = 250
0.5(500) = 250
250 = 250 ....... third equation works
All three equations of the original system are true for the c, s and 'a' values mentioned.
Therefore, the answers are confirmed.

Other approaches are possible. Feel free to explore other options.

Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
Not the only way to assign variables, but.....

PERSON TYPE PRICE HOW MANY COST Children 4 0.5(x+y) Students 6 x Adults 8 y TOTAL 750 4600

And then to fill the COSTs

PERSON TYPE PRICE HOW MANY COST Children 4 0.5(x+y) 2(x+y) Students 6 x 6x Adults 8 y 8y TOTAL 750 4600

-----------work with this system.
Answer by ikleyn(52782)   (Show Source): You can put this solution on YOUR website!
.
Formulate a system of equations for the situation below and solve.
A theater has a seating capacity of 750 and charges $4 for children, $6 for students, and $8 for adults.
At a certain screening with full attendance, there were half as many adults as children
and students combined. The receipts totaled $4600. How many children attended the show?
~~~~~~~~~~~~~~~~~~~~~~~~

If a person is able to read and make simplest logical conclusions from the written text, he (or she) can immediately conclude that the number of adults was one third of 750, i.e. 250, while the combined number of children and students was 500. After that, the problem can be simply reduced to one unknown and one equation. Let x be the number of children (the major unknown in the problem). Then the number of students is (500-x). After that, we can write the total money equation 4x + 6*(500-x) + 8*250 = 4600. Simplify and find x 4x + 3000 - 6x = 4600 - 2000 -2x = 4600 - 2000 - 3000 -2x = -400 x = = 200. ANSWER. 200 children.

Solved (using single equation in one unknown).

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Formulate a system of equations for the situation below and solve. A theater has a seating capacity of 750 and charges $4 for children, $6 for students, and $8 for adults. At a certain screening with full attendance, there were half as many adults as children and students combined. The receipts totaled $4600. How many children attended the show? ______children Let number of children, students, and adults, be C, S, and A, respectively Then we get: You requested the above!! 2A + A = 750 ----- Substituting 2A for C + S in eq (i) 3A = 750 Number of adults, or C + S = 2(250) --- Substituting 250 for A in eq (iii) C + S = 500 S = 500 - C -- eq (iv) 4C + 6S + 8A = 4,600 ---- eq (ii) 4C + 6(500 - C) + 8(250) = 4,600 ---- Substituting 500 - C, and 250, for S and A, respectively, in eq (ii) 4C + 3,000 - 6C + 2,000 = 4,600 4C - 6C = 4,600 - 5,000 - 2C = - 400 Number of children, or

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

Probably a formal algebraic solution is what you were supposed to find. You have received several responses that show different ways to do that, using either three variables or just one.

While you should be able to solve the problem using formal algebra, you will get probably better mental exercise and better experience in problem solving by solving the whole problem informally, using logical reasoning and mental arithmetic.

As one of the tutors points out, careful reading of the problem followed by logical reasoning shows that 1/3 of the tickets were adult tickets and 2/3 were the combined children and student tickets. 1/3 of the total 750 is 250; and 250 adult tickets at $8 each makes $2000 in sales. So the other 500 tickets, at $4 or $6 each, combine for a total of $2600 in sales.

To find the number of child and student tickets sold, you can do the following, again using only logical reasoning and some mental arithmetic.

If all 500 remaining tickets had been child tickets, the total sales for those tickets would have been $2000; if all 500 had been student tickets, that total would have been $3000. The actual total sales from those 500 tickets was $2600.

Use mental arithmetic to see that the actual total of $2600 is 600/1000 = 3/5 of the way from $2000 to $3000; that means 3/5 of those 500 tickets were the higher-priced student tickets. So 2/5 of those 500 tickets were the children's tickets.

ANSWER: # of children = 2/5 of 500 = 200

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