Problem with an algebra problem : [?

Question

Gosh, these get me every time.....

Drama Club sold 311 tickets for a play. Student tickets cost $.50 and nonstudent tickets cost $1.50. If total receipts were $385.50, find how many tickets of each type were sold.


If you could show the steps, that would be very much appreciated!! Thanks!

Answer 1

Let S = student tickets sold
Let N = non-student tickets sold

"Drama Club sold 311 tickets for a play."

This tells me:

S + N = 311

"Student tickets cost $.50 and nonstudent tickets cost $1.50. If total receipts were $385.50"

This tell me:

0.50 * S + 1.50 * N = 385.50

You now have a system of equations.

Solve the first for N

N = 311 - S

Substitute into the second equation:

0.50 * S + 1.50 * (311-S) = 385.50

0.50S + 466.5 - 1.5S = 385.50

-S + 466.5 = 385.5
-S = -81
S = 81

Substitute this value back into the first equation:

81 + N = 311

N = 230

They sold 81 Student tickets and 230 non-student tickets

Answer 2

Drama Club sold 311 tickets for a play. Student tickets cost $.50 and nonstudent tickets cost $1.50. If total receipts were $385.50, find how many tickets of each type were sold.

This is a typical value problem
One equation is the number of tickets
The other is the value of the tickets

S + N = 311
0.50S + 1.5N = 385.50

Using elimination
S + N = 311
-S - 3N = -771

-2N = -460
N = 230

S = 311 - 230 = 81

81 Student & 230 Nonstudent

Value = .5(81) + 1.50(230)
= 40.50 + 345.00
= 385.50

Answer 3

Okay you will have two equations to solve it out

S = Student tickets
N = Non-student tickets

First equation is S + N = 311 because altogether the tickets added up to 311.
Second equation is 0.5S + 1.5N = 385.5 because the student ticket were $0.50 and the non-student tickets were $1.50 adding up to a total of $385.50.

The substitution method would probably be the easiest way to solve this problem.

I would chose the first equation to put in the substitution "format"
S + N = 311 --> S = 311 - N

You will substitute that equation into the second equation
0.5(311 - N) + 1.5N = 385.5

Use the distribution method
155.5 - 0.5N +1.5N = 385.5

Simplify
155.5 + 1N = 385.5

Subtract 155.5 on each side
N = 230

Plug in N in the first equation
S = 311 - 230

Solve
S = 81

Student tickets sold = 81
Non-student tickets sold = 230

Answer 4

The two equations are:

X + Y = 311 and .50X + 1.5Y = 385.50

use the first one to solve for either X or Y, and plug the value into the second

X=311-Y plugged into the second = .50(311-Y) + 1.5Y = 385.50

155.50 - .5Y + 1.5Y = 385.50
155.50 + Y = 385.50 same as Y = 230

plug the value of Y back into the first equation to get X
X = 311-230 = 81

proof:
(81 x .50) + (230 x 1.50) = 40.50 + 345.00 = 385.50

Answer 5

a = student ticket
b = nonstudent ticket

311 = a + b

a = 311 - b
b = 311 - a



385.50 = .5a + 1.5b
385.50 = .5a + 1.5(311-a)
385.50 = .5a + 466.5 - 1.5a
-81 = .5a - 1.5a
-81 = -1a
81 = 1a
81 = a

Recall that a = 311 - b
81 = 311 - b
-230 = -b
230 = b

So 81 student tickets and 230 non student tickets were sold.

Answer 6

Let the no of students tickets sold be x
Therefore,the no of non-students tickets sold were 311-x
cost of x students tickets
=$0.50x
Cost of 311-x non-students tickets
=$1.50(311-x)
=$466,50-1.5x
According to the problem,
0.50x+466.50-1.5x=385.50
-x=385.50-466.50= -81
x-81
Therefore,no of student tickets sold were 81
No of non-students tickets sold were
311-81
=230

Answer 7

To solve, you will need to set up a system of equations.

let x stand for the number of student tickets
let y stand for the number of nonstudent tickets

311 = x + y
385.50 = .50x + 1.50y

From there you can use the substitution method to say that:
311 - y = x

3.85.50 = .50(311 - y) + 1.50y
From here you can just distribute and solve for y.
Then plug your y-value into the original equation and solve for x.

Hope this helps!

Answer 8

if s is student tickets and n is non-student tickets:

s + n = 311
0.5s +1.5n = 385.5

so s = 311 -n
and 0.5(311-n) +1.5n = 385.5
therefore 155.5 -0.5n+1.5n =385.5
therefore n = 230
therefore s = 311 - 230 = 81

therefore 230 nonstudent tickets and 81 student tickets were sold.

Answer 9

s = student tickets
n = nonstudent tickets

s + n = 311
.5s + 1.5n = 385.50

times first equation by -1.5
(s + n = 311) -1.5

-1.5s + -1.5n = -466.5

add two eq together
.5s + 1.5n = 385.50
-1.5s + -1.5n = -466.5

equals
-1s = -81
s= 81

now solve for n
s + n = 311
81 + n = 311
n = 230

Answer 10

let S = number of ticket sold to students
311 - S =number of ticket sold to non-students

$0.50 S + $1.50 (311 - S ) = $385.5 . . . . cancel the $
0.50 S + 466.5 - 1.5 S = 385.5
S = 81 students
311-S = 230 . . . . non-students