Michelle sold tickets for the basketball game. Each adult ticket costs 3$ and each student ticket costs 1.50. There were 105 tickets sold for a total of 250$. How many of each type of ticket were sold?
x = adult tkts
y = student tkts
3x = value of adult tkts
1.5y = value of student tkts
x + y = 105
3x + 1.5y = 250
solve these equation together
for your answer
post back if you need more help
How do you solve the equations? ):
(1) x + y = 105
(2) 3x + 1.5y = 250
multiply equation (1) by -3
-3 (x + y = 105) = -3x - 3y = -315
add the two equations together
-3x - 3y = -315
3x + 1.5y = 250
0 - 1.5y = - 65
-1.5y = -65
y = 43.33 student tkts
x + y = 105
x + 43.33 = 105
x = 61.67 student tkts
are you sure you typed this problem right?
because there were 61.67 adult tkts
and 43.33 student tkts
the number of tkts should not be a fraction, but I checked this 4 times
To solve this problem, we can use a system of equations. Let's define two variables:
Let A represent the number of adult tickets sold.
Let S represent the number of student tickets sold.
Based on the given information, we can write two equations:
Equation 1: A + S = 105
This equation represents the total number of tickets sold, which is 105.
Equation 2: 3A + 1.50S = 250
This equation represents the total amount of money collected from ticket sales, which is $250.
To solve this system of equations, we can use the substitution method or the elimination method. Let's use the substitution method in this case.
From Equation 1, we can solve for A:
A = 105 - S
Substitute this value of A into Equation 2:
3(105 - S) + 1.50S = 250
Now, simplify and solve for S:
315 - 3S + 1.50S = 250
-1.50S = 250 - 315
-1.50S = -65
S = -65 / -1.50
S ≈ 43.33 (approx.)
Since the number of tickets must be a whole number (cannot have a fraction of a ticket), we can round this to the nearest whole number. In this case, we have three options: 43 student tickets, 44 student tickets, or 43 adult tickets.
Let's calculate the corresponding number of adult tickets for each case:
Case 1: If S = 43, then A = 105 - 43 = 62
In this case, 62 adult tickets and 43 student tickets were sold.
Case 2 (with approx. values): If S = 44, then A = 105 - 44 = 61
In this case, 61 adult tickets and 44 student tickets were sold.
Case 3: If A = 43, then S = 105 - 43 = 62
In this case, 43 adult tickets and 62 student tickets were sold.
Therefore, the possible combinations of tickets sold are:
1. 62 adult tickets and 43 student tickets
2. 61 adult tickets and 44 student tickets
3. 43 adult tickets and 62 student tickets.
These are the possible solutions based on the given information.