Tickets for the senior play at the local elementary school cost $3 for children, $5 for students, and $8 for adults. 800 tickets were sold and the total in ticket sales was $4750. The number of adult tickets sold was 50 more than two times the number of children tickets. Find the number of each type of ticket using a system of three equations.
a = 50 + 2 c
a + s + c = 800
8 a + 5 s + 3 c = 4750
so
50 + 2 c + s + c = 800
or
s + 3 c = 750 that is just in s and c
then also
8 a + 5 s + 3 c = 4750
8 (50 + 2 c) + 5 s + 3 c = 4750
400 + 16 c + 5 s + 3 c = 4750
5 s + 19 c = 4350 that is also just in s and c
So you have
s + 3 c = 750
5 s + 19 c = 4350
multiply the first equation by 5
5 s + 15 c = 3750
5 s + 19 c = 4350
------------------------- subtract
-4 c = - 600
c = 150
Your turn
To solve this problem using a system of three equations, we can assign variables to the unknowns. Let's use:
- Let's say the number of children's tickets sold is "x".
- The number of student tickets sold is "y".
- The number of adult tickets sold is "z".
Now, we can set up three equations based on the given information:
Equation 1: The total number of tickets sold is 800.
x + y + z = 800
Equation 2: The total ticket sales amount is $4750.
3x + 5y + 8z = 4750
Equation 3: The number of adult tickets sold is 50 more than two times the number of children tickets.
z = 2x + 50
Now we have a system of three equations. We can solve this system using elimination or substitution method.
Let's start with the elimination method:
From Equation 2, which states the total ticket sales amount, we can get rid of decimals by multiplying the equation by 100:
300x + 500y + 800z = 47500
Now we have the system of equations:
x + y + z = 800
300x + 500y + 800z = 47500
z = 2x + 50
From Equation 1, we can rearrange it to express "x" in terms of "z":
x = 800 - y - z
Substituting this expression for "x" in Equation 3, we get:
z = 2(800 - y - z) + 50
z = 1600 - 2y - 2z + 50
3z + 2z = 1650 - 2y
5z = 1650 - 2y
Now, we can substitute this equation for "z" in Equation 2:
300x + 500y + 800(1650 - 2y) = 47500
300x + 500y + 1320000 - 1600y = 47500
300x - 1100y = -1245000
Next, substitute the expression for "x" in terms of "z" obtained earlier into this equation:
300(800 - y - z) - 1100y = -1245000
240000 - 300y - 300z - 1100y = -1245000
-1400y - 300z = -1489000
14y + 3z = 1489 (Divided by -100 to simplify)
Now we have a system of two equations:
5z = 1650 - 2y
14y + 3z = 1489
We can now solve this system of equations using substitution or elimination method.