All of the 5th grade teachers and students from Almond went on a field trip to an archaeology museum. Tickets were $8.50 each for teachers and $2.00 each for students, and the group paid $29.00 in total.
A few weeks later, the same group visited a natural history museum where the tickets cost $34.00 each for teachers and $8.50 each for students, and the group paid $119.00 in total.
Find the number of teachers and students on the field trips.
teachers --- t
students --- s
from first case:
8.5t + 2s = 29
from the 2nd case:
34t + 8.5s = 119
System of 2 equations in 2 unknowns.
Just your favourite method of solving.
To solve this problem, let’s assign some variables:
Let t be the number of teachers.
Let s be the number of students.
From the information given, we know that:
8.50t + 2.00s = 29.00 (equation 1) -- this represents the first field trip
and
34.00t + 8.50s = 119.00 (equation 2) -- this represents the second field trip
To solve this system of equations, we can use the method of substitution or elimination.
Method 1: Substitution
1. Rearrange equation 1 to solve for t:
8.50t = 29.00 - 2.00s
t = (29.00 - 2.00s) / 8.50 (equation 3)
2. Substitute equation 3 into equation 2:
34.00((29.00 - 2.00s) / 8.50) + 8.50s = 119.00
3. Now we can solve for s:
Multiply both sides of equation 2 by 8.50 to eliminate decimals:
34.00(29.00 - 2.00s) + 8.50s = 1015.50
Expand and simplify:
986 - 68s + 8.50s = 1015.50
-68s + 8.50s = 1015.50 - 986
-59.50s = 29.50
Divide both sides by -59.50:
s = -29.50 / -59.50
s = 0.5
4. Substitute the value of s back into equation 3 to find t:
t = (29.00 - 2.00(0.5)) / 8.50
t = (29.00 - 1.00) / 8.50
t = 28.00 / 8.50
t = 3.29
Since we cannot have fractional parts of teachers or students, we can assume that the number of teachers and students are:
t ≈ 3 teachers
s ≈ 0 students (or none)
This result may not make sense in the context of the problem, so let's try Method 2: Elimination.
Method 2: Elimination
1. Multiply equation 1 by 4 to eliminate decimals:
34.00t + 8.00s = 116.00 (equation 4)
2. Subtract equation 1 (multiplied by 34) from equation 2 (multiplied by 8) to eliminate t:
34.00t + 8.50s - (34.00t + 8.00s) = 119.00 - 116.00
8.50s - 8.00s = 3.00
0.50s = 3.00
3. Solve for s:
s = 3.00 / 0.50
s = 6 students
4. Substitute the value of s back into equation 1 to find t:
8.50t + 2.00(6) = 29.00
8.50t + 12.00 = 29.00
8.50t = 29.00 - 12.00
8.50t = 17.00
t = 17.00 / 8.50
t = 2 teachers
So, the number of teachers and students on the field trips is:
t = 2 teachers
s = 6 students