Let's break down the problem systematically to answer each part.
a) What equation would represent the cost of going to a movie on Friday?
On Friday, we have 10 adults and 25 children, with a total cost of $670. We can express this as:
\[
10a + 25c = 670
\]
Response: 10a + 25c = 670
b) What equation would represent the cost of going to a movie on Sunday?
On Sunday, we have 5 adults and 10 children, with a total cost of $290. We can express this as:
\[
5a + 10c = 290
\]
Response: 5a + 10c = 290
c) What would be the best algebraic method to solve given how the equations are set up?
The equations are structured nicely for elimination, especially since both equations can easily be manipulated to eliminate one of the variables. You can multiply the Sunday equation by 2, to line them up and eliminate the 'c' variable or manipulate them further to eliminate 'a'.
Response: elimination because the equations are already lined up nicely - you could just multiply the Sunday equation by -2 to eliminate the 'a' variable
d) What is the cost of an adult ticket?
Now, let’s solve the system of equations.
We have the following two equations:
- \( 10a + 25c = 670 \) (1)
- \( 5a + 10c = 290 \) (2)
To eliminate \( c \), we can first simplify equation (2):
Dividing equation (2) by 5:
\[
a + 2c = 58 \quad (3)
\]
Now we can multiply equation (3) by -10 to align with equation (1):
\[
-10a - 20c = -580 \quad (4)
\]
Now, we can add equation (1) and equation (4):
\[
(10a + 25c) + (-10a - 20c) = 670 - 580
\]
This reduces to:
\[
5c = 90
\]
Dividing by 5 gives:
\[
c = 18
\]
Now that we know \( c \), plug it back into equation (3):
\[
a + 2(18) = 58
\]
\[
a + 36 = 58
\]
\[
a = 22
\]
Therefore, the cost of an adult ticket is 22.
Response: 22