Let \( a \) represent the cost of an adult ticket and \( c \) represent the cost of a children's ticket.
a) The equation for the cost of going to a movie on Friday:
On Friday, 10 adults and 25 children went to the movie, and the total cost was $670. The equation can be set up as follows:
\[ 10a + 25c = 670 \]
b) The equation for the cost of going to a movie on Sunday:
On Sunday, 5 adults and 10 children went to the movie, and the total cost was $290. The equation can be established as:
\[ 5a + 10c = 290 \]
c) The best algebraic method to solve the equations:
Since we have a system of two equations, we can use the substitution method or the elimination method to solve for the variables \( a \) and \( c \). For this case, elimination might be more straightforward since both equations can be adjusted to eliminate one of the variables easily.
d) Solving the equations:
We have the following two equations:
- \( 10a + 25c = 670 \) (1)
- \( 5a + 10c = 290 \) (2)
First, let's simplify equation (2) by dividing everything by 5:
\[ a + 2c = 58 \quad (3) \]
Now we can solve equation (3) for \( a \):
\[ a = 58 - 2c \]
Next, substitute \( a \) from equation (3) into equation (1):
\[ 10(58 - 2c) + 25c = 670 \]
Distributing the 10:
\[ 580 - 20c + 25c = 670 \]
Combine like terms:
\[ 580 + 5c = 670 \]
To isolate \( c \), subtract 580 from both sides:
\[ 5c = 90 \]
Now divide by 5:
\[ c = 18 \]
Now, substitute \( c = 18 \) back into equation (3) to find \( a \):
\[ a = 58 - 2(18) = 58 - 36 = 22 \]
Summary:
- The cost of an adult ticket (\( a \)) is $22.
- The cost of a children's ticket (\( c \)) is $18.
Thus, the final answers are:
- Cost of adult ticket: \( \boxed{22} \)
- Cost of children's ticket: \( \boxed{18} \)