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A movie theater charges different prices for children and adults. On Friday, 10 adults and 25 children went to the movie and it...Question
A movie theater charges different prices for children and adults. On Friday, 10 adults and 25 children went to the movie and it cost $670. On Sunday 5 adults and 10 children went and it cost $290.
a) What equation would represent the cost of going to a movie on Friday? 10a + 25c = 670, 5a + 10c = 290
b) What equation would represent the cost of going to a movie on Sunday? 10a + 25c = 670, 5a + 10c = 290
c) What would be the best algebraic method to solve given how the equations are set up? Substitution because one of the equations already has a variable solved for, Elimination because the equations are already lines 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? 18, 58, 29, 22
a) What equation would represent the cost of going to a movie on Friday? 10a + 25c = 670, 5a + 10c = 290
b) What equation would represent the cost of going to a movie on Sunday? 10a + 25c = 670, 5a + 10c = 290
c) What would be the best algebraic method to solve given how the equations are set up? Substitution because one of the equations already has a variable solved for, Elimination because the equations are already lines 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? 18, 58, 29, 22
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Answered by
GPT-4o mini
Let's break down the problem step-by-step.
### a) Equation representing the cost of going to a movie on Friday:
The cost for Friday can be modeled with the equation:
- **10a + 25c = 670**
Here, \(a\) represents the cost of an adult ticket, and \(c\) represents the cost of a child ticket.
### b) Equation representing the cost of going to a movie on Sunday:
The cost for Sunday can be modeled with the equation:
- **5a + 10c = 290**
### c) Best algebraic method to solve the system:
To solve the system of equations:
- From the two equations, we have:
1. **10a + 25c = 670** (Equation 1)
2. **5a + 10c = 290** (Equation 2)
A good method to solve would be **Elimination** because the second equation can be easily manipulated. You can multiply the second equation by -2 to align the coefficients of \(a\):
Multiply Equation 2 by -2:
- **-10a - 20c = -580** (Equation 3)
Now we can add Equation 1 and Equation 3:
- \( (10a + 25c) + (-10a - 20c) = 670 - 580 \)
- \( 5c = 90 \)
- \( c = 18 \)
Now, substitute \(c = 18\) back into one of the original equations (let's use Equation 2):
\[ 5a + 10(18) = 290 \]
\[ 5a + 180 = 290 \]
\[ 5a = 110 \]
\[ a = 22 \]
### d) Cost of an adult ticket:
From our calculations, we find the cost of an adult ticket is:
- **22**
Thus, to summarize:
- **a)** 10a + 25c = 670
- **b)** 5a + 10c = 290
- **c)** Elimination method is preferable.
- **d)** The cost of an adult ticket is **22**.
### a) Equation representing the cost of going to a movie on Friday:
The cost for Friday can be modeled with the equation:
- **10a + 25c = 670**
Here, \(a\) represents the cost of an adult ticket, and \(c\) represents the cost of a child ticket.
### b) Equation representing the cost of going to a movie on Sunday:
The cost for Sunday can be modeled with the equation:
- **5a + 10c = 290**
### c) Best algebraic method to solve the system:
To solve the system of equations:
- From the two equations, we have:
1. **10a + 25c = 670** (Equation 1)
2. **5a + 10c = 290** (Equation 2)
A good method to solve would be **Elimination** because the second equation can be easily manipulated. You can multiply the second equation by -2 to align the coefficients of \(a\):
Multiply Equation 2 by -2:
- **-10a - 20c = -580** (Equation 3)
Now we can add Equation 1 and Equation 3:
- \( (10a + 25c) + (-10a - 20c) = 670 - 580 \)
- \( 5c = 90 \)
- \( c = 18 \)
Now, substitute \(c = 18\) back into one of the original equations (let's use Equation 2):
\[ 5a + 10(18) = 290 \]
\[ 5a + 180 = 290 \]
\[ 5a = 110 \]
\[ a = 22 \]
### d) Cost of an adult ticket:
From our calculations, we find the cost of an adult ticket is:
- **22**
Thus, to summarize:
- **a)** 10a + 25c = 670
- **b)** 5a + 10c = 290
- **c)** Elimination method is preferable.
- **d)** The cost of an adult ticket is **22**.
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