Answer :
To determine the maximum number of ride tickets [tex]\( r \)[/tex] Alana can buy while satisfying the given constraints, we need to consider the system of inequalities:
1. [tex]\( r + f \geq 16 \)[/tex] (Alana buys at least 16 tickets in total)
2. [tex]\( 4r + 2f \leq 40 \)[/tex] (Alana spends at most \$40 on tickets)
Let's go through the steps to solve these inequalities:
1. Analyze and Simplify the Budget Inequality:
We can divide the second inequality by 2 to simplify it:
[tex]\[ 4r + 2f \leq 40 \implies 2r + f \leq 20 \][/tex]
2. Identify Constraints:
We now have two inequalities:
[tex]\[ r + f \geq 16 \][/tex]
[tex]\[ 2r + f \leq 20 \][/tex]
3. Testing Options:
Let's test each option to ensure it satisfies both inequalities.
- For [tex]\( r = 4 \)[/tex]:
[tex]\[ r + f \geq 16 \implies 4 + f \geq 16 \implies f \geq 12 \][/tex]
[tex]\[ 2r + f \leq 20 \implies 2 \cdot 4 + f \leq 20 \implies 8 + f \leq 20 \implies f \leq 12 \][/tex]
Both [tex]\( f \geq 12 \)[/tex] and [tex]\( f \leq 12 \)[/tex] are satisfied, so [tex]\( f = 12 \)[/tex]. Thus, [tex]\( r = 4 \)[/tex] is a valid option.
- For [tex]\( r = 6 \)[/tex]:
[tex]\[ r + f \geq 16 \implies 6 + f \geq 16 \implies f \geq 10 \][/tex]
[tex]\[ 2r + f \leq 20 \implies 2 \cdot 6 + f \leq 20 \implies 12 + f \leq 20 \implies f \leq 8 \][/tex]
Here, [tex]\( f \geq 10 \)[/tex] and [tex]\( f \leq 8 \)[/tex] are conflicting, so [tex]\( r = 6 \)[/tex] is not a valid option.
- For [tex]\( r = 10 \)[/tex]:
[tex]\[ r + f \geq 16 \implies 10 + f \geq 16 \implies f \geq 6 \][/tex]
[tex]\[ 2r + f \leq 20 \implies 2 \cdot 10 + f \leq 20 \implies 20 + f \leq 20 \implies f \leq 0 \][/tex]
Here, [tex]\( f \geq 6 \)[/tex] and [tex]\( f \leq 0 \)[/tex] are conflicting, so [tex]\( r = 10 \)[/tex] is not a valid option.
- For [tex]\( r = 12 \)[/tex]:
[tex]\[ r + f \geq 16 \implies 12 + f \geq 16 \implies f \geq 4 \][/tex]
[tex]\[ 2r + f \leq 20 \implies 2 \cdot 12 + f \leq 20 \implies 24 + f \leq 20 \implies f \leq -4 \][/tex]
Here, [tex]\( f \geq 4 \)[/tex] and [tex]\( f \leq -4 \)[/tex] are conflicting, so [tex]\( r = 12 \)[/tex] is not a valid option.
Therefore, after testing all given options, the maximum number of ride tickets Alana can buy while satisfying the given inequalities is [tex]\( \boxed{4} \)[/tex].
1. [tex]\( r + f \geq 16 \)[/tex] (Alana buys at least 16 tickets in total)
2. [tex]\( 4r + 2f \leq 40 \)[/tex] (Alana spends at most \$40 on tickets)
Let's go through the steps to solve these inequalities:
1. Analyze and Simplify the Budget Inequality:
We can divide the second inequality by 2 to simplify it:
[tex]\[ 4r + 2f \leq 40 \implies 2r + f \leq 20 \][/tex]
2. Identify Constraints:
We now have two inequalities:
[tex]\[ r + f \geq 16 \][/tex]
[tex]\[ 2r + f \leq 20 \][/tex]
3. Testing Options:
Let's test each option to ensure it satisfies both inequalities.
- For [tex]\( r = 4 \)[/tex]:
[tex]\[ r + f \geq 16 \implies 4 + f \geq 16 \implies f \geq 12 \][/tex]
[tex]\[ 2r + f \leq 20 \implies 2 \cdot 4 + f \leq 20 \implies 8 + f \leq 20 \implies f \leq 12 \][/tex]
Both [tex]\( f \geq 12 \)[/tex] and [tex]\( f \leq 12 \)[/tex] are satisfied, so [tex]\( f = 12 \)[/tex]. Thus, [tex]\( r = 4 \)[/tex] is a valid option.
- For [tex]\( r = 6 \)[/tex]:
[tex]\[ r + f \geq 16 \implies 6 + f \geq 16 \implies f \geq 10 \][/tex]
[tex]\[ 2r + f \leq 20 \implies 2 \cdot 6 + f \leq 20 \implies 12 + f \leq 20 \implies f \leq 8 \][/tex]
Here, [tex]\( f \geq 10 \)[/tex] and [tex]\( f \leq 8 \)[/tex] are conflicting, so [tex]\( r = 6 \)[/tex] is not a valid option.
- For [tex]\( r = 10 \)[/tex]:
[tex]\[ r + f \geq 16 \implies 10 + f \geq 16 \implies f \geq 6 \][/tex]
[tex]\[ 2r + f \leq 20 \implies 2 \cdot 10 + f \leq 20 \implies 20 + f \leq 20 \implies f \leq 0 \][/tex]
Here, [tex]\( f \geq 6 \)[/tex] and [tex]\( f \leq 0 \)[/tex] are conflicting, so [tex]\( r = 10 \)[/tex] is not a valid option.
- For [tex]\( r = 12 \)[/tex]:
[tex]\[ r + f \geq 16 \implies 12 + f \geq 16 \implies f \geq 4 \][/tex]
[tex]\[ 2r + f \leq 20 \implies 2 \cdot 12 + f \leq 20 \implies 24 + f \leq 20 \implies f \leq -4 \][/tex]
Here, [tex]\( f \geq 4 \)[/tex] and [tex]\( f \leq -4 \)[/tex] are conflicting, so [tex]\( r = 12 \)[/tex] is not a valid option.
Therefore, after testing all given options, the maximum number of ride tickets Alana can buy while satisfying the given inequalities is [tex]\( \boxed{4} \)[/tex].