8. Ivy earns \( \$ 25 \) an hour tutoring students and \( \$ 12 \) an hour working at the library. She wants to earn at least \( \$ 150 \) in a week, but is not able to work for more than 8 hours. (i) Write down an inequality to represent the above information. Define the variables used. (ii) State all the possible combinations of hours that Ivy could tutor or work at the library.

To solve this problem, we will define the variables and then set up the inequalities based on the conditions provided. ### Part (i): Define Variables and Write Inequalities Let: - \( x \) = number of hours Ivy tutors students - \( y \) = number of hours Ivy works at the library From the problem, we have the following conditions: 1. Ivy earns \( \$ 25 \) per hour tutoring, so her earnings from tutoring can be represented as \( 25x \). 2. Ivy earns \( \$ 12 \) per hour working at the library, so her earnings from the library can be represented as \( 12y \). 3. Ivy wants to earn at least \( \$ 150 \), which gives us the inequality: \[ 25x + 12y \geq 150 \] 4. Ivy cannot work more than 8 hours in total, which gives us the inequality: \[ x + y \leq 8 \] Thus, the inequalities representing the situation are: 1. \( 25x + 12y \geq 150 \) 2. \( x + y \leq 8 \) ### Part (ii): Possible Combinations of Hours To find the possible combinations of hours that Ivy could tutor or work at the library, we need to solve the system of inequalities. We will analyze the inequalities to find the feasible region. 1. **From the first inequality** \( 25x + 12y \geq 150 \): - Rearranging gives us \( y \geq \frac{150 - 25x}{12} \). 2. **From the second inequality** \( x + y \leq 8 \): - Rearranging gives us \( y \leq 8 - x \). Next, we will find the intersection points of these lines to determine the feasible region. Let's solve the equations: 1. \( 25x + 12y = 150 \) 2. \( x + y = 8 \) We can solve these equations simultaneously. ### Solving the System of Equations 1. From \( x + y = 8 \), we can express \( y \) in terms of \( x \): \[ y = 8 - x \] 2. Substitute \( y \) into the first equation: \[ 25x + 12(8 - x) = 150 \] Simplifying this: \[ 25x + 96 - 12x = 150 \] \[ 13x + 96 = 150 \] \[ 13x = 54 \] \[ x = \frac{54}{13} \approx 4.15 \] 3. Substitute \( x \) back to find \( y \): \[ y = 8 - \frac{54}{13} = \frac{104 - 54}{13} = \frac{50}{13} \approx 3.85 \] Now we have one intersection point: \( \left( \frac{54}{13}, \frac{50}{13} \right) \). ### Finding the Boundary Points Next, we need to check the boundary points where \( x \) and \( y \) are non-negative and satisfy both inequalities. 1. **When \( x = 0 \)**: - From \( 25(0) + 12y \geq 150 \) → \( 12y \geq 150 \) → \( y \geq 12.5 \) (not possible since \( y \) must be ≤ 8). 2. **When \( y = 0 \)**: - From \( 25x + 12(0) \geq 150 \) → \( 25x \geq 150 \) → \( x \geq 6 \). - From \( x + 0 \leq 8 \) → \( x \leq 8 \). - Thus, \( x \) can be between \( 6 \) and \( 8 \) (i.e., \( 6 \leq x \leq 8 \)). ### Summary of Possible Combinations The possible combinations of hours that Ivy could tutor or work at the library are: - \( (x, y) \) such that: - \( 6 \leq x \leq 8 \) - \( y = 8 - x \) This means: - If \( x = 6 \), then \( y = 2 \). - If \( x = 7 \), then \( y = 1 \). - If \( x = 8 \), then \( y = 0 \). Thus, the possible combinations of hours are: - \( (6, 2) \) - \( (7, 1) \) - \( (8, 0) \)