# Step-by-Step Solution ## 1. Understanding the Problem - Let \( x \) be the amount of water used (in HCF). - Let \( C(x) \) be the cost (in dollars). - The relationship is linear: \( C(x) = mx + b \). Given: - \( C(21) = 37.41 \) - \( C(7) = 93.21 \) - Find \( C(28) \) (the cost for 28 HCF). --- ## 2. Set Up the System of Equations We have: 1. \( m \cdot 21 + b = 37.41 \) 2. \( m \cdot 7 + b = 93.21 \) --- ## 3. Solve for \( m \) and \( b \) Subtract equation 2 from equation 1: \[ (m \cdot 21 + b) - (m \cdot 7 + b) = 37.41 - 93.21 \] \[ m \cdot (21 - 7) = -55.80 \] \[ m \cdot 14 = -55.80 \] \[ m = \frac{-55.80}{14} = -3.9857 \] Now substitute \( m \) into one of the original equations (use \( C(21) = 37.41 \)): \[ -3.9857 \cdot 21 + b = 37.41 \] \[ -83.700 + b = 37.41 \] \[ b = 37.41 + 83.700 = 121.11 \] --- ## 4. Write the Cost Function \[ C(x) = -3.9857x + 121.11 \] --- ## 5. Calculate the Cost for 28 HCF \[ C(28) = -3.9857 \cdot 28 + 121.11 \] \[ C(28) = -111.5996 + 121.11 \] \[ C(28) = 9.5104 \] --- ## 6. Final Answer \[ \boxed{C(28) \approx \$9.51} \] --- ### Summary Table | HCF Used | Cost (\$) | |----------|----------| | 7 | 93.21 | | 21 | 37.41 | | 28 | 9.51 |

# Step-by-Step Solution ## 1. Understanding the Problem We need to determine the monthly cost of water based on a linear function, which means the relationship between the cost and the amount of water used is a straight line. - Let \( x \) represent the amount of water used (in HCF). - Let \( C(x) \) denote the cost (in dollars). - Since the relationship is linear, we can express it as: \[ C(x) = mx + b \] where \( m \) is the slope and \( b \) is the y-intercept. ### Given Information: - Using 21 HCF costs $37.41: \( C(21) = 37.41 \) - Using 7 HCF costs $93.21: \( C(7) = 93.21 \) - We need to find \( C(28) \). --- ## 2. Set Up the System of Equations Using the given information, we can set up two equations based on the linear function: 1. From \( C(21) = 37.41 \): \[ 21m + b = 37.41 \] 2. From \( C(7) = 93.21 \): \[ 7m + b = 93.21 \] --- ## 3. Solve for \( m \) and \( b \) To eliminate \( b \), subtract the second equation from the first: \[ (21m + b) - (7m + b) = 37.41 - 93.21 \] This simplifies to: \[ 14m = -55.80 \] Solving for \( m \): \[ m = \frac{-55.80}{14} \approx -3.9857 \] Next, substitute \( m \) back into one of the original equations to find \( b \). Using the first equation: \[ -3.9857 \cdot 21 + b = 37.41 \] Calculating the left side: \[ -83.700 + b = 37.41 \] Solving for \( b \): \[ b = 37.41 + 83.700 = 121.11 \] --- ## 4. Write the Cost Function Now that we have \( m \) and \( b \), we can write the cost function: \[ C(x) = -3.9857x + 121.11 \] --- ## 5. Calculate the Cost for 28 HCF To find the cost for 28 HCF, substitute \( x = 28 \) into the cost function: \[ C(28) = -3.9857 \cdot 28 + 121.11 \] Calculating the cost: \[ C(28) = -111.5996 + 121.11 = 9.5104 \] --- ## 6. Final Answer Therefore, the cost for using 28 HCF of water is approximately: \[ \boxed{C(28) \approx \$9.51} \] --- ### Summary Table | HCF Used | Cost (\$) | |----------|-----------| | 7 | 93.21 | | 21 | 37.41 | | 28 | 9.51 | This table summarizes the costs associated with different amounts of water usage, confirming that as the amount of water used increases, the cost decreases in this specific scenario.