Caleb runs a farm stand that sells strawberries and grapes. Yesterday Caleb sold 28 pounds of strawberries and 37 pounds of grapes for a total revenue of \( \$ 183.50 \). Today he sold 14 pounds of strawberries and 16 pounds of grapes for a total revenue of \( \$ 85.50 \). Determine the price of each pound of strawberries and the price of each pound of grapes. Answer Attempt 1 out of 10 Each pound of strawberries costs \( \$ \square \) and each pound of grapes costs \( \$ \square \).
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Let \( s \) be the price per pound of strawberries and \( g \) be the price per pound of grapes. We can create a system of equations based on the information provided. From Caleb's sales yesterday: 1. \( 28s + 37g = 183.50 \) (Equation 1) From today's sales: 2. \( 14s + 16g = 85.50 \) (Equation 2) Now, let's solve these equations. First, simplify Equation 2: Multiply Equation 2 by 2 to make it easier to eliminate \( s \): \[ 28s + 32g = 171 \] (Equation 3) Now, subtract Equation 1 from Equation 3: \[ 28s + 32g - (28s + 37g) = 171 - 183.50 \] This simplifies to: \[ -5g = -12.50 \] Now, solve for \( g \): \[ g = \frac{12.50}{5} = 2.50 \] So, each pound of grapes costs \( \$2.50 \). Next, we can substitute \( g \) back into either Equation 1 or Equation 2 to find \( s \). Using Equation 1: \[ 28s + 37(2.50) = 183.50 \] This simplifies to: \[ 28s + 92.50 = 183.50 \] Now subtract 92.50 from both sides: \[ 28s = 91 \] Dividing by 28 gives: \[ s = \frac{91}{28} \approx 3.25 \] So, each pound of strawberries costs \( \$3.25 \). Final Answer: Each pound of strawberries costs \( \$3.25 \) and each pound of grapes costs \( \$2.50 \).