Answer:
Drew burns 300 calories per hour playing basketball and 600 calories per hour canoeing.
Explanation:
To solve this problem, we need to use a system of equations. Let's define two variables:
b = number of calories Drew burns per hour playing basketball
c = number of calories Drew burns per hour canoeing
Using these variables, we can set up two equations based on the information given:
Equation 1: 1b + 2c = 1500 (calories burned on Friday)
Equation 2: 2b + 3c = 2400 (calories burned on Saturday)
To solve for b and c, we can use elimination or substitution. Let's use elimination:
Multiply Equation 1 by 2:
2b + 4c = 3000
Subtract Equation 2 from this new equation:
(2b + 4c) - (2b + 3c) = 4c - 3c
c = 600
Now that we know c = 600, we can substitute this value into either Equation 1 or Equation 2 to solve for b. Let's use Equation 1:
1b + 2(600) = 1500
1b + 1200 = 1500
1b = 300
b = 300
Therefore, Drew burns 300 calories per hour playing basketball and 600 calories per hour canoeing.