To determine a viable solution for the number of robotics kits (y) and chemistry sets (x) Cindy can purchase with her $50.00 winnings, we can set up an equation based on the given information:
Each robotics kit costs $10.00, so if she buys y kits, the total cost for robotics kits is: \[ 10y \]
Each chemistry set costs $8.00, so if she buys x sets, the total cost for chemistry sets is: \[ 8x \]
The total amount she spends should equal her winnings of $50.00: \[ 10y + 8x = 50 \]
Now, we can look for integer solutions \( (x, y) \) where both \( x \) (number of chemistry sets) and \( y \) (number of robotics kits) are non-negative integers.
To find viable solutions, we can rearrange the equation for x: \[ 8x = 50 - 10y \] \[ x = \frac{50 - 10y}{8} \]
For \( x \) to be an integer, \( 50 - 10y \) must be divisible by 8. Let's simplify \( 50 - 10y \):
- When \( y = 0 \): \( x = \frac{50 - 10(0)}{8} = \frac{50}{8} = 6.25 \) (not an integer)
- When \( y = 1 \): \( x = \frac{50 - 10(1)}{8} = \frac{40}{8} = 5 \)
- When \( y = 2 \): \( x = \frac{50 - 10(2)}{8} = \frac{30}{8} = 3.75 \) (not an integer)
- When \( y = 3 \): \( x = \frac{50 - 10(3)}{8} = \frac{20}{8} = 2.5 \) (not an integer)
- When \( y = 4 \): \( x = \frac{50 - 10(4)}{8} = \frac{10}{8} = 1.25 \) (not an integer)
- When \( y = 5 \): \( x = \frac{50 - 10(5)}{8} = \frac{0}{8} = 0 \)
Thus, the viable integer solutions (x, y) where both variables are non-negative integers are:
- \( (5, 1) \) meaning 5 chemistry sets and 1 robotics kit.
- \( (0, 5) \) meaning 0 chemistry sets and 5 robotics kits.
In conclusion, the viable solutions are:
- 5 chemistry sets and 1 robotics kit (x=5, y=1).
- 0 chemistry sets and 5 robotics kits (x=0, y=5).
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