To determine which of the labeled points is a viable solution for the inequality \( 25z + 15y < 200 \), we need to substitute each point into the inequality and see if it satisfies it.
Let’s denote \( z \) (number of lawns mowed) as \( x \) and \( y \) as the number of dogs walked. So, we are checking the inequality \( 25x + 15y < 200 \).
Checking each point:
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Point (20, 11.5): \[ 25(20) + 15(11.5) = 500 + 172.5 = 672.5 , (not , a , solution) \]
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Point (2, 3): \[ 25(2) + 15(3) = 50 + 45 = 95 , (< 200 , (is , a , solution)) \]
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Point (10, 10): \[ 25(10) + 15(10) = 250 + 150 = 400 , (not , a , solution) \]
Conclusion:
The only point that satisfies the inequality \( 25z + 15y < 200 \) is (2, 3). Thus, the viable solution in this context is (2, 3).