Let x be the number of hardbacks and y be the number of paperbacks that Kiani purchases.
The two-variable inequality for the budget constraint is:
12x + 5y ≤ 60
The constraints are:
x ≥ 0 (non-negative number of hardbacks)
y ≥ 0 (non-negative number of paperbacks)
Now, we substitute the given points into the inequality to determine which one represents a viable solution.
1) (6,6):
12(6) + 5(6) = 72 + 30 = 102 > 60
This point does not satisfy the inequality, so it is not a viable solution.
2) (-2,6):
12(-2) + 5(6) = -24 + 30 = 6 ≤ 60
This point satisfies the inequality and the constraints. It represents a viable solution.
3) (1,4.5):
12(1) + 5(4.5) = 12 + 22.5 = 34.5 ≤ 60
This point satisfies the inequality and the constraints. It represents a viable solution.
4) (2,4):
12(2) + 5(4) = 24 + 20 = 44 ≤ 60
This point satisfies the inequality and the constraints. It represents a viable solution.
Therefore, the points (−2,6), (1,4.5), and (2,4) all represent viable solutions.
Kiani is shopping at a bookstore with a budget of $60. Hardbacks cost $12 and paperbacks cost $5. Write a two-variable inequality and determine the constraints of the problem. Which of the following points represents a viable solution?(1 point)
Responses
(6,6)
(−2,6)
(1,4.5)
(2,4)
5 answers
you can only choose one
Apologies for the confusion. In that case, the point that represents a viable solution is (2,4).
This graph shows the inequality 25x+15y≥200 , and represents Lina’s earnings for lawn mowing and dog walking where x represents the number of lawns mowed and y represents the number of dogs walked. Which of the labeled points is a viable solution in this context?
(1 point)
Responses
None of the points are a viable solution.
(2,3)
(10,10)
(20,11.5)
(1 point)
Responses
None of the points are a viable solution.
(2,3)
(10,10)
(20,11.5)
To determine which of the labeled points is a viable solution, we substitute the x and y values into the inequality 25x + 15y ≥ 200.
1) (2,3):
25(2) + 15(3) = 50 + 45 = 95 < 200
This point does not satisfy the inequality and is not a viable solution.
2) (10,10):
25(10) + 15(10) = 250 + 150 = 400 ≥ 200
This point satisfies the inequality and is a viable solution.
3) (20,11.5):
25(20) + 15(11.5) = 500 + 172.5 = 672.5 ≥ 200
This point satisfies the inequality and is a viable solution.
Therefore, the viable solution is (10,10).
1) (2,3):
25(2) + 15(3) = 50 + 45 = 95 < 200
This point does not satisfy the inequality and is not a viable solution.
2) (10,10):
25(10) + 15(10) = 250 + 150 = 400 ≥ 200
This point satisfies the inequality and is a viable solution.
3) (20,11.5):
25(20) + 15(11.5) = 500 + 172.5 = 672.5 ≥ 200
This point satisfies the inequality and is a viable solution.
Therefore, the viable solution is (10,10).