Answer:
(0,2)
(1.5,3)
Explanation:
To determine the reasonable solutions for this inequality we simply substitute the values of x and y into the equation and check whether the condition on the right hand side is met. If the condition is met then the set of values are reasonable solutions otherwise if the condition is violated the set of values are not reasonable solutions;
The first alternative cannot be a reasonable solution since the quantity of broccoli cannot be negative.
For the second alternative, we substitute x with 0 and y with 2 in the left hand side;
1.10(0)+2.50(2) = 5. 5 is less than 10 hence (0,2) is a reasonable solution.
For the third alternative, we substitute x with 3 and y with 2.5 in the left hand side;
1.10(3)+2.50(2.5) = 9.55. 9.55 is less than 10 hence (3,2.5) is a reasonable solution but it will not be applicable in this case since we can not purchase a fraction of a can.
For the fourth alternative, we substitute x with 2 and y with 4 in the left hand side;
1.10(2)+2.50(4) = 12.2. 12.2 is greater than 10 hence (2,4) is not a reasonable solution.
For the fifth alternative, we substitute x with 0.5 and y with 3.78 in the left hand side;
1.10(0.5)+2.50(3.78) = 10. 10 is not less than 10 hence (0.5,3.78) is not a reasonable solution.
For the sixth alternative, we substitute x with 1.5 and y with 3 in the left hand side;
1.10(1.5)+2.50(3) = 9.15. 9.15 is less than 10 hence (1.5,3) is a reasonable solution.