Consider the inequality representing the total amount spent:
[tex]\[ 1.10x + 2.50y < 10 \][/tex]
We need to check which given pairs [tex]\((x, y)\)[/tex] satisfy this inequality.
1. For the pair [tex]\((-1, 4)\)[/tex]:
[tex]\[
1.10(-1) + 2.50(4) = -1.10 + 10 = 8.90
\][/tex]
Since [tex]\(8.90 < 10\)[/tex], [tex]\((-1, 4)\)[/tex] is a valid solution.
2. For the pair [tex]\((0, 2)\)[/tex]:
[tex]\[
1.10(0) + 2.50(2) = 0 + 5 = 5.00
\][/tex]
Since [tex]\(5.00 < 10\)[/tex], [tex]\((0, 2)\)[/tex] is a valid solution.
3. For the pair [tex]\((3, 2.5)\)[/tex]:
[tex]\[
1.10(3) + 2.50(2.5) = 3.30 + 6.25 = 9.55
\][/tex]
Since [tex]\(9.55 < 10\)[/tex], [tex]\((3, 2.5)\)[/tex] is a valid solution.
4. For the pair [tex]\((2, 4)\)[/tex]:
[tex]\[
1.10(2) + 2.50(4) = 2.20 + 10 = 12.20
\][/tex]
Since [tex]\(12.20 \not< 10\)[/tex], [tex]\((2, 4)\)[/tex] is not a valid solution.
5. For the pair [tex]\((0.5, 3.78)\)[/tex]:
[tex]\[
1.10(0.5) + 2.50(3.78) = 0.55 + 9.45 = 10.00
\][/tex]
Since [tex]\(10.00 \not< 10\)[/tex], [tex]\((0.5, 3.78)\)[/tex] is not a valid solution.
6. For the pair [tex]\((1.5, 3)\)[/tex]:
[tex]\[
1.10(1.5) + 2.50(3) = 1.65 + 7.50 = 9.15
\][/tex]
Since [tex]\(9.15 < 10\)[/tex], [tex]\((1.5, 3)\)[/tex] is a valid solution.
So, the reasonable solutions are:
[tex]\[
(-1, 4), (0, 2), (3, 2.5), (1.5, 3)
\][/tex]
