Question

Which ordered pair is in the solution set of 8x+16y>32
?

Answer 1

To determine which ordered pair is in the solution set of the inequality \( 8x + 16y > 32 \), we can first simplify the inequality:

\[ 8x + 16y > 32 \]

Dividing every term by 8 gives:

\[ x + 2y > 4 \]

Now we want to find an ordered pair \((x, y)\) that satisfies this inequality.

Let's try some examples:

1. **(0, 3)**:
- Substitute \(x = 0\) and \(y = 3\):
\[
0 + 2(3) = 0 + 6 = 6 > 4 \quad \text{(True)}
\]
Thus, \((0, 3)\) is in the solution set.

2. **(1, 1)**:
- Substitute \(x = 1\) and \(y = 1\):
\[
1 + 2(1) = 1 + 2 = 3 > 4 \quad \text{(False)}
\]
Thus, \((1, 1)\) is not in the solution set.

3. **(4, 0)**:
- Substitute \(x = 4\) and \(y = 0\):
\[
4 + 2(0) = 4 + 0 = 4 > 4 \quad \text{(False)}
\]
Thus, \((4, 0)\) is not in the solution set.

4. **(2, 2)**:
- Substitute \(x = 2\) and \(y = 2\):
\[
2 + 2(2) = 2 + 4 = 6 > 4 \quad \text{(True)}
\]
Thus, \((2, 2)\) is in the solution set.

From these calculations, \((0, 3)\) and \((2, 2)\) are valid solutions.

If you have multiple specific ordered pairs you'd like to test, please provide them, and I can determine if they satisfy the given inequality!