To determine which ordered pairs make both of the given inequalities true, let's check each pair step-by-step.
The inequalities we need to consider are:
1. [tex]\( y < 5x + 2 \)[/tex]
2. [tex]\( y \geq \frac{1}{2}x + 1 \)[/tex]
Let's analyze each ordered pair:
### Ordered Pair: [tex]\((-1, 3)\)[/tex]
- Check inequality [tex]\( y < 5x + 2 \)[/tex]:
[tex]\[
y = 3, \quad x = -1
\][/tex]
Substituting the values into [tex]\( y < 5x + 2 \)[/tex]:
[tex]\[
3 < 5(-1) + 2
\][/tex]
[tex]\[
3 < -5 + 2
\][/tex]
[tex]\[
3 < -3 \quad \text{(This is false)}
\][/tex]
- Check inequality [tex]\( y \geq \frac{1}{2}x + 1 \)[/tex]:
[tex]\[
y = 3, \quad x = -1
\][/tex]
Substituting the values into [tex]\( y \geq \frac{1}{2}x + 1 \)[/tex]:
[tex]\[
3 \geq \frac{1}{2}(-1) + 1
\][/tex]
[tex]\[
3 \geq -0.5 + 1
\][/tex]
[tex]\[
3 \geq 0.5 \quad \text{(This is true)}
\][/tex]
Since the first inequality is not satisfied, the ordered pair [tex]\((-1, 3)\)[/tex] does not make both inequalities true.
### Ordered Pair: [tex]\((0, 2)\)[/tex]
- Check inequality [tex]\( y < 5x + 2 \)[/tex]:
[tex]\[
y = 2, \quad x = 0
\][/tex]
Substituting the values into [tex]\( y < 5x + 2 \)[/tex]:
[tex]\[
2 < 5(0) + 2
\][/tex]
[tex]\[
2 < 0 + 2
\][/tex]
[tex]\[
2 < 2 \quad \text{(This is false)}
\][/tex]
- Check inequality [tex]\( y \geq \frac{1}{2}x + 1 \)[/tex]:
[tex]\[
y = 2, \quad x = 0
\][/tex]
Substituting the values into [tex]\( y \geq \frac{1}{2}x + 1 \)[/tex]:
[tex]\[
2 \geq \frac{1}{2}(0) + 1
\][/tex]
[tex]\[
2 \geq 0 + 1
\][/tex]
[tex]\[
2 \geq 1 \quad \text{(This is true)}
\][/tex]
Since the first inequality is not satisfied, the ordered pair [tex]\((0, 2)\)[/tex] does not make both inequalities true.
### Conclusion
After checking both ordered pairs, we can see that neither [tex]\( (-1, 3) \)[/tex] nor [tex]\( (0, 2) \)[/tex] makes both inequalities true. Therefore, none of these ordered pairs satisfy both inequalities simultaneously.
