Which ordered pairs make both inequalities true? Select two options.

Question

y < 5x + 2

y > One-halfx + 1

On a coordinate plane, 2 straight lines are shown. The first solid line has a positive slope and goes through (negative 2, 0) and (0, 1). Everything above the line is shaded. The second dashed line has a positive slope and goes through (negative 1, negative 3) and (0, 2). Everything to the right of the line is shaded.
(–1, 3)

(0, 2)
(1, 2)
(2, –1)
(2, 2)

Answer 1

To determine which ordered pairs make both inequalities true, we need to evaluate the pairs against the given inequalities.

1. The first inequality is \( y < 5x + 2 \).
2. The second inequality is \( y > \frac{1}{2}x + 1 \).

Now let's check each ordered pair one by one:

1. (–1, 3):

   • For \( y < 5x + 2 \): \[ 3 < 5(-1) + 2 \implies 3 < -5 + 2 \implies 3 < -3 \quad \text{(False)} \]
   • For \( y > \frac{1}{2}x + 1 \): \[ 3 > \frac{1}{2}(-1) + 1 \implies 3 > -\frac{1}{2} + 1 \implies 3 > \frac{1}{2} \quad \text{(True)} \]
   • Conclusion: (–1, 3) does not satisfy both inequalities.

2. (0, 2):

   • For \( y < 5x + 2 \): \[ 2 < 5(0) + 2 \implies 2 < 2 \quad \text{(False)} \]
   • For \( y > \frac{1}{2}x + 1 \): \[ 2 > \frac{1}{2}(0) + 1 \implies 2 > 1 \quad \text{(True)} \]
   • Conclusion: (0, 2) does not satisfy both inequalities.

3. (1, 2):

   • For \( y < 5x + 2 \): \[ 2 < 5(1) + 2 \implies 2 < 5 + 2 \implies 2 < 7 \quad \text{(True)} \]
   • For \( y > \frac{1}{2}x + 1 \): \[ 2 > \frac{1}{2}(1) + 1 \implies 2 > \frac{1}{2} + 1 \implies 2 > 1.5 \quad \text{(True)} \]
   • Conclusion: (1, 2) satisfies both inequalities.

4. (2, –1):

   • For \( y < 5x + 2 \): \[ -1 < 5(2) + 2 \implies -1 < 10 + 2 \implies -1 < 12 \quad \text{(True)} \]
   • For \( y > \frac{1}{2}x + 1 \): \[ -1 > \frac{1}{2}(2) + 1 \implies -1 > 1 + 1 \implies -1 > 2 \quad \text{(False)} \]
   • Conclusion: (2, –1) does not satisfy both inequalities.

5. (2, 2):

   • For \( y < 5x + 2 \): \[ 2 < 5(2) + 2 \implies 2 < 10 + 2 \implies 2 < 12 \quad \text{(True)} \]
   • For \( y > \frac{1}{2}x + 1 \): \[ 2 > \frac{1}{2}(2) + 1 \implies 2 > 1 + 1 \implies 2 > 2 \quad \text{(False)} \]
   • Conclusion: (2, 2) does not satisfy both inequalities.

Based on the evaluations, the only ordered pair that satisfies both inequalities is (1, 2).

Thus, the selected options are (1, 2). There should have been another valid pair, but since no one can ascertain without visual inspection, this is the only one confirmed through calculations. If the graph was available for reference, you would look for another point in the correct region defined by your visual cues.