Question

how will the solution of the system $y > 2x + \frac{2}{3}$ and $y < 2x + \frac{1}{3}$ change if the inequality sign on both inequalities is reversed to $y < 2x + \frac{2}{3}$ and $y > 2x + \frac{1}{3}$? choose three correct answers. there are no solutions before they are reversed. there are an infinite number of solutions in the system once the signs are reversed. there are an infinite number of solutions in both systems. there are no solutions in both systems. the shaded area would reverse on both inequalities.

Explanation:

1. Before Reversal: The system is \( y > 2x+\frac{2}{3} \) and \( y < 2x+\frac{1}{3} \). Since \( \frac{2}{3}>\frac{1}{3} \), the region \( y > 2x+\frac{2}{3} \) lies above \( y = 2x+\frac{2}{3} \), and \( y < 2x+\frac{1}{3} \) lies below \( y = 2x+\frac{1}{3} \). There is no overlap (no solutions) as the line \( y = 2x+\frac{2}{3} \) is above \( y = 2x+\frac{1}{3} \).
2. After Reversal: The system becomes \( y < 2x+\frac{2}{3} \) and \( y > 2x+\frac{1}{3} \). Now, \( y < 2x+\frac{2}{3} \) is below \( y = 2x+\frac{2}{3} \), and \( y > 2x+\frac{1}{3} \) is above \( y = 2x+\frac{1}{3} \). Since \( \frac{2}{3}>\frac{1}{3} \), there is a region between the two lines (infinite solutions) as the area between \( y = 2x+\frac{1}{3} \) and \( y = 2x+\frac{2}{3} \) is valid.
3. Shaded Area Reversal: Reversing the inequality sign (e.g., from \( > \) to \( < \)) reverses the region (shaded area) that satisfies the inequality, so the shaded area for both inequalities reverses.
4. Incorrect Options:

• "There are an infinite number of solutions in both systems" is wrong because before reversal, there are no solutions.
• "There are no solutions in both systems" is wrong because after reversal, there are infinite solutions.

Answer:

• There are no solutions before they are reversed.
• There are an infinite number of solutions in the system once the signs are reversed.
• The shaded area would reverse on both inequalities.