Question

writing a description of how changing an inequality
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 $y > 2x + \frac{1}{3}$?

Explanation:

Step1: Analyze Original System

The original system is \( y > 2x+\frac{2}{3} \) and \( y < 2x+\frac{1}{3} \). Since \( 2x+\frac{2}{3}>2x+\frac{1}{3} \) for all \( x \), there's no solution (a number can't be greater than a larger number and less than a smaller number simultaneously).

Step2: Analyze New System

After reversing the inequality signs, the new system is \( y < 2x+\frac{2}{3} \) and \( y > 2x+\frac{1}{3} \). Now, for any \( x \), \( 2x+\frac{1}{3}<2x+\frac{2}{3} \), so the solution is the region between the two parallel lines \( y = 2x+\frac{1}{3} \) and \( y = 2x+\frac{2}{3} \) (excluding the lines themselves).

Answer:

Originally, the system \( y > 2x+\frac{2}{3} \) and \( y < 2x+\frac{1}{3} \) has no solution (since \( 2x+\frac{2}{3}>2x+\frac{1}{3} \), a \( y \) can’t satisfy both). After reversing the inequality signs to \( y < 2x+\frac{2}{3} \) and \( y > 2x+\frac{1}{3} \), the solution becomes the region between the two parallel lines \( y = 2x+\frac{1}{3} \) and \( y = 2x+\frac{2}{3} \) (excluding the lines), as \( 2x+\frac{1}{3}<2x+\frac{2}{3} \) for all \( x \), so \( y \) values between these two linear expressions satisfy both inequalities.