To address how the solution changes when the inequalities are reversed, let's examine the initial and the modified systems of inequalities step-by-step.
### Initial System of Inequalities
The initial system of inequalities is:
1. [tex]\( y > 2x + \frac{2}{3} \)[/tex]
2. [tex]\( y < 2x + \frac{1}{3} \)[/tex]
#### Interpret the Initial System
- The first inequality, [tex]\( y > 2x + \frac{2}{3} \)[/tex], represents the region above the line [tex]\( y = 2x + \frac{2}{3} \)[/tex].
- The second inequality, [tex]\( y < 2x + \frac{1}{3} \)[/tex], represents the region below the line [tex]\( y = 2x + \frac{1}{3} \)[/tex].
Since [tex]\( y = 2x + \frac{2}{3} \)[/tex] and [tex]\( y = 2x + \frac{1}{3} \)[/tex] are parallel lines (both have the same slope of 2), the solution to the initial system is the region that lies above the line [tex]\( y = 2x + \frac{2}{3} \)[/tex] and below the line [tex]\( y = 2x + \frac{1}{3} \)[/tex].
### Modified System of Inequalities
When we reverse the inequality signs, the system becomes:
1. [tex]\( y < 2x + \frac{2}{3} \)[/tex]
2. [tex]\( y > 2x + \frac{1}{3} \)[/tex]
#### Interpret the Modified System
- The first inequality, [tex]\( y < 2x + \frac{2}{3} \)[/tex], now represents the region below the line [tex]\( y = 2x + \frac{2}{3} \)[/tex].
- The second inequality, [tex]\( y > 2x + \frac{1}{3} \)[/tex], now represents the region above the line [tex]\( y = 2x + \frac{1}{3} \)[/tex].
Thus, the solution to the modified system is the region that lies below the line [tex]\( y = 2x + \frac{2}{3} \)[/tex] and above the line [tex]\( y = 2x + \frac{1}{3} \)[/tex].
### Conclusion
By reversing the inequality signs in the original system, the solution becomes the area between the parallel lines [tex]\( y = 2x + \frac{2}{3} \)[/tex] and [tex]\( y = 2x + \frac{1}{3} \)[/tex]. Therefore, the solution to the modified system is:
[tex]\[
\boxed{\text{The solution is the area between the parallel lines } y = 2x + \frac{1}{3} \text{ and } y = 2x + \frac{2}{3}.}
\][/tex]
