To determine whether there are any solutions to the system of inequalities
[tex]\[
\begin{array}{l}
y < 7x + 2 \\
y > 7x - 3
\end{array}
\][/tex]
we need to analyze both inequalities and their implications. Let's break this down step-by-step:
1. Understanding the inequalities:
- The first inequality is [tex]\( y < 7x + 2 \)[/tex].
- The second inequality is [tex]\( y > 7x - 3 \)[/tex].
2. Consider the graphical representation:
- Both inequalities represent regions on a Cartesian plane bounded by the lines [tex]\( y = 7x + 2 \)[/tex] and [tex]\( y = 7x - 3 \)[/tex].
- These lines have the same slope but different intercepts, which means they are parallel and do not intersect.
3. Implications of parallel lines with different intercepts:
- Parallel lines never meet. The line [tex]\( y = 7x + 2 \)[/tex] is always 5 units above the line [tex]\( y = 7x - 3 \)[/tex], as [tex]\( (7x + 2) - (7x - 3) = 5 \)[/tex].
- Since there is a consistent vertical distance of 5 units between these lines, any point [tex]\( (x, y) \)[/tex] that could potentially satisfy both inequalities would have to lie in between these two lines.
4. Checking the feasibility of simultaneous satisfaction:
- For any point to lie between these lines, it must satisfy [tex]\( 7x - 3 < y < 7x + 2 \)[/tex].
- However, if we think about potential values of [tex]\( y \)[/tex], it asserts that [tex]\( y \)[/tex] cannot be simultaneously less than [tex]\( 7x + 2 \)[/tex] and greater than [tex]\( 7x - 3 \)[/tex] while maintaining these strict inequalities because [tex]\( 7x + 2 \)[/tex] is always greater than [tex]\( 7x - 3 \)[/tex].
5. Conclusion:
- The two inequalities create regions that do not overlap. There is no [tex]\( y \)[/tex] value that can fulfill both [tex]\( y < 7x + 2 \)[/tex] and [tex]\( y > 7x - 3 \)[/tex] at the same time.
- Therefore, there are no solutions that satisfy both inequalities simultaneously.
Thus, the statement that there are no solutions to this system of inequalities is True.
So, the answer is:
A. True
