Yes, a system of linear inequalities can indeed have no solutions. This situation occurs when the inequalities are contradictory, meaning that the regions defined by the inequalities do not overlap, leaving no common area that satisfies all the conditions simultaneously.
What It Looks Like Graphically
-
Finding No Solutions: Consider a system with two inequalities:
- \( y < x + 1 \) (a line with a slope of 1 and a y-intercept of 1, shaded below)
- \( y > -x + 2 \) (a line with a slope of -1 and a y-intercept of 2, shaded above)
Graphically, the line defined by \( y = x + 1 \) divides the plane such that for any point below this line, the inequality \( y < x + 1 \) holds true. The line defined by \( y = -x + 2 \) divides the plane such that for any point above this line, the inequality \( y > -x + 2 \) holds true.
-
No Overlapping Region: When plotted, these lines will show that the area below the line \( y = x + 1 \) (shaded) and the area above the line \( y = -x + 2 \) (shaded) do not overlap.
-
Verification: If you try to find a point that satisfies both inequalities, you will find that it's impossible because any point below the line \( y = x + 1 \) will not be able to satisfy \( y > -x + 2 \) and vice versa.
Visual Representation:
- The line \( y = x + 1 \) will tilt upwards while the line \( y = -x + 2 \) will tilt downwards. The regions will be shaded in such a way that one is entirely above the other with no intersection.
Conclusion
In conclusion, a system of linear inequalities can have no solutions when the inequalities define regions that do not overlap. Graphically, this situation manifests as non-overlapping shaded regions on a graph.
1 answer