In which quadrant is there no solution for y 2x + 1 and y 1/2*x -

Question

In which quadrant is there no solution for y ≥ 2x + 1 and y > 1/2*x - 1?

A. I
B. II
C. III
D. IV
E. II and IV

dotsoftme · Answer

To find the quadrant that contains no solutions for the given system of inequalities, we can analyze the region where the shaded areas of both inequalities overlap.
The system is:
 
 $y \ge 2x + 1$ 
 $y > \frac{1}{2}x - 1$  
Step 1: Analyze Inequality 1 ($y \ge 2x + 1$)
 
  Boundary Line:  The line $y = 2x + 1$ has a $y$-intercept at $(0, 1)$ and a positive slope of $2$. It passes through Quadrant III, Quadrant II, and Quadrant I. 
  Shading:  Since $y$ is  greater than or equal to , we shade  above  the line. This region covers parts of Quadrants I, II, and III.  
Step 2: Analyze Inequality 2 ($y > \frac{1}{2}x - 1$)
 
  Boundary Line:  The dashed line $y = \frac{1}{2}x - 1$ has a $y$-intercept at $(0, -1)$ and a shallower positive slope of $\frac{1}{2}$. It passes through Quadrant IV, Quadrant III, and Quadrant I. 
  Shading:  Since $y$ is  greater than , we shade  above  this line as well. This region covers Quadrants I, II, III, and a small upper portion of Quadrant IV.  
Step 3: Find the Overlap (The Solution Set)
Let's check how the combined shaded region interacts with each quadrant:
 
  Quadrant I:  Both lines enter Quadrant I, and shading above both lines completely covers a massive portion of this quadrant. (Contains solutions) 
  Quadrant II:  The region above $y \ge 2x + 1$ fully covers Quadrant II. Since Quadrant II is also entirely above the line $y > \frac{1}{2}x - 1$, the overlap covers almost all of Quadrant II. (Contains solutions) 
  Quadrant III:  To the left of the $y$-axis, the region above $y \ge 2x + 1$ is also comfortably above the lower line $y > \frac{1}{2}x - 1$. (Contains solutions) 
  Quadrant IV:  Quadrant IV lies entirely  below  the first line ($y \ge 2x + 1$). Because the first inequality restricts our solution set exclusively to regions above $y = 2x + 1$,  no solutions can exist in Quadrant IV , even though the second inequality passes through it.  
Conclusion
There is absolutely no point $(x, y)$ in Quadrant IV that satisfies both inequalities simultaneously.
 Correct Answer:D. IV

Nav_124 · Answer

https://gmatclub.com/forum/download/file.php?id=88576 GMAT-Club-Forum-4ry7hpl6.png

AdarshSambare · Answer

We need points that satisfy both:
 
 y ≥ 2x + 1 
 y > 1⁄2x − 1  
The solution region is above both lines.
Check Quadrant IV (x > 0, y < 0):
If x > 0, then 2x + 1 > 1.
So the first inequality requires:
y ≥ 2x + 1 > 1
This means y must be positive.
But in Quadrant IV, y is negative.
Therefore, no point in Quadrant IV can satisfy the inequalities.
 Answer: D. IV .

AKAV46 · Answer

Two inequalities should be tested at the same time. To the left and up of the intersected point you see I, II, III quadrants. So the answer is IV

In which quadrant is there no solution for y 2x + 1 and y > 1/2*x -

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