In the rectangular coordinate system shown above, which quadrant, if

Attachment:

The attachment Untitled.png is no longer available

In the rectangular coordinate system shown above, which quadrant, if any, contains no point (x,y) that satisfies the inequality \(2x - 3y\leq{- 6}\) ?

(A) None
(B) I
(C) II
(D) III
(E) IV

Sol: The inequality can be 2x=3y=-6 can be re-written in Y intercept form as y=2x/3 +2



Notice that the slop of the line is positive and hence it will definitely pass through Quad 1 and 3.
If the slope was negative then the line will definitely pass through Quad 2 and 4.



Now for x=0, y= 2 that means line will have to pass through Quad 2 as well.
Hence No point in Quad 4 will satisfy the given equation.

Guys, I am still not clear with the solution. By any way can you simplify it further ?

Hi kshitij89 - For every point lying on the line segment \(2x - 3y =-6\), the x and y coordinates are such that subtracting 3 times the y coordinate from 2 times the x coordinate is equal to -6. Examples are (3,4), (6,6) etc.

For any other point not lying on this line segment, this difference of 2x and 3y is either less than -6 or greater than -6.

The question asks us to find the location of points for which \(2x - 3y <= -6\) is not true i.e. points for which the difference of 2x and 3y is not less than or equal to 6. For finding such points, we need to first plot its equivalent line segment in the X-Y coordinate system.

Let's see how we can plot the line. We need two points for plotting the line segment.We know that for all points on the X-axis, their Y coordinate is 0 and vice versa. So, putting x =0 in the equation of the line segment, we get the value of y = 2 and for y =0, we get the value of x = -3. Now, we have the x and the y - intercept for the line segment. Using this information, we can plot the line segment as shown below:



Now, we will need some test case to establish that on which side of the line the points do not satisfy the inequality \(2x - 3y <= -6\). The best way is to test for the intersection point of X & Y axis i.e. (0, 0). If we put x =0 and y =0 in the inequality, we get \(0 <=-6\) which is not true. So, we can say with certainty that the side of the line which contains ( 0, 0) does not satisfy the inequality.

This would mean that the area on the left side of the line segment \(2x - 3y =6\) satisfy the inequality \(2x - 3y <= -6\) and the area on the right side of the line segment do not satisfy the inequality. Since the question talks about such regions in terms of quadrants, we can observe that area on the left side of the line segment passes through Q- I, II & III only. So, we can say for sure that Q- IV does not contain any point that satisfy the inequality.

Hope this is clear. Let me know if you still have trouble in understanding of the solution.

Regards
Harsh

Hi Harsh,

Thanks for your reply.

The explanation was crisp and clear but I am not sure if I will be able to solve similar questions with different nos.

Can you please provide examples of similar questions to further test my understanding ?

Regards
Kshitij

Hi kshitij89,

How are your overall graphing "skills"? Are you comfortable with the basic concepts, formulas, drawing graphs, etc.? If so, then you'll probably handle the concept on Test Day just fine.

"Graphing", as a category, is relatively rare on the GMAT - you'll likely see just 1-2 graphing questions on Test Day and they will probably be considerably easier than this one.

Unless you've already mastered all of the 'big' categories (Algebra, Arithmetic, Formulas, Broader Geometry, Ratios, DS, etc.), then this nitpick category really isn't worth the extra time.

GMAT assassins aren't born, they're made,
Rich

Hi kshitij89 - what you need is more practice to get yourself comfortable with the graphical method of solving inequalities. To begin with, I would suggest you to plot lines on the X-Y coordinate system and find out points which lie on either side of the line. You may further extend this exercise to plotting of 2 lines in the X-Y coordinate system and finding points which satisfy corresponding inequalities of both the lines.

Once you get used to it, you will prefer using the graphical method for solving inequalities. Getting comfortable with the X-Y coordinate system will also strengthen your understanding of the Coordinate Geometry section

For your practice, you may refer the following posts which uses graphical method for solving questions on inequalities and coordinate geometry:

in-the-xy-plane-region-r-consists-of-all-the-points-x-y-102233.html
in-the-xy-plane-region-a-consists-of-all-the-points-x-y-154784.html?hilit=inequalities%20inequalities%20graph
point-x-y-is-a-point-within-the-triangle-what-is-the-139419.html
set-t-consists-of-all-points-x-y-such-that-x-2-y-2-1-if-15626.html
in-the-coordinate-plane-rectangular-region-r-has-vertices-a-104869.html

Hope it helps!

Regards
Harsh

1. Rewrite the equation (Y=mx+b) --> y=>2/3*x+2
2. Set x=0 and then y=0 --> (0, =>2); (-3<=, 0) so we have now two points the coordinate plane
3. Draw the line (see attachment) --Y you can see that Quadrant IV is not involved there (E)

Let’s rewrite the inequality:

-3y ≤ -2x - 6

y ≥ (2/3)x + 2

We see that the graph of the inequality y ≥ (2/3)x + 2 consists of the line y = (2/3)x + 2, which is a positively sloped line with a y-intercept of 2 and the region above this line. Thus, the one quadrant that would not satisfy this inequality is quadrant IV.

Answer: E

Rewrite the inequality in slope-intercept form for the inequality's boundary line

\(2x – 3y ≤ –6\)

\(-3y ≤ -2x - 6\)

Because dividing by -3, switch the sign:

\(y \geq \frac{2}{3} x + 2\)

Slope is positive.
Positive slope always runs through Quadrants I and III

Eliminate answers B and D

Find the intercepts to graph the line and/or to assess signs

Set x equal to 0 to find y-intercept of the inequality's boundary line

\(y \geq \frac{2}{3} (0) + 2\)

\(y \geq 2\)

Set y equal to 0 to find x-intercept of boundary line

\(0 \geq \frac{2}{3} x + 2\)

\(-2 \geq \frac{2}{3} x\)

\((\frac{3}{2})(-2) \geq{x}\)

\(x ≤ - 3\)

Intercepts of the inequality's boundary line are (0,2) and (-3,0)

Assess
I graphed the intercepts, connected the points and extended the line. See diagram.


No point in Quadrant IV satisfies the inequality

You don't have to graph.
(I graph often. For me, it is quick. This problem took just under a minute.)

Quadrants II has positive y-values and negative x-values.
-- y-intercept is positive
-- x-intercept is negative
Those signs fit: The boundary line runs through Quadrant II.
Eliminate Answer C

At this point, we know the graphed inequality's boundary line runs through Quadrants I, II, and III.
All straight lines pass through at least two, and at most three, quadrants.
(Most straight lines pass through three quadrants. Never four.)

By POE, Quadrant IV contains no points that satisfy the inequality.

AND/OR

Quadrant IV has no positive y-values and no negative x-values.
-- y-intercept is positive. The x-intercept is negative.
-- Quadrant IV's (x, y) values are exactly the opposite: (+x, -y)
No point in Quadrant IV satisfies this inequality

Answer

is there any article on graphical representation of inequality and how to find the shaded region available on club?

I think you might find the following useful:
https://gmatclub.com/forum/graphic-appr ... 68037.html
https://gmatclub.com/forum/quick-way-to ... 76255.html
https://gmatclub.com/forum/solving-quad ... 70528.html

Hope it helps.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.