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

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Math Expert
Joined: 02 Sep 2009
Posts: 43349
In the rectangular coordinate system shown above, which quadrant, if a [#permalink]

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17 Dec 2017, 22:20
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Difficulty:

55% (hard)

Question Stats:

59% (01:21) correct 41% (00:59) wrong based on 53 sessions

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In the rectangular coordinate system shown above, which quadrant, if any, contains no point (x,y) that satisfies the inequality 2x – 3y ≤ –6?

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

[Reveal] Spoiler:
Attachment:

2017-12-18_1011_001.png [ 3.25 KiB | Viewed 600 times ]
[Reveal] Spoiler: OA

_________________
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Joined: 22 May 2016
Posts: 1259
In the rectangular coordinate system shown above, which quadrant, if a [#permalink]

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18 Dec 2017, 00:43
Bunuel wrote:

In the rectangular coordinate system shown above, which quadrant, if any, contains no point (x,y) that satisfies the inequality 2x – 3y ≤ –6?

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

[Reveal] Spoiler:
Attachment:
The attachment 2017-12-18_1011_001.png is no longer available

Attachment:

quadrants.png [ 8.25 KiB | Viewed 418 times ]

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

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.

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

[Reveal] Spoiler:
E

_________________

At the still point, there the dance is. -- T.S. Eliot
Formerly genxer123

Manager
Joined: 31 Jul 2017
Posts: 142
Location: Malaysia
WE: Consulting (Energy and Utilities)
In the rectangular coordinate system shown above, which quadrant, if a [#permalink]

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18 Dec 2017, 01:43
genxer123 wrote:
Bunuel wrote:

In the rectangular coordinate system shown above, which quadrant, if any, contains no point (x,y) that satisfies the inequality 2x – 3y ≤ –6?

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

[Reveal] Spoiler:
Attachment:
2017-12-18_1011_001.png

Attachment:

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

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.

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

[Reveal] Spoiler:
E

Hi genxer123,

The question asks if any (x,y) of the 4-Quadrants will not satisfy $$\frac{-x}{3} + \frac{y}{2} ≤ 1$$ equation. In IV quadrant (6,-4) will definitely satisfy the equation. Can you please clarify my doubt or am I missing something here??
Math Expert
Joined: 02 Aug 2009
Posts: 5537
Re: In the rectangular coordinate system shown above, which quadrant, if a [#permalink]

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18 Dec 2017, 01:54
rahul16singh28 wrote:
Bunuel wrote:

In the rectangular coordinate system shown above, which quadrant, if any, contains no point (x,y) that satisfies the inequality 2x – 3y ≤ –6?

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

[Reveal] Spoiler:
Attachment:
2017-12-18_1011_001.png

Hi genxer123,

The question asks if any (x,y) of the 4-Quadrants will not satisfy \frac{-x}{3 + y/2 < 1} equation. In IV quadrant (6,-4) will definitely satisfy the equation. Can you please clarify my doubt or am I missing something here??

HI
2x – 3y ≤ –6....
Right Hand Side is -6 or NEGATIVE..
when will the LHS be POSITIVE..
when 2x is positive or x is positive and -3y is also positive or -y is positive means y is negative
so y- NEGATIVE and x-POSITIVE
THIS is exactly what is there in quad IV
so LHS can never be ≤-6
E
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

BANGALORE/-

Manager
Joined: 31 Jul 2017
Posts: 142
Location: Malaysia
WE: Consulting (Energy and Utilities)
Re: In the rectangular coordinate system shown above, which quadrant, if a [#permalink]

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18 Dec 2017, 01:59
chetan2u wrote:
rahul16singh28 wrote:
Bunuel wrote:

In the rectangular coordinate system shown above, which quadrant, if any, contains no point (x,y) that satisfies the inequality 2x – 3y ≤ –6?

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

[Reveal] Spoiler:
Attachment:
2017-12-18_1011_001.png

Hi genxer123,

The question asks if any (x,y) of the 4-Quadrants will not satisfy \frac{-x}{3 + y/2 < 1} equation. In IV quadrant (6,-4) will definitely satisfy the equation. Can you please clarify my doubt or am I missing something here??

HI
2x – 3y ≤ –6....
Right Hand Side is -6 or NEGATIVE..
when will the LHS be POSITIVE..
when 2x is positive or x is positive and -3y is also positive or -y is positive means y is negative
so y- NEGATIVE and x-POSITIVE
THIS is exactly what is there in quad IV
so LHS can never be ≤-6
E

Hi Chetan2u,

Can't the equation 2x – 3y ≤ –6, be written as $$\frac{−x}{3}+\frac{y}{2} ≤ 1$$. Apologies if its a stupid question
Math Expert
Joined: 02 Aug 2009
Posts: 5537
In the rectangular coordinate system shown above, which quadrant, if a [#permalink]

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18 Dec 2017, 02:03
1
KUDOS
Expert's post
rahul16singh28 wrote:
Hi Chetan2u,

Can't the equation 2x – 3y ≤ –6, be written as $$\frac{−x}{3}+\frac{y}{2} ≤ 1$$. Apologies if its a stupid question

Where you have gone wrong
$$\frac{-x}{3 + y/2 < 1}$$
when you divide both sides by -6 , CHANGE the sign
2x – 3y ≤ –6 will become $$\frac{2x}{-6}-\frac{3y}{-6}>\frac{-6}{-6}.......\frac{-x}{3}+\frac{y}{2}>1$$

a simpler look x<-1 will become -x>1
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

BANGALORE/-

VP
Joined: 22 May 2016
Posts: 1259
Re: In the rectangular coordinate system shown above, which quadrant, if a [#permalink]

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18 Dec 2017, 11:31
1
KUDOS
rahul16singh28 wrote:
Bunuel wrote:

In the rectangular coordinate system shown above, which quadrant, if any, contains no point (x,y) that satisfies the inequality 2x – 3y ≤ –6?

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

[Reveal] Spoiler:
Attachment:
2017-12-18_1011_001.png

Hi genxer123,

The question asks if any (x,y) of the 4-Quadrants will not satisfy $$\frac{-x}{3} + \frac{y}{2} ≤ 1$$ equation. In IV quadrant (6,-4) will definitely satisfy the equation. Can you please clarify my doubt or am I missing something here??

rahul16singh28 , I think chetan2u has answered your question (you simply forgot to switch the sign when you divided by -6). I appreciate your polite and gracious manner.
_________________

At the still point, there the dance is. -- T.S. Eliot
Formerly genxer123

Re: In the rectangular coordinate system shown above, which quadrant, if a   [#permalink] 18 Dec 2017, 11:31
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