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In the rectangular coordinate system shown above, which [#permalink]

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10 Feb 2010, 12:06

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E

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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? (the quadrants are the standard quadrants in a co-ordinate system, I can't really draw it out here)

1. In the rectangular coordinate system shown above, which quadrant, if any, contains no point ( x, y ) that satisfies the inequality 2x-3y≤−6? (the quadrants are the standard quadrants in a co-ordinate system, I can't really draw it out here) (A) None (B) Ι (C) ΙI (D) ΙII (E) IV

Please explain your answer. I will post the OA soon.

\({2x-3y}\leq{-6}\) --> \(y\geq{\frac{2}{3}x+2\). Thi inequality represents ALL points, the area, above the line \(y={\frac{2}{3}x+2\). If you draw this line you'll see that the mentioned area is "above" IV quadrant, does not contains any point of this quadrant.

Else you can notice that if \(x\) is positive, \(y\) can not be negative to satisfy the inequality \(y\geq{\frac{2}{3}x+2\), so you can not have positive \(x\), negative \(y\). But IV quadrant consists of such \((x,y)\) points for which \(x\) is positive and \(y\) negative. Thus answer must be E.

1. In the rectangular coordinate system shown above, which quadrant, if any, contains no point ( x, y ) that satisfies the inequality 2x-3y≤−6? (the quadrants are the standard quadrants in a co-ordinate system, I can't really draw it out here) (A) None (B) Ι (C) ΙI (D) ΙII (E) IV

Please explain your answer. I will post the OA soon.

\({2x-3y}\leq{-6}\) --> \(y\geq{\frac{2}{3}x+2\). Thi inequality represents ALL points, the area, above the line \(y={\frac{2}{3}x+2\). If you draw this line you'll see that the mentioned area is "above" IV quadrant, does not contains any point of this quadrant.

Else you can notice that if \(x\) is positive, \(y\) can not be negative to satisfy the inequality \(y\geq{\frac{2}{3}x+2\), so you can not have positive \(x\), negative \(y\). But IV quadrant consists of such \((x,y\)) points for which \(x\) is positive and \(y\) negative. Thus answer must be E.

Answer: E.

Thats great! I understand it now. I always seem to have problems with co-ordinate geometry. The official answer is E.

1. In the rectangular coordinate system shown above, which quadrant, if any, contains no point ( x, y ) that satisfies the inequality 2x-3y≤−6? (the quadrants are the standard quadrants in a co-ordinate system, I can't really draw it out here) (A) None (B) Ι (C) ΙI (D) ΙII (E) IV

Please explain your answer. I will post the OA soon.

Let us represent the equation in the intercept form: x/a + y/b = 1 x/(-3) + y/(2) >= 1 Now plot these intercepts on the chart and you will find that the quadrant not being touched is IV.

Draw the equation line by determining to points, the easiest points are (0,y) and (x,0). Those points are (-3,0) and (0,2). Draw the line through the 2 points. Now pick one point and see if the inequality should be shaded to the left or to the right. I picked (-3,0), the inequality states that x<=-3, so shade to the left. you'll clearly see that the only quadrant that has no shading is IV.

1. In the rectangular coordinate system shown above, which quadrant, if any, contains no point ( x, y ) that satisfies the inequality 2x-3y≤−6? (the quadrants are the standard quadrants in a co-ordinate system, I can't really draw it out here) (A) None (B) Ι (C) ΙI (D) ΙII (E) IV

Please explain your answer. I will post the OA soon.

Hello can someone please tell me whats wrong with this approach ?

y >= -2/3 x + 6

Thus slope is negative

lets find x-intercept ... substitute y=0 we get , x <= 9 . To find y-intercept substitute x=0, we get y >=6

Thus it should be 3rd quadrant... but this is wrong ...

In a rectangular coordinat system which quadrant, if any, contains no point (x,y) that satifies the inequality 2x - 3y <= -6?

A) None B) 1st C) 2nd D)3rd E) 4th

How to solve this guys any idea?

2x - 3y <= -6 => -2x + 3y >= 6 ---- (1)

equate the both sides of the inequality: -2x + 3y = 6 =>x/(-3) + y/2 = 1

comparing with x/a + y/b = 1 x intercept is -3 and y intercept is 2 ie the line passes thru (-3,0) and (0,2)

This line obviously passes through the 1st, 2nd and 3rd quadrant and going back to the original inequality (1) we understand the points that satisfy this inequality would be above this line which means there is no chance that any of these points would lie in quadrant 4. So the ans is E
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Re: In the rectangular coordinate system shown above, which [#permalink]

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05 Jan 2013, 01:14

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loki wrote:

In the rectangular coordinate system shown above, which quadrant, if any, contains no point ( x, y ) that satisfies the inequality 2x-3y≤−6? (the quadrants are the standard quadrants in a co-ordinate system, I can't really draw it out here)

A. None B. Ι C. ΙI D. ΙII E. IV

Try I: 2(1) - 3(10) = -28 Yes! Try II: 2(-10) - 3(1) = -17 Yes! Try III: 2(-10) - 3(-1) = -17 Yes! Try IV: Since y is negative and x is positive... 2x will always be positive and -3y will always be positive = No!

Let convert this in to an equation 3y = 2x + 6 ------> y = \(\frac{2x}{3}\) + 2

We know that equation of any line is y = mx + c ------where m = slope and C = y intercept

So in our case slope = \(\frac{2}{3}\) and Y intercept = 2

With above values following line can be drawn

We can notice that in the original inequality 3y≥ 2x + 6 as the value of Y will go up, the corresponding value of X will also go up to satisfy the inequality and hence the line drawn above will also continue to incline in the same direction in 1st quadrant. We can deduce that this line will never pass thru IV quadrant. Hence Choice E is the answer.
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Re: In the rectangular coordinate system shown above, which [#permalink]

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08 Oct 2013, 08:34

loki wrote:

In the rectangular coordinate system shown above, which quadrant, if any, contains no point ( x, y ) that satisfies the inequality 2x-3y≤−6? (the quadrants are the standard quadrants in a co-ordinate system, I can't really draw it out here)

A. None B. Ι C. ΙI D. ΙII E. IV

I'm not sure if this is the right approach but the way i did it was:

2x-3y≤−6, so 2x-3y≤0 so it the subtraction of both HAS to be negative. Therefore, the only way for this to NOT happen is if 2x>0 and -3y<0 because then both will be positive and it is impossible to have a negative number as a result.

In the rectangular coordinate system shown above, which [#permalink]

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07 Aug 2014, 03:54

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kostyan5 wrote:

Here's my approach, which i think is easiest.

Draw the equation line by determining to points, the easiest points are (0,y) and (x,0). Those points are (-3,0) and (0,2). Draw the line through the 2 points. Now pick one point and see if the inequality should be shaded to the left or to the right. I picked (-3,0), the inequality states that x<=-3, so shade to the left. you'll clearly see that the only quadrant that has no shading is IV.

This is a perfect approach for this kind of problems. I'd just fine tune this method by ALWAYS picking (0,0) to see if the inequality lies to the left or the right side of the line. I think this will create the most efficient method for such problems.
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In the rectangular coordinate system shown above, which [#permalink]

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07 Aug 2014, 08:48

Given Inequality : 2x-3y≤−6

we can rewrite it as 3y ≥ 2x + 6 => y ≥ (2/3)x + 2 . Now from the line equation y = mx + c , where m is slope of the line and C is the intercept of the line we can deduce m= 2/3 => Positive slope. Also from line equation (x/a) + (y/b) = 1 , where a and b are x and y intercepts respectively, we can deduce "a = -3" and "b=2" .

Hence, if you draw the line in coordinate plane we can observe it as passing in 1st, 2nd and 3rd quadrants.

The inequality mentioned will be satisfied for all the points on the line and above it, which means atleast one point from 1st , 2nd and 3rd coordinate satisy this inequality. The only (x,y) pairs which do not satisy are points in the 4th quadrant.

Re: In the rectangular coordinate system shown above, which [#permalink]

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14 Sep 2015, 08:26

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Re: In the rectangular coordinate system shown above, which [#permalink]

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01 Jul 2016, 06:43

\({2x-3y}\leq{-6}\) --> \(y\geq{\frac{2}{3}x+2\). Thi inequality represents ALL points, the area, above the line \(y={\frac{2}{3}x+2\). If you draw this line you'll see that the mentioned area is "above" IV quadrant, does not contains any point of this quadrant.

Hi,

Upon substitution of x = -1 in the above equality, we get the point (-1, 4/3). This point is below the above stated line.

Can you explain if i am missing something.

Thanks

gmatclubot

Re: In the rectangular coordinate system shown above, which
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01 Jul 2016, 06:43

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