In the rectangular coordinate system which quadrant, if any, contains no point ( x, y ) that satisfies the inequality 2x − 3y ≤ − 6 ?
(A) None
(B) Ι
(C) ΙI
(D) ΙII
(E) IV
OA is E.
My Response to the ques:
Assuming x=0, then -3y<= -6 or y>= 2.
Assuming y=0, x<= -3.
Therefore a line with coordinates (X<= -3, Y>=2) should be located in quadrant 2 only.
Having done this much, I got stuck. I did not really comprehend what the question stem meant.
PS- Coordinate Geometry
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2x - 3y <= -6
x = 0==> -3y <= -6 ==> y>=2
y = 0==> 2x <= -6 ==> x <= -3
Since, any line drawn using the above points will result lines passing 1st, 2nd & 3rd quadrant.
So, Answer {E}
x = 0==> -3y <= -6 ==> y>=2
y = 0==> 2x <= -6 ==> x <= -3
Since, any line drawn using the above points will result lines passing 1st, 2nd & 3rd quadrant.
So, Answer {E}
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Let us first trace the region depicted by 2x − 3y ≤ − 6pareekbharat86 wrote:In the rectangular coordinate system which quadrant, if any, contains no point ( x, y ) that satisfies the inequality 2x − 3y ≤ − 6 ?
(A) None
(B) Ι
(C) ΙI
(D) ΙII
(E) IV
2x − 3y ≤ − 6
−3y ≤ -2x − 6
3y ≥ 2x + 6 ... multiplying both sides by (-1)
y ≥ (2/3)x + 2
Consider the line y = (2/3)x + 2
Its y intercept = 2 (put x = 0 in the above equation)
Its x intercept = -3 (put y = 0 in the above equation)
We need a region which is greater than y = (2/3)x + 2
(since the original equation states greater than equal to...)
Original question: which quadrant contains no point that satisfies the inequality ...
The points in the 4th quadrant will not be a part of the equation y ≥ (2/3)x + 2 and thus the answer is E
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For those who are unfamiliar with the location of the quadrants in the x-y coordinate plane, they are shown herepareekbharat86 wrote:In the rectangular coordinate system which quadrant, if any, contains no point ( x, y ) that satisfies the inequality 2x − 3y ≤ −6 ?
(A) None
(B) Ι
(C) ΙI
(D) ΙII
(E) IV
Another approach here is the recognize how the coordinates of points look in each quadrant.
Quadrant I: (positive, positive)
Quadrant II: (negative, positive)
Quadrant III: (negative, negative)
Quadrant IV: (positive, negative)
From here, we can check each quadrant.
For example, let's see if it's possible for a point in QUADRANT I (where the x and y coordinates are both positive) to satisfy the inequality 2x − 3y ≤ −6
Well, how about x = 1 and y = 10.
When we plug these in, we get 2(1) − 3(10) ≤ −6
Simplify to get -28 ≤ −6 [PERFECT it works]
ELIMINATE B
Is it possible for a point in QUADRANT II (where the x is negative and y is positive) to satisfy the inequality 2x − 3y ≤ −6?
Well, how about x = -10 and y = 10.
When we plug these in, we get 2(-10) − 3(10) ≤ −6
Simplify to get -50 ≤ −6 [PERFECT it works]
ELIMINATE C
Is it possible for a point in QUADRANT III (where the x is negative and y is negative) to satisfy the inequality 2x − 3y ≤ −6?
Well, how about x = -10 and y = -1.
When we plug these in, we get 2(-10) − 3(-1) ≤ −6
Simplify to get -17 ≤ −6 [PERFECT it works]
ELIMINATE D
Is it possible for a point in QUADRANT IV (where the x is positive and y is negative) to satisfy the inequality 2x − 3y ≤ −6?
NO.
When we plug these in, we get 2(positive) − 3(negative) ≤ −6
Simplify to get positive - negative ≤ −6
Since a positive number minus a negative number will always be positive, we get . . .
POSITIVE ≤ −6
Since this is IMPOSSIBLE, the correct answer must be E
Cheers,
Brent