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10 Jun 2020, 05:47
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Vesper2018
Region = a 2-dimensional space.
01 Oct 2020, 00:58
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metallicafan
Although there are too good solutions, here's my take.
If you are good with drawing lines, this question would be easier to solve.
Region R for \(2x + 3y =< 6\) would lie below the line \(2x + 3y = 6\), the side on which origin(as it satisfies the condition) lies. For the condition to be satisfied the most important thing is that the line must be parallel(equal slopes) to \(2x + 3y = 6\) and region is clearly defined i.e. either <6 or >6. Let's see what we have.
(1) \(3r + 2s = 6\)
Without doing much math we can eliminate options A and D. Because the coefficients of r and s are different from that of line \(2x + 3y = 6\) - 3 for r/2 for x and 2 for s/3 for y - the two lines would intersect. Thus, suggesting that a part of line \(3r + 2s = 6\) would be on the other side of the line \(2x + 3y = 6\). Hence some points would satisfy the condition and some would not.
OR
If we write the two line equation in the form y = mx + c, we can say that the two lines have different slopes - -\(\frac{2}{3}\) and -\(\frac{3}{2}\). \(y = -\frac{2}{3}x + 2\) and \(s = -\frac{3}{2}r + 3\), suggesting that the two lines intersect.
In any case, we need not to check points.
(Note: RHS of the equation i.e. 6 makes no sense here as you can see the slopes are different. Had the value been <6 again then it would not have been again sufficient.)
INSUFFICIENT.
(2) \(r=< 3\) and \(s=< 2\)
Again, applying the similar theory we see that some points would not satisfy(r = 3 and s = 2) the condition and some would(r = -3 and y = -2).
Eliminate option B.
INSUFFICIENT.
Combined together statement 1 and 2.
Refer screenshot:
Attachment:
Region R .png [ 710.17 KiB | Viewed 3699 times ]
The triangular region with 3 different colored dots invalidate the condition and rest satisfy the condition.
Answer E.
03 Oct 2020, 09:06
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BrentGMATPrepNow
BrentGMATPrepNow why did you change inequality sign to equality ? in which cases is it good idea to make such changes
