2x+3y=6
stem1:3r+2s=6
if (r,s) is in the line 2x+3y=6 then 2r+3y=6
we have 2 equations we can solve for (r,s), but more importantly, we know that since they don't have the same slope (convert to y=mx+b if you can't quickly see that; or realize that the 2 equations only have 1 solution, not infinite) they only intersect at 1 point.
=> (r,s) can fall in the region, but not necessarily => insuff

stem2:
we're given values, straightforward suff

answer is B.

JC,

I am not sure I understand your explanation on (1) Can you please re explain?

What's the solution for this ? OA seems to be E

I agree with B.

1. is not sufficient. Notice that you're told that 3r + 2s = 6. In the stem however, we are told that 2x + 3y = 6. If the given point were on this line, then we'd have 2r + 3s = 6. This is not necessarily the case, though. You can see the fine difference between the two equations in that their coefficients are reversed. This is basically a trick to test your attention, but do not be fooled!

2. We're given numeric values for both r and s, which could be easily replaced in the initial equation to check for a fit. Indeed, we don't have a match, so we can safely say that the given point is not on the line.

I agree with B.

Choice 1 isn't possible all by itself; two lines which aren't parallel on a plane always intersect.

In choice 2, we can determine if the point is in the region or not.

original poster mistakenly posted statement 2. Should read

2) r equal to or less than 3; s equal to or less than 2.

OA E is correct. If anyone has explanation for this answer that's better than the OG, please share. Thanks

netigen wrote:In the xy-plane, region R consists of all the points (x, y) such that 2x + 3y = 6. Is the point
(r, s) in region R ?
(1) 3r + 2s = 6
(2) r = 3 and s = 2

I'm not sure if the problem is copied wrong or this is just a different version of the same problem

in OG-12, Q121:
In the xy plane, region R consists of all the points (x, y) such that 2x + 3y <= 6. Is the point (r,s) in the Region R ?
1) 3r + 2s = 6
2) r <= 3 and s <= 2

I've been wrestling with "2x + 3y = 6" being an "region".

deborah83 wrote:

netigen wrote:In the xy-plane, region R consists of all the points (x, y) such that 2x + 3y = 6. Is the point
(r, s) in region R ?
(1) 3r + 2s = 6
(2) r = 3 and s = 2

I'm not sure if the problem is copied wrong or this is just a different version of the same problem

in OG-12, Q121:
In the xy plane, region R consists of all the points (x, y) such that 2x + 3y <= 6. Is the point (r,s) in the Region R ?
1) 3r + 2s = 6
2) r <= 3 and s <= 2

I've been wrestling with "2x + 3y = 6" being an "region".

Guys can someone solve the correct question ???? as deborah83 mentioned.


In the xy plane, region R consists of all the points (x, y) such that 2x + 3y <= 6. Is the point (r,s) in the Region R ?
1) 3r + 2s = 6
2) r <= 3 and s <= 2

Can someone explain this question better? Explanation with a diagram is much appreciated.

can someone plzz explain this question


OA in the book is E

searched for an answer / explanation on this one and did not find.... if you know two lines have different slopes, you know they intersect. so if the intersect, how do you then figure out if they are in the same region? More importantly, WHAT is the region they are talking about?!!!

is it the region under or above the point?

Hi,

My attempt at understanding the question:-

The eq. of the line is 2x + 3y <= 6,

so x/3+y/2 <=1,

so x=3 and y=2 is the x and y intercept from which we can draw the line as in fig. The region represented by the line is the region internal to the line towards the origin. Can be easily verified by using (0,0).

Now x,y=(3,2) represent a point outside this region. But anything less than (3,2) may lie inside the region also. So statement (B) also is insufficient.

statement (A) represents another line with some part in the given region as the line is non parallel. But nothing can be derived from the two. Hence the choice (E). [/img]