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shankhamala28
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Doubt from Thursdays with RON - Mar 17, 2011 session

by shankhamala28 Sat Sep 14, 2013 5:04 pm

This is the following question on which I have a doubt from March 17, 2011 session:

In the xy-plane, does the line of the equation y= 3x + 2 contain the point (r,s) ?

1. (3r+2-s)(4r+9-s)
2. (4r-6-s)(3r+2-s)

I got this question correct considering that A & B are insufficient and combining both A&B, only 3r+2-s satisfies the eqn. y = 3x+2.

There were a couple of variations of this question. One of them is the following one:
1. (3r+2-s)(4r+9-s)
2. (2r-5-s)(3r+2-s)

In this case, both A & B are insufficient.
Now, we combine both A & B, we get the following pair of the equations:
1---> 3r+2-s or 4r+9-s
2---> 2r-5-s or 3r+2-s

Ron, you mentioned that atleast one of these in 1 or one of these in 2 must be equal to zero.
Therefore, you said the following combinations are true:
a. 4r+9=s
3r+2=s
or
b. 3r+2=s
2r-5=s
or
c. 3r+2=s
3r+2=s
and checked separately for the below pair.
d. 4r+9=s
2r-5=s
I understood how d will stand true.
But, I am not clear why a and b should be true (it's clear to me why c is true).

As in, when we solved statements 1 and 2, we found it insufficient because only one of them satisfied the equation, i.e 3r+2-s = 0 satisfied y = 3x + 2 whereas 4r+9-s=0 did not. Similarly, for st. 2.

Doesn't the same logic holds true for the pair of equations in a & b (only 1 satisfies whereas the other doesn't). If so, then why is it True?

I am probably missing out some basic concept, not sure, but some explanation from your side will be of great help for me.
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Re: Doubt from Thursdays with RON - Mar 17, 2011 session

by RonPurewal Sun Sep 15, 2013 7:40 am

I'll present a simpler case that's less clumsy to talk about. The concept is the same.

Is z = 0?
(1) (x - 1)(z) = 0
(2) (x - 2)(z) = 0

ok, so, if you have one of the individual statements, then it's an "OR" type situation. i.e.,
statement 1 alone --> x = 1 OR z = 0 (or both). They don't both have to be true.
statement 2 alone --> x = 2 OR z = 0 (or both). They don't both have to be true.
In either case z ≠ 0 is possible, if x has the requisite value.

If you COMBINE the two statements, BOTH of them have to be true. That means one of the blue things has to be true, AND one of the orange things has to be true. There are theoretically four possible combinations:

* x = 1 AND x = 2
This is impossible.

* x = 1 AND z = 0
This is possible.

* z = 0 AND x = 2
This is possible.

* z = 0 AND z = 0
This is possible.

In all three possible cases, z = 0.

The key here is that you're not differentiating adequately between "AND" and "OR".
shankhamala28
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Re: Doubt from Thursdays with RON - Mar 17, 2011 session

by shankhamala28 Sun Sep 15, 2013 2:59 pm

RonPurewal Wrote:If you COMBINE the two statements, BOTH of them have to be true. That means one of the blue things has to be true, AND one of the orange things has to be true. There are theoretically four possible combinations:

* x = 1 AND x = 2
This is impossible.

* x = 1 AND z = 0
This is possible.

* z = 0 AND x = 2
This is possible.

* z = 0 AND z = 0
This is possible.

In all three possible cases, z = 0.


As you stated above, then the same holds true in the previous example as well. Both 4r+9=s AND 3r+2=s have to be true.
But the question asks does the equation y= 3x + 2 contain the point (r,s). For 4r + 9 = s, values for (r,s) do not satisfy the equation y = 3x + 2.
Yes, 4r+9=s and 3r+2=s are possible and could be valid but it may not be true for the equation y = 3x + 2.
If it's an "AND", shouldn't we consider that both should be true.
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Re: Doubt from Thursdays with RON - Mar 17, 2011 session

by RonPurewal Mon Sep 16, 2013 8:58 am

shankhamala28 Wrote:As you stated above, then the same holds true in the previous example as well. Both 4r+9=s AND 3r+2=s have to be true.
But the question asks does the equation y= 3x + 2 contain the point (r,s). For 4r + 9 = s, values for (r,s) do not satisfy the equation y = 3x + 2.
Yes, 4r+9=s and 3r+2=s are possible and could be valid but it may not be true for the equation y = 3x + 2.
If it's an "AND", shouldn't we consider that both should be true.


If they're both true, then both are true. I.e., neither is false.

It's like solving a normal system of 2 equations in 2 variables. If both 4r + 9 = s and 3r + 2 = s are true, then you're only going to have the single pair of coordinates for which both of those are true (i.e., r = -7 and s = -19).
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Re: Doubt from Thursdays with RON - Mar 17, 2011 session

by shankhamala28 Mon Sep 16, 2013 11:10 pm

Thanks Ron. It's crystal clear to me now. :)
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Re: Doubt from Thursdays with RON - Mar 17, 2011 session

by RonPurewal Tue Sep 17, 2013 6:26 am

shankhamala28 Wrote:Thanks Ron. It's crystal clear to me now. :)


You're welcome.