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.