Hello my friend, function is like equation
In the xy plane does the line with equation y = 3x + 2 contain the point ( r , s) ?
1. ( 3r+2-s)(4r+9-s) = 0.
2. (4r-6-s)(3r+2-s) =0.
y is
s and
x is
r, now rewrite everything --> s=3r+2 OR 3r+2-s=0

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I know you smile too
st(1) BECOMES ridiculously simple ( 3r+2-s)(4r+9-s) = 0 is Not Sufficient, because we have also (4r+9-s) can be 0
st(2) heh, (4r-6-s)(3r+2-s) =0; you know everything here --> (3r+2-s) =0 ? Not Sufficient
may be Combined ? st(1&2): both ( 3r+2-s)(4r+9-s) = 0 AND (4r-6-s)(3r+2-s) =0 can be zero if one of common factors is 0 (zero). Only possible with the ( 3r+2-s) = 0 why not Sufficient?
IOM
C
Cheers
earnest10 wrote: