sinhap07 wrote:
Hi Bunuel
Not very sure how you got x>1 region on the numberline.
From stmt 1, I got a-ax>0 or a(1-x)>0 or a(x-1)<0 and on the number line ---(-)----1-----(+)-------- ie x<1.
Pls clarify.
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0
(1) a + b > 0
(2) a - b > 0
(1) a + b > 0
(2) a - b > 0
Given:\(b=-ax\) . Question: is\(x>0\)
(1)\(a+b>0\) -->\(a-ax>0\) -->\(a(1-x)>0\) --> either\(a>0\) and\(1-x>0\) , so\(x<1\) OR \(a<0\) and\(1-x<0\) , so\(x>1\) . Not sufficient.
(2)\(a-b>0\) -->\(a+ax>0\) -->\( a(1+x)>0\)
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