kawan84 wrote:
Hi Bunnel,
I've been struggling a lot in inequalities.. I've a question pertaining to solution provided by saxenaashi ...
1) x >=3 ==> x - 3 = 2x -3 ====> x = 0 ; discard[Understood this part]2) 3/2 < x < 3 ===> -(x-3) = 2x -3 ====> -x + 3 = 2x -3 ===> x =2 ( accepted solution )[Did not understand, Why -ve sign is here as the x range is positive]3) x < 3/2 ===> -(x-3) = -(2x -3) ====> x = 3 (discarded the solution)[Understood]
Another thing is 3/2(inclusive) is not at all considered in the range in any of the three checkpoints.
What am i missing here and how to solve these questions effectively
Absolute value properties:When \(x
\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);
When \(x
\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).
SOLUTION:We have two transition points for \(|x-3|=|2x-3|\): \(x=\frac{3}{2}\) and \(x=3\). Thus three ranges to check:
1. \(x<\frac{3}{2}\);
2. \(\frac{3}{2}\leq{x}\leq{3}\);
3. \(3<x\)
Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them.
1. When \(x<\frac{3}{2}\), then \(x-3\) is negative and \(2x-3\) is negative too, thus \(|x-3|=-(x-3)\) and \(|2x-3|=-(2x-3)\).
Therefore for this range \(|x-3|=|2x-3|\) transforms to \(-(x-3)=-(2x-3)\) --> \(x=0\). This solution is OK, since \(x=0\) is in the range we consider (\(x<\frac{3}{2}\)).
2. When \(\frac{3}{2}\leq{x}\leq{3}\), then \(x-3\) is negative and \(2x-3\) is positive, thus \(|x-3|=-(x-3)\) and \(|2x-3|=2x-3\). Therefore for this range \(|x-3|=|2x-3|\) transforms to \(-(x-3)=2x-3\) --> \(x=2\). This solution is OK, since \(x=2\) is in the range we consider (\(\frac{3}{2}\leq{x}\leq{3}\)).
3. When \(3<x\), then \(x-3\) is positive and \(2x-3\) is positive too, thus \(|x-3|=x-3\) and \(|2x-3|=2x-3\).
Therefore for this range \(|x-3|=|2x-3|\) transforms to \(x-3=2x-3\) --> \(x=0\). This solution is
NOT OK, since \(x=0\) is
NOT in the range we consider (\(3<x\)).
Thus \(|x-3|=|2x-3|\) has two solutions \(x=0\) and \(x=2\).
Hope it's clear.
P.S. Though for this particular question I still suggest another approach shown in my post here:
http://gmatclub.com/forum/is-x-0-1-x-3- ... l#p1048512Would like to ask question on the part highlighted in red in your post, specifically on determining x - 3 is negative. Since the range is \(\frac{3}{2}\leq{x}\leq{3}\), then if i plug in 3/2 in x - 3, it will result in negative number, however if i plug in 3 in x - 3, it will result in 0 or non-negative, and therefore x - 3 can be negative or non-negative. In your post x - 3 must be negative, how do we eliminate the non-negative one ? Or do we simply ignore non-negative ?