Amogh wrote:
Bunuel,
when you say as -3<= X and as x< \(\frac{2}{3}\) then -3<=x<=\(\frac{2}{3}\) does this mean that the range of x that satisfies the condition x<\(\frac{2}{3}\) is all those points on the number line which satisfy both inequalities.
Also you say ranges from A and B give us solution as -3<= x<= \(\frac{7}{5}\) . Can you please explain this to me. Does this mean that the final solution to this problem would be all those ranges that have a solution to their respective conditions i.e the conditions x<\(\frac{2}{3}\) , \(\frac{2}{3}\)<=x<\(\frac{5}{2}\) and x>=\(\frac{5}{2}\).
Also if for example ranges from A and B were -5<=x<2 and 4<=x<10, then would the final solution be both these ranges or do they have to have an overlap?
I know this is a supremely dumb and elementary question, but unfortunately inequalities and modulus happen to be my weakest areas. I cannot express in words my gratitude to you for having put up the GMAT Math Book. That thing is my Math Bible. You've made math so easy to understand.
Best wishes and many many thanks !!!
Amogh.
We consider three ranges:
A. \(x<\frac{2}{3}\);
B. \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\);
C. \(x>\frac{5}{2}\).
In range A we get: \(-3\leq{x}<\frac{2}{3}\);
In range B we get: \(\frac{2}{3}\leq{x}\leq\frac{7}{5}\);
In range C there is no solution.
So, the given inequality holds true for \(-3\leq{x}<\frac{2}{3}\) and \(\frac{2}{3}\leq{x}\leq\frac{7}{5}\). Now, we can combine these two ranges and write: \(-3\leq{x}\leq\frac{7}{5}\).
Next, the solution set is \(-3\leq{x}\leq\frac{7}{5}\), means that any x from \(-3\leq{x}\leq\frac{7}{5}\) will satisfy \(|3x-2|\leq|2x-5|\).
If for some question we had -5<=x<2 from one range and 4<=x<10 from another, then the solution will be both ranges (no need for overlap).
Hope it's clear.
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