What is the solution set for |3x-2|\leq|2x-5|

One way to solve is to square both the terms of course , but what is other way of solving it.

First you should determine the check points (key points): \frac{2}{3} and \frac{5}{2}. Hence we'll have three ranges to check:

A. x<\frac{2}{3} --> -3x+2\leq-2x+5 --> -3\leq{x}, as x<\frac{2}{3}, then -3\leq{x}<\frac{2}{3};

B. \frac{2}{3}\leq{x}\leq\frac{5}{2} --> 3x-2\leq-2x+5 --> -x\leq\frac{7}{5}, as \frac{2}{3}\leq{x}\leq\frac{5}{2} , then \frac{2}{3}\leq{x}\leq\frac{7}{5};

C. x>\frac{5}{2} --> 3x-2\leq2x-5 --> x\leq{-3}, as x>\frac{5}{7}, then in this range we have no solution;

Ranges from A and B give us the solution as: -3\leq{x}\leq\frac{7}{5}.

Thank you , it helps greatly.

Question : What was normal way of doing it back in school? I am wondering how I used to solve them ?

Working out questions from your post and other guys notes on the site in the mean time.

Thanks a lot Bunuel.

@ Bunuel:

I didn't pick your strategy.

My way of doing is simple but it is giving different answer.

|3x-2| <= |2x-5|
We will consider two scenarios
3x-2 <= 2x-5 & -(3x-2) <= 2x-5
x<= -3 & x >= 7/5

So the range will be x<= -3 & x >= 7/5

Kindly let me know where did I go wrong???

If you plug the numbers from the ranges you got, you'll see that the inequality doesn't hold true.

As for the solution: we have two absolute values |3x-2| and |2x-5|. |3x-2| changes sign at \frac{2}{3} and |2x-5| changes sign at \frac{5}{2}.

---(I)---\frac{2}{3}---(II)---\frac{5}{2}---(III)---

We got three ranges as above. We should expand given inequality in these ranges and see what we'll get.

Hope it's clear.

Thanks for the reply Bunue!

But there is one more issue.

In range A, you have changed the signs of both modules, in range B you have done it only for 2x+5 and in third case you haven't changed the signs. What's the logic behind this one??

I got your point. I'm not "changing" the signs, I'm expanding the absolute values in each range.

In range A, when x<\frac{2}{3}: |3x-2|=-3x+2 and |2x-5|=-2x+5, so we get -3x+2\leq-2x+5.

In range B, when \frac{2}{3}\leq{x}\leq\frac{5}{2}: |3x-2|=3x-2 and |2x-5|=-2x+5, so we get 3x-2\leq-2x+5.

In range C, when x>\frac{5}{2}: |3x-2|=3x-2 and |2x-5|=2x-5, so we get 3x-2\leq2x-5.