Inequalities - Challenging and tricky One

GMATMadeeasy · **Posted:** Sat Jan 16, 2010 6:00 am


		00:00



	GMAT Timer...

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.

Bunuel · **Posted:** Sat Jan 16, 2010 11:54 am

Hussain15 wrote:

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.

Bunuel · **Posted:** Fri Mar 11, 2011 12:51 pm

fluke wrote:

Bunuel wrote:

Hussain15 wrote:

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.

My slow brain is just not getting it!!! How both LHS and RHS get negated for the range x<2/3 and only the RHS gets negated for the range B and nothing for range C. What if there were 4 or 5 ranges; what is the deciding factor as to what gets negated.

Could you please help me understand?

Absolute value properties:
When x\leq{0} then |x|=-x, or more generally when some \ expression\leq{0} then |some \ expression|\leq{-(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|\leq{some \ expression}. For example: |5|=5;

In range A, when x<\frac{2}{3}: 3x-2<0 so |3x-2|=-(3x-2) and 2x-5<0 so |2x-5|=-(2x-5), and we get -3x+2\leq-2x+5.

In range B, when \frac{2}{3}\leq{x}\leq\frac{5}{2}: 3x-2>0 so |3x-2|=3x-2 and 2x-5<0 so |2x-5|=-(2x-5), so we get 3x-2\leq-2x+5.

For more check: math-absolute-value-modulus-86462.html

Hope it's clear.

Bunuel · **Posted:** Sat Jan 16, 2010 6:34 am

GMATMadeeasy wrote:

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}.

GMATMadeeasy · **Posted:** Sat Jan 16, 2010 8:15 am

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.

Hussain15 · **Posted:** Sat Jan 16, 2010 9:49 am

@ 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???

Bunuel · **Posted:** Sat Jan 16, 2010 11:04 am

Hussain15 wrote:

@ 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.

Hussain15 · **Posted:** Sat Jan 16, 2010 11:39 am

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??

prashantbacchewar · **Posted:** Sat Sep 18, 2010 6:08 am

Thanks Bunuel you posted excellent approach to deal with inequalities. I think you should write somthing on inequalities as well.
Kudos to you

puneet3085 · **Posted:** Fri Mar 11, 2011 10:22 am

Bunuel wrote:

GMATMadeeasy wrote:

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}.

that's a lovely and uncomplicated way....

fluke · **Posted:** Fri Mar 11, 2011 12:42 pm

Bunuel wrote:

Hussain15 wrote:

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.

My slow brain is just not getting it!!! How both LHS and RHS get negated for the range x<2/3 and only the RHS gets negated for the range B and nothing for range C. What if there were 4 or 5 ranges; what is the deciding factor as to what gets negated.

Could you please help me understand?

thesfactor · **Posted:** Sat Mar 19, 2011 12:44 pm

Great solution bunuel, kudos!

Inequalities - Challenging and tricky One

Who is online

Main Navigation

GMAT Resources

Partners

MBA Resources