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Re: What is the solution set for |3x-2|<=|2x-5| [#permalink]
02 Nov 2012, 08:12

Thanx i just needed to clarify it is just by substitution of a value from the range that you decide the sign with which inequality should be multiplied.

Re: What is the solution set for |3x-2|<=|2x-5| [#permalink]
18 Dec 2012, 00:33

Ans: to solve this we see that there are 4 cases, when both are –ve, both +ve, and alternately –ve and +ve..therefore we get the solution set as -3<equal to x<equal 7/5. _________________

Re: What is the solution set for |3x-2|<=|2x-5| [#permalink]
12 May 2013, 05:41

In addition to the critical points that each absolute value expression has ( 2/3 and 5/2) there is another 2 critical points that comes from their interaction together that are ( 2/3 and 7/5) those last 2 critical points you can get by equating the 2 absolute values :

/3x-2/ = /2x-5/, thus 2 scenarios either 3x-2 = 2x-5 thus x = 7/5 or 3x-2 = -(2x-5) thus x = -3

you then test each region by substituting values from each region into the original inequality /3x-2/ = /2x-5/ , when u do this u ll end up with -3<=x<=7/5

Re: What is the solution set for |3x-2|<=|2x-5| [#permalink]
08 Oct 2014, 21:51

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Re: What is the solution set for |3x-2|<=|2x-5| [#permalink]
22 Oct 2015, 10:54

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 \(x>\frac{5}{2}\), then in this range we have no solution;

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

Hi Bunuel, I've one question regarding your solution. I'm using the same method as you...

A. \(-3\leq{x}\) can be <2/3 and >2/3 how do you limit the range of first expression \(-3\leq{x}\) ? -2 ok, 100 not... or we can just say by the soltion x=3 and is in the given range x>2/3. _________________

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What is the solution set for |3x-2|<=|2x-5| [#permalink]
22 Oct 2015, 11:08

BrainLab wrote:

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 \(x>\frac{5}{2}\), then in this range we have no solution;

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

Hi Bunuel, I've one question regarding your solution. I'm using the same method as you...

A. \(-3\leq{x}\) can be <2/3 and >2/3 how do you limit the range of first expression \(-3\leq{x}\) ? -2 ok, 100 not... or we can just say by the soltion x=3 and is in the given range x>2/3.

Let me try to answer.

For case A, we are assuming that x<2/3 . This condition alongwith the solution of x \(\geq\)-3, gives the total range as \(2/3 > x \geq-3\).

Re: What is the solution set for |3x-2|<=|2x-5| [#permalink]
22 Oct 2015, 11:47

Thanks Engr2012, I think I've mixed it up with such kind of questions, where we have clear roots....

|x+3|−|4−x|=|8+x|. How many solutions does the equation have?

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions: a) x<−8. −(x+3)−(4−x)=−(8+x) --> x=−1 is not within that range and so not valid b) −8≤x<−3. −(x+3)−(4−x)=(8+x) --> x=−15 is not within that range and so not valid c) −3≤x<4. (x+3)−(4−x)=(8+x) --> x=9 is not within that range and so not valid d) x≥4. (x+3)+(4−x)=(8+x) --> x=−1 is not within that range and so not valid _________________

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Re: What is the solution set for |3x-2|<=|2x-5| [#permalink]
23 Oct 2015, 01:24

Let me try to answer.

For case A, we are assuming that x<2/3 . This condition alongwith the solution of x \(\geq\)-3, gives the total range as \(2/3 > x \geq-3\).

Hope this helps.[/quote]

Hi Engr2012, I've one more question; If we would have been asked to find not the ranges but the solutions for this example.. what would be the eanswer here ? Or by such kind of inequalities we are talking always about ranges like 0<x<5 and not x=5 _________________

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Share some Kudos, if my posts help you. Thank you !

What is the solution set for |3x-2|<=|2x-5| [#permalink]
23 Oct 2015, 03:36

BrainLab wrote:

Hi Engr2012, I've one more question; If we would have been asked to find not the ranges but the solutions for this example.. what would be the eanswer here ? Or by such kind of inequalities we are talking always about ranges like 0<x<5 and not x=5

What example are you talking about? Is it the question |3x-2|<=|2x-5| or |x+3|−|4−x|=|8+x| ? If it is 1st, then your question is not clear. It is an inequality and as such needs to have a range (often) of values. As for the 2nd example, refer to x-3-4-x-8-x-how-many-solutions-does-the-equation-148996.html#p1193962

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