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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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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
What is the solution set for |3x-2|<=|2x-5| [#permalink]
17 Dec 2015, 03:44
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.
Hi dear math experts, I'm just trying to find an approach that I understand and can apply for such kind of questions. If we have something like the expression below, we can just test 2 cases and it's a valid approach for this type. |2x - 1| = |4x + 9| Solution: \(x =-\frac{4}{3} or-5\) \(2x-1=4x+9\) or \(2x-1=-(4x+9)\) Both solutions satisfy this equation
but as we saw in the example of Hussain15 it's not enough to solve absolute value inequalities . I think the most important point is here not to change the sign of an inequality and just expand the the absolute values . I've tried to solve it this way taking the approach from Hussain15 into sonsideration:
We have 4 cases:
Case 1: \(3x-2 ≤ 2x-5, x ≤ -3\) Case 2: \(3x-2 ≤ -2x+5, x ≤ \frac{7}{5}\) Case 3: \(-3x-2 ≤ 2x-5, x ≥ \frac{7}{5}\) Case 4: \(-3x+2 ≤ -2x+5, x ≥ -3\)
After inserting the values from the above ranges in the expression one can see that the expression hold true ONLY in Cases 4 and 2 \(−3≤x≤\frac{7}{5}\). The second important point I've observed here is that while testing 4 cases we see a pattern - we have actually 2 different values with the interchanginng inequality signs.
Would it be a valid approach speaking generally about such kind of questions ? (Offcourse I'll try it out myself while solving other inequalities, but I don't have this experience yet) _________________
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Re: What is the solution set for |3x-2|<=|2x-5| [#permalink]
17 Dec 2015, 05:12
BrainLab 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.
Hi dear math experts, I'm just trying to find an approach that I understand and can apply for such kind of questions. If we have something like the expression below, we can just test 2 cases and it's a valid approach for this type. |2x - 1| = |4x + 9| Solution: \(x =-\frac{4}{3} or-5\) \(2x-1=4x+9\) or \(2x-1=-(4x+9)\) Both solutions satisfy this equation
but as we saw in the example of Hussain15 it's not enough to solve absolute value inequalities . I think the most important point is here not to change the sign of an inequality and just expand the the absolute values . I've tried to solve it this way taking the approach from Hussain15 into sonsideration:
We have 4 cases:
Case 1: \(3x-2 ≤ 2x-5, x ≤ -3\) Case 2: \(3x-2 ≤ -2x+5, x ≤ \frac{7}{5}\) Case 3: \(-3x-2 ≤ 2x-5, x ≥ \frac{7}{5}\) Case 4: \(-3x+2 ≤ -2x+5, x ≥ -3\)
After inserting the values from the above ranges in the expression one can see that the expression hold true ONLY in Cases 4 and 2 \(−3≤x≤\frac{7}{5}\). The second important point I've observed here is that while testing 4 cases we see a pattern - we have actually 2 different values with the interchanginng inequality signs.
Would it be a valid approach speaking generally about such kind of questions ? (Offcourse I'll try it out myself while solving other inequalities, but I don't have this experience yet)
Your method seems to be fine, I will mention mine below. But as GMAT questions comes with options, I would say that you also need to use the options to your favor in such questions. This question is asking for different discrete values or range of x. Your method is similar to the one mentioned below but is more time consuming.
You can clearly see that with 2 of the 5 options as
1) -3<x<5 2)-1<x<5
1 of these options can be eliminated by looking at x=-2.
Coming back to the analytical way to solve such questions, you are asked about the range of values of x satisfying |3x-2|<=|2x-5|. Bunuel has mentioned the most straightforward solution at what-is-the-solution-set-for-3x-2-2x-89266.html#p675126
You need to look at 3 intervals : \(x<2/3\), \(2/3 \leq x < 5/2\) and \(x \geq 5/2\).
Out of the above 3 intervals, the last one will give you no solution while the ranges from 1 and 2 will give you the desired solution of \(−3≤x≤\frac{7}{5}\) _________________
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