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Re: What is the solution set for |3x-2|<=|2x-5|  [#permalink]

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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.
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Re: What is the solution set for |3x-2|<=|2x-5|  [#permalink]

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Answer: -3 <= x <= 7/5

More detailed solution for similar problems: http://burnoutorbreathe.blogspot.com/2012/12/how-to-get-solution-for-absolute-values.html
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Re: What is the solution set for |3x-2|<=|2x-5|  [#permalink]

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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.
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Re: What is the solution set for |3x-2|<=|2x-5|  [#permalink]

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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 draw the number line as follows

......-3..........2/3........7/5..............5/2...............

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

Hope this helps
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Re: What is the solution set for |3x-2|<=|2x-5|  [#permalink]

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Bunuel 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]

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BrainLab wrote:
Bunuel 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 $$2/3 > x \geq-3$$.

Hope this helps.
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Re: What is the solution set for |3x-2|<=|2x-5|  [#permalink]

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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]

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1
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]

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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

The 2nd inequality does not have any solutions.

Hope this helps.
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What is the solution set for |3x-2|<=|2x-5|  [#permalink]

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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]

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BrainLab 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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Re: What is the solution set for |3x-2|<=|2x-5|  [#permalink]

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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.

How do we solve using square?

(Sorry, I am sort of confused.)

Would greatly appreciate, if you could please brief it once. _________________
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Re: What is the solution set for |3x-2|<=|2x-5|  [#permalink]

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rekhabishop 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.

How do we solve using square?

(Sorry, I am sort of confused.)

Would greatly appreciate, if you could please brief it once. One more way to solve:
(3x-2)^2<=(2x-5)^2
(3x-2)^2-(2x-5)^2<=0
a2-b2=(a+b)(a-b)
similarily (x+3)(5x-7)<=0
two solutions: x=-3,x=7/5

Now on a number line plot -3 and 7/5. Everything to extreme right mark '+' and alternate between them.
___+___(-3)______-________(7/5)__________+__________

We are looking for the equation to be less than or equal to zero.

-3<=x<=7/5 is the solution
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Re: What is the solution set for |3x-2|<=|2x-5|  [#permalink]

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Bunuel 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: $$-3\leq{x}\leq\frac{7}{5}$$.

excellent ..! very didactic post ...
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Re: What is the solution set for |3x-2|<=|2x-5|  [#permalink]

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Bunuel 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

when evaluating between 2/3 and 7/5, you have selected 7/5, and when evaluating between -3 and 2/3, you have selected -3 as the range for final solution...
Is this because, you have already set the range of x, as x is less than 2/3 for condition A, and have set the range of x, as x is less than 5/2 and greater than 2/3 for condition B, or you have taken the help of number line when setting the range for final solution that is "x is greater than or equal to -3 and smaller than or equal to 7/5"

please shed some light on this issue ...

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Re: What is the solution set for |3x-2|<=|2x-5|  [#permalink]

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Bunuel 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: $$-3\leq{x}\leq\frac{7}{5}$$.

Can you please explain me the reason of choosing these three ranges? Why didn't you use x>=2/3? Thanks in advance. I am really poor in this topic, I read the GMAT math book for this topic also, but finding it so hard to understand, please kindly help. Veritas Prep GMAT Instructor V
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Re: What is the solution set for |3x-2|<=|2x-5|  [#permalink]

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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.

Solving the question using number line (as requested in a pm)

$$|3x-2|\leq|2x-5|$$

$$3 * |x-\frac{2}{3}|\leq 2*|x-\frac{5}{2}|$$

$$3 * |x-\frac{4}{6}|\leq 2*|x-\frac{15}{6}|$$

--------------------------- (4/6) ----------------------------(15/6)----------------

Assuming just "equal to" for now.
Three times the distance from 4/6 is equal to two times the distance from 15/6

Divide the distance of 11/6 into 5 parts to get 11/30 each part. x would be a point 22/30 away from 4/6 i.e. x = 42/30 = 7/5
On the left of 7/5, "three times the distance from 4/6" will be less than "two times the distance from 15/6"

On the left side of 4/6, there will be another such point x such that thrice the distance from 4/6 is equal to twice the distance from 15/6
$$3*(4/6 - x) = 2(15/6 - x)$$
$$x = -3$$
On the right of -3, "three time distance from 4/6" will be less than "two times the distance from 15/6"

So the required range is $$-3 \leq x \leq 7/5$$
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Re: What is the solution set for |3x-2|<=|2x-5|  [#permalink]

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What I don't understand is: why can I sometimes square both sides and get the correct solutions and other times I have to use this range approach? How do I know when the range approach is required?
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Re: What is the solution set for |3x-2|<=|2x-5|  [#permalink]

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Bunuel 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, how can you conclude (without plugging in the numbers) that range C has no solution? Thanks!
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Re: What is the solution set for |3x-2|<=|2x-5|  [#permalink]

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Bunuel 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! I haven't understood the basics. How do you get the range? Re: What is the solution set for |3x-2|<=|2x-5|   [#permalink] 19 May 2019, 18:32

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