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realprakhar
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What is the solution set for |3x - 2| <= |2x - 5|? 11 Sep 2025, 04:54
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As both the sides of the equation > 0 , we can simply square both the sides and then solve the quadractic eq. which gives the answer as B
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What is the solution set for |3x - 2| <= |2x - 5|? 13 Oct 2025, 10:47
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"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"

How did you get these 2 points?
KarishmaB



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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What is the solution set for |3x - 2| <= |2x - 5|? 28 Oct 2025, 10:24
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Hi, where am i doing wrong?

|3x-2|<=|2x-5|

as, |x|=root x^2 we can write:-

(3x-2)^2<= (2x-5)^2
(3x-2)^2-(2x-5)^2<=0
[3x-2+2x-5][3x-2-2x+5]<=0
(5x-7)(x+3)<=0

from here,

x<=7/5; x<=-3
Bunuel

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

A. \(x<\frac{2}{3}\)

For this range, \(3x - 2 < 0\) and \(2x - 5 < 0\), so we get:

\(-(3x-2)\leq-(2x-5)\);

\(-3\leq{x}\)
Since we consider \(x < \frac{2}{3}\) range, we'd get:

\(-3\leq{x}<\frac{2}{3}\).
B. \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\)

For this range, \(3x - 2 \geq 0\) and \(2x-5\leq 0\), so we get:

\(3x-2\leq-(2x-5)\)

\(x\leq\frac{7}{5}\)
Since we consider \(\frac{2}{3}\leq{x}\leq\frac{5}{2}\) range, we'd get:

\(\frac{2}{3}\leq{x}\leq\frac{7}{5}\).
C. \(x>\frac{5}{2}\)

For this range, \(3x - 2 > 0\) and \(2x-5>0\), so we get:

\(3x-2\leq2x-5\)

\(x\leq{-3}\)
Since we consider \(x>\frac{5}{2}\) range, there is no solution in this range.

The solution from ranges A and B is:

\(-3 \leq x \leq \frac{7}{5}\).­
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What is the solution set for |3x - 2| <= |2x - 5|? 28 Oct 2025, 10:31
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Azeem123
Hi, where am i doing wrong?

|3x-2|<=|2x-5|

as, |x|=root x^2 we can write:-

(3x-2)^2<= (2x-5)^2
(3x-2)^2-(2x-5)^2<=0
[3x-2+2x-5][3x-2-2x+5]<=0
(5x-7)(x+3)<=0

from here,

x<=7/5; x<=-3

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All is good except the last part.

(5x - 7)(x + 3) ≤ 0 gives -3 ≤ x ≤ 7/5.

The roots are -3 and 7/5. Because the inequality is "≤", the solution lies between the roots: -3 ≤ x ≤ 7/5.


Inequalities

Theory:

Questions:

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.­
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