If |2x + 5| = |3x − 2|, which of the following is a possible value of

2x+5 = 3x-2
x= 7
2x+5 = -3x+2
x=-3/5

Two ways..
I. Open the MOD..
|2x+5|=|3x-2|....
Square both sides...
\((2x+5)^2=(3x-2)^2.......4x^2+20x+25=9x^2-12x+4....5x^2-32x-21=0.........(5x+3)(x-7)=0\)..
So x can be \(-\frac{3}{5}\) or 7...

II. Substitution..
Start with the central value..
\(x=-\frac{3}{5}\)..
\(|2*\frac{-3}{5}+5|=|3*\frac{-3}{5}-2|........|5-\frac{6}{5}|=|\frac{-9}{5}-2|=\frac{19}{5}\)... Correct

C

After plugging in Answers A and E because the math was quick, and finding neither to be a solution, I took the case approach.
+LHS = +RHS
+LHS = -RHS

|2x+5|=|3x−2|

Case 1:
2x + 5 = 3x - 2
7 = x

Case 2:
2x + 5 = -(3x-2)
2x + 5 = -3x + 2
5x = -3
x = -3/5

Answer C

Hi adkikani

When you open the mod take care to solve for both positive and negative situations i.e \(±\)

Here \(|2x+5|=|3x-2|\). Now lets decide to open the LHS mod

so we will have \(2x+5=±|3x-2| => 2x+5=|3x-2|\) and \(2x+5=-|3x-2|\)

Case 1: \(|3x-2|=2x+5\), again we have a mod so we will have further two cases \(3x-2 =±(2x+5)\)

so \(3x-2=2x+5 => x=7\)

and \(3x-2=-(2x+5) => x= -\frac{3}{5}\)

Case 2: \(-|3x-2|=2x+5\) or \(|3x-2|=-2x-5\), again we have a mod so we will have further two cases \(3x-2=±(-2x-5)\)

so \(3x-2=-2x-5 => x=-\frac{3}{5}\)

and \(3x-2 = 2x+5 => x=7\)

So finally we have \(x=7\) or \(\frac{-3}{5}\)

Hence Option C

Alternatively as Chetan had explained, you can square both sides to remove the mod. the only challenge in this method can be, if the equation is complex then it will lead to further complexity.

niks18

Thanks for your explanation. However I really found it a bit intimidating.

Is rahul16singh28 solution technically correct too?

He simply opened LHS modulus to take two cases of positive and negative terms of RHS.
But interestingly he probably assumed that LHS will be always be positive
ie no scope for -2x-5

Let me know your inputs

Hi adkikani,

To be fail-safe, in my opinion, when we remove mod we need to consider both positive and negative scenarios for each expression

For this question squaring works best because the equation is linear and has only 1 variable x. Squaring should result in a simple quadratic equation. If the equation had been any other polynomial or had contained more than one variable, then squaring method would have been cumbersome.

Finally, you can always assume two scenarios for variables -

Case 1: if 2x+5>0 or x>-5/2, then |3x-2|=2x+5, solve this to get two values of x that satisfy x>-5/2

Case 2: if 2x+5<0 or x<-5/2, then |3x-2|=-(2x+5).....{because if x<0, then |x|=-x}, solve this to get two values of x that satisfy x<-5/2

then combine the values of x derived from the above two cases.

Would request other experts to chip-in with their thoughts

Method 1:

If I get this question in the test, I would just substitute options in the equation. That's the fastest way.

|2x+5|=|3x−2|
Try the integer values very quickly first. They don't satisfy

|2x + 25/5| = |3x - 10/5|
Now try 3/5, then -3/5.

-3/5 satisfies. Answer (C)


Method 2: |2x+5|=|3x−2|

2 * |x + 5/2| = 3 * |x - 2/3|

____________________ ( - 15/6) _________________________ ( 0 ) ____________ ( 4/6 )_____________________

This is a distance of 19 between them. Twice the distance from -15/6 should be equal to thrice the distance from 4/6.

So the point will be \(\frac{-15}{6} + (\frac{19}{6}*\frac{3}{5})\)

We get \(\frac{-3}{5}\)

VeritasPrepKarishma





I believe you also sensed that having 5 in denominator will help you ease the addition/ subtraction of fractions.
Super quick, thanks a ton.



I totally could not understand method 2. Can you elaborate?

These links explain you what absolute value is:
https://youtu.be/oqVfKQBcnrs
https://anaprep.com/algebra-inequalitie ... in-action/
https://anaprep.com/algebra-the-why-beh ... questions/

In this question, we say that x is a point such that twice its distance from -5/2 is equal to thrice its distance from 2/3.

So x will divide the distance between the two points in the ratio 3:2. Hence, x will be at a distance 3/5th to the right from -15/6.

Does this help?

First we can solve for when both 2x+5 and 3x−2 are positive:

|2x + 5| = |3x − 2|

2x + 5 = 3x - 2

-x = -7

x = 7

Next we can solve for when 2x+5 is positive and 3x−2 is negative.

|2x + 5| = |3x − 2|

2x + 5 = -(3x - 2)

2x + 5 = -3x + 2

5x = -3

x = -3/5

Answer: C

I am not an expert but would like to share my thoughts here.

When we have modulus on both side of the equality (=) sign and when we are trying the cases approach, we only need to consider two cases. First, both sides have same sign. Second, different sign on both sides.

One may try the other two cases as well, but the results will be the same as above.

Hope that helps. Let me know if you want more details.

Cheers,
Kabir

This question is very easy and my tips for such question is that you first equate LHS and RHS and do the same but with negative of RHS.

easy one,

We have 2x+5 = 3x -2 (1) and 2x+5 = -3x + 2 (2)

Solve (1) we have x =7, eliminate because x=7 does not exist in 5 given answers.

Solve (2) we have x= -3/5, answer C shows x= -3/5

=> Answer C is correct.

How does everyone seem to know that we only need to test the original LHS equal to negative RHS? Seems to me that there are 4 distinct combinations that can be solved to find the answer.

original LHS = orignial RHS
negative LHS = original RHS
original LHS = negative RHS
negative LHS = negative RHS

In the 4 options..
a) original LHS = orignial RHS <=> negative LHS = negative RHS...Both the cases are same
b) negative LHS = original RHS <=> original LHS = negative RHS ...Both these cases are also same

Ahhh yes, I see it now. Thank so much!

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