If |x - 2| = |x + 3|, x could equal

|x - 2| = |x + 3|

We see both the sides are positive, hence we can square both the sides and we get

X^2 + 4 -4x = X^2 + 9 -6x

10x= -5

X= -1/2

Answer is B.

Since there are two Equations. Lets solve by Critical Value

|x - 2| = |x + 3| Critical Value will be 2, -3 hence we Have 3 Solutions

Solution No 1 When X < -3 ( RHS Will be Negative )

There fore x-2 = - (x + 3 ) => 2x = -1 => x = -1/2 which is not in line with x< -3 , hence rejected

Solution No 2 When -3 < X <= 2 ( LHS will be Negative )

-(x-2) = x+3 => 2x= -1 => x = -1/2 which fits in the range Hence Solution.

We can check by substituting in |x - 2| = |x + 3| Therefore 5/2= 5/2
Solution 3 when X > 2 both sides positive

x-2 = x+3 No Value hence rejected

Bunuel, in one of your post , you ave explained if we have mod on both side of the equation, we can safely square on both side
(x-2)^2= (x+3)^2
by solving we get x= -1/2 .
Please correct me if i missed something.

..

Hi sudhir,

yes, whenever you have ONLY mods on both sides..
example |a+b-2|>|a-b-4|... you can square both sides to fet your answer..
and you have correctly found your answer here

Given |x - 2| = |x + 3|

The best way to get rid of the modulus is to square on both sides.

\((x - 2)^2\) = \((x+3)^2\)

=> \(x^2\) + 4 + 4x = \(x^2\) + 9 + 6x

=> x = -1/2.

Correct answer is B...

In the first condition in which X is less than -3, why isn't the (x-2) term negative as well? The expression (x-2) is negative for all values where X is less than 2.

Yes you are right.

If \(x < -3\), then \(|x - 2| = -(x - 2)\) and \(|x + 3| = -(x + 3)\);

If \(-3 \leq x \leq 2\), then \(|x - 2| = -(x - 2)\) and \(|x + 3| = x + 3\);

If \(x > 2\), then \(|x - 2| = x - 2\) and \(|x + 3| = x + 3\).

With absolute value equations this simple, for me, the method below is easy and fast (less than a minute).

The only way the two sides will be equal is if quantities inside the absolute value brackets are 1) equal or 2) equal but with opposite signs.

So |A| = B or -B*

1. Remove the absolute value bars

2. LHS = RHS or LHS = -RHS

3. Set up the two equations

Case 1: x - 2 = x + 3, OR

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

4. Solve

Case 1:
x - 2 = x + 3
-2 = 3. Not a solution

CASE 2:
x - 2 = -(x + 3)
x - 2 = -x - 3
2x = -1
x = -\(\frac{1}{2}\)

5. Check x = -\(\frac{1}{2}\) in equation. --> |-\(\frac{5}{2}\)| = |\(\frac{5}{2}\)|

Correct. x = -\(\frac{1}{2}\). Answer B

Hope it helps

*and |B| = A or -A. Technically there are four cases. Case 3 is RHS = LHS. Case 4 is RHS = - LHS. These two cases are identical to Cases 1 and 2.
Case 3: +RHS = +LHS, (x + 3) = (x - 2) (= Case 1)
Case 4: +RHS = -LHS, (x + 3) = (- x + 2). Multiply both sides by (-1) --> (-x - 3) = (x - 2) (= Case 2)

STRATEGY: As with all GMAT Problem Solving questions, we should immediately ask ourselves, Can I use the answer choices to my advantage?
In this case, we can easily test the answer choices.
Now we should give ourselves about 20 seconds to identify a faster approach.
In this case, we can also solve the equation. HOWEVER, since the equation involves absolute values, we may end up solving two different equations (x - 2 = x + 3 as well as x - 2 = -(x + 3) before we arrive at the correct answer.
Given this, I think it may be faster to test the answer choices.


A. -5
Plug in this value to get: |(-5) - 2| = |(-5) + 3|
Simplify: |-7| = |-2|
Doesn't work.

B. -1/2 (aka -0.5)
Plug in this value to get: |(-0.5) - 2| = |(-0.5) + 3|
Simplify: |-2.5| = |2.5|
Works!

Answer: B

Given that |x - 2| = |x + 3| and we need to find the value of x

Let's solve this using three methods

Method 1: Substitution

Let's take values in each answer choice and substitutive in the equation and check which one satisfies the equation

A. -5 Put x = -5 in |x - 2| = |x + 3| and check if it satisfies the equation
=> |-5 - 2| = |-5 + 3| => |-8| = |-2|
=> 8 = 2 => FALSE

B. -1/2 Put x = -1/2 in |x - 2| = |x + 3| and check if it satisfies the equation
=> |-1/2 - 2| = |-1/2 + 3| => |-5/2| = |5/2|
=> 5/2 = 5/2 => TRUE

We don't need to check further as the question has asked for only one value

Method 2: Algebra

|x - 2| = |x + 3|
Square both the sides we get
\((|x-2|)^2 = (|x+3|)^2\)
=> \((x-2)^2 = (x+3)^2\)
=> \(x^2 -2*x*2 + 2^2 = x^2 + 2*x*3 + 3^2\)
=> \(x^2 -2*x*2 + 2^2 - x^2 - 2*x*3 - 3^2\) = 0
=> -4x + 4 - 6x - 9 = 0
=> -10x = 5
=> x = \(\frac{-5}{10}\) = \(\frac{-1}{2}\)

Method 3: Graphical Method

|x - 2| = Distance between x and 2
|x + 3| = | x - (-3)| = Distance between x and -3



So, for Distance between x and 2 = Distance between x and -3
x has to be at the mid-point of -3 and 2
=> x = \(\frac{-3+2}{2}\) = \(\frac{-1}{2}\)

So, Answer will be B
Hope it helps!

Watch the following video to learn How to Solve Absolute Value Problems

If −3≤x≤2−3≤x≤2 , then |x−2|=−(x−2)|x−2|=−(x−2) and |x+3|=x+3|x+3|=x+3;
can you please help me understand it? because if we chose any value between -3 to 2 for x then |x-2| can be positive, negative, or 0. for example if we take X= -3 then |x-2| will be negative i.e. -x+2. but if we take x=2, then |x-2| will be 0.

do we then not consider the 0 case?

|x - 2| = |x + 3|, we will put the options as answers and then check for their correctness

A. -5=> |-7| = |-2| =>incorrect
B. -1/2 => |-5/2| =|5/2| => correct
C. 1/2
D. 5
E. No real solution