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

Hello M7MBA.

Well, you can plug in the values and see which option is the correct.

Also, we can solve it by expanding the absolute values |x - 2| and |x + 3| as follows:

Each expression between the absolute values is equal to zero when x=2 and x=-3. Then, we have the following cases:

Case 1: x<-3.

- If x<-3 then x+3<0.
- If x<-3 then x<2 and therefore x-2<0.

Hence $$\left|x-2\right|=\left|x+3\right|\ \ \Leftrightarrow\ \ \ -\left(x-2\right)=-\left(x+3\right)\ \ \ \Leftrightarrow\ \ -x+2=-x-3\ \ \ \Leftrightarrow\ \ 2=-3\ \ \ False.$$ In this case we didn't get a real solution.

Case 2: -3 < x < 2.

- If -3<x then x+3>0.
- If x < 2then x-2<0.

Hence $$\left|x-2\right|=\left|x+3\right|\ \ \Leftrightarrow\ \ \ -\left(x-2\right)=x+3\ \ \ \Leftrightarrow\ \ -x+2=x+3\ \ \ \Leftrightarrow\ \ 2x=-1\ \ \ \Leftrightarrow\ \ \ x=-\frac{1}{2}.$$ This implies that the correct answer is the option [spoiler]C=-1/2[/spoiler].

Finally

Case 3: x > 2

- If x > 2 then x-2 > 0.
- If x > 2 then x > -3 and therefore x+3 > 0.

Hence $$\left|x-2\right|=\left|x+3\right|\ \ \Leftrightarrow\ \ \ x-2=x+3\ \ \ \Leftrightarrow\ \ -2=3\ \ False$$ I hope it helps you.

M7MBA wrote:If |x - 2| = |x + 3|, x could equal

A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution

Case 1: signs unchanged
x-2 = x+3
-2 = 3
Since the resulting equation is invalid, Case 1 is not possible.

Case 2: signs changed in ONE of the absolute values
x-2 = -x-3
2x = -1
x = -1/2.

The correct answer is B.

M7MBA wrote:If |x - 2| = |x + 3|, x could equal

A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution

Alternate approach:
|a-b| = the distance between a and b.
|a+b| = |a-(-b)| = the distance between a and -b.

|x - 2| = |x + 3| implies the following:
The distance between x and 2 is equal to the distance between x and -3.
In other words, x must be HALFWAY BETWEEN -3 AND 2.
Halfway between two numbers = the AVERAGE of the two numbers:
(-3+2)/2 = -1/2.

the correct answer is B.

IMPORTANT: Plug x = -1/2 into the ORIGINAL equation, to make sure it isn't an extraneous root.

The warning above must be heeded when only one side of an equation is an absolute value.
But when BOTH sides of an equation are absolute values, any solution that we derive will be guaranteed to work, so the step above is unnecessary.
Since both sides of |x - 2| = |x + 3| are absolute values, we do not need to check the validity of x=-1/2.

M7MBA wrote:If |x - 2| = |x + 3|, x could equal

A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution

By scanning the answer choices, we see that when x = -1/2, we have

|-1/2 - 2| = |-5/2| = 5/2 and |-1/2 + 3| = |5/2| = 5/2

Alternate Solution:

We consider two cases for this absolute value question.

Case 1. Drop the absolute value symbols and solve for x.

x – 2 = x + 3

0 = 5

This doesn’t work.

Case 2. (x – 2) is positive and (x + 3) is negative.

x – 2 = -(x + 3)

x – 2 = -x – 3

2x = -1

x = -½

When x = -1/2, we see that |x - 2| = |x + 3| becomes |(-1/2) - 2| = |(-1/2) + 3|, or 5/2 = 5/2, which is a true statement. Thus, the solution for this equation is x = -1/2.

Answer: B