When we have a DS problem with absolute values for ex:

|x + 3| = 4x – 3

Here it is clear that we have to try:
1- The Nonnegative option: |x + 3|>=0 ---> x+3=4x-3
2-The Negative option: |x + 3|<0 -->-x-3=4x-3

But when we have 2 absolute values:

|x – 3| = |2x – 3|

What do we have to try?

When both are nonnegative, when both are negative, when one is negative and the other nonnegative, and when one is nonnegative and the other negative?

Do we have to try all the 4 cases?

Thanks

Even |x| = |y|, there are only 2 distinct cases. Out of the four you mentioned, two of them are equivalent to each other and the same applies for the remaining two.

Consider all four:
x = y
-x = -y (Dividing both sides by -1 yields x = y)

x = -y
-x = y (Multiplying both sides by -1 yields x = -y)

givemeanid's explanation is correct. One thing I wanted to note is that for

|x + 3| = 4x – 3

the two options are:

1- The Nonnegative option: (x + 3)>=0 ---> x+3=4x-3
2-The Negative option: (x + 3)<0 -->-x-3=4x-3

I just changed Luci's absolute values to parentheses, in cases 1 and 2, because absolute values are never negative. (I'm sure that was just a typo, Luci.)

Another thing to notice is that we need to check the two possibilities to see which one is a genuine solution.

Let's check (1) first. We have

x+3=4x-3
6=3x
2=x

If we substitute this back into |x + 3| = 4x – 3, we get

|2 + 3| = 4(2) – 3
5=5

This makes sense, so (1) is a genuine solution.


Now let's check (2). We have
-x-3=4x-3
0=5x
0=x

If we substitute this back into |x + 3| = 4x – 3, we get

|0 + 3| = 4(0) – 3
3=-3

This does NOT make sense, so (2) is not a genuine solution.

-Jad