[x+5]=3x-7 check for extraneous solutions?

I suspect you meant to use "absolute value" symbols (unofficially called "the pipes")

|x+5| = 3x - 7

The concept of "absolute value" is one of distance. In number fields, it is simple the distance from zero.

With straight numbers, we interpret that as the value of the number, without the sign.

|7| = 7, because the distance from 0 to 7, is 7.
|-7| = 7, because the distance from 0 to -7 is also 7
(in mathematics, there are no negative distances)

Thus, with numbers, the pipes have one of two effects:

1) if the value inside the pipes is positive, then the pipes have no effect, and you can simply remove them.

2) if the value inside the pipes is negative, then the pipes have the same effect as a multiplication by -1. You can replace them with a negative sign in front of the "inside".

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|x+5| = 3x - 7

The value inside the pipes is x+5
when is this value negative? when x is less than -5 (less = more negative).

Therefore, you solve the problem this way.

You write (yes, this is part of the answer):

if x >= -5
(if x is greater than or equal to -5), then the "absolute value" has no effect and can be removed, leaving us with:

x + 5 = 3x - 7
12 = 2x
x = 6 (which is valid, since it is greater than -5)
We verify:
|6+5| = 3(6) - 7
|11| = 18-7
11 = 11

if x < -5, then the "absolute value" has the same effect as a multiplication by -1, leaving us with
-(x + 5) = 3x - 7
-x - 5 = 3x - 7
+2 = 4x
1/2 = x
which is rejected, because it is not less than -5

Just to make sure, we check
|0.5 + 5| = 3(0.5) - 7
5.5 = -5.5
not true.

Therefore, the problem only has one solution: x = 6

12 = 2x
x = 6