You should use the absolute value definition, such that:

`|a| = {(a, a>= 0),(-a, a< 0):}`

Reasoning by analogy, yields:

`|x - 2| = {(x - 2, x - 2>= 0),(2 - x, x - 2< 0):}`

`|x - 2| = {(x - 2, x >= 2),(2 - x, x< 2):}`

You need to solve the equation `x - 2 = 5` , for `x >= 2` , such that:

`x - 2 = 5 => x = 2 + 5 => x = 7 >= 2`

You need to solve the equation `2 - x = 5` , for `x < 2` , such that:

`2 - x = 5 => -x = 5 - 2 => -x = 3 => x = -3 < 2`

**Hence, evaluating the solutions to the absolute value equation, yields `x = -3, x = 7` .**

Solve |x-2|=5:

(1) The argument for the absolute value (modulus) can be either 5 or -5 if the answer is 5 as |5|=|-5|=5:

x-2=5 ==> x=7

x-2=-5 ==> x=-3

(2) Alternatively, |x-2|=5 asks for the points that are 5 units from 2. (In general, |x-c|=L asks for the points L units from c.)

Looking on a number line it is easy to find that both 7 and -3 are 5 units from 2.