How do you solve # |a+6|>12#?
Solve it as a normal equation, then examine the 'inequality' boundaries.
In general, a one-sided inequality can be solved by first solving the equality, as that will give you the “boundary” of the solution. In this case |a+6|=12. ONLY the 'a' varies, so we can remove the 6 from the inequality by normal subtraction. |a| = 12 – 6 = 6.
So now we know that our 'boundary' is when a = 6 and when a = -6. Anything between those boundaries will not satisfy the inequality (the resulting values will be between 0 and 11), so our inequality solution is |a| > 6 . This may also be written as a
#a+6 > 12 to a >6.
So, a is outside the closed interval