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> 6
First you isolate the absolute value by multiplying
both sides by 2
|a + 5| > 12
Then you use whichever of these two rules applies:
Rules for getting rid of absolute value bars in inequalities:
----------------------------------------------
Rule for < or ≦
For |mx+b| < c
If c is positive the inequality becomes
-c < mx+b < c
Then you solve that with x in the middle.
If c is not positive there is no solution.
It's the same for ≦
--------------------------------------
Rule for > or ≧
For |mx+b| > c
If c is positive the inequality becomes
mx+b < -c OR mx+b > c
Then you solve each of those for x and write "OR" between them
If c is not positive the solution is "ALL REAL NUMBERS"
It's the same for ≧
--------------------------------------------
Back to your problem
|a + 5| > 12
To get rid of the absolute value bars
Since 12 is positive the inequality becomes
a+5 < -12 OR a+5 > 12
a < -17 OR a > 7
<=====)------(=====>
-17 0 7
(-∞,-17) ᑌ (7,∞)
Edwin