|a + 4| < 7
-7 < a + 4 < 7
Subtracting 4 throughout:
-11 < a < 3
3 > a > -11, (C) IMO!
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|a + 4| < 7

We will have to take two cases

Case 1: Whatever is inside the modulus is >= 0
=> a+4 >= 0 => a >= -4
=> |a+4| = a+4 (as |X| = X when X >= 0)
=> a+4 < 7
=> a < 7-4
=> a < 3
But condition was a >=-4
=> So solution will be the intersection of two of them (As shown in the image below)
=> -4 <= a < 3



Case 2: Whatever is inside the modulus is < 0
=> a+4 < 0 => a < -4
=> |a+4 | = -(a+4 ) (as |X| = -X when X < 0)
=> -(a+4 ) < 7
=> -a - 4 < 7
=> -a < 7 + 4
=> -a < 11
=> a > -11 (multiplying both sides with negative sign reverses the inequality sign)
But the condition was a < -4
=> So solution will be the intersection of two of them (As shown in the image below)
=> -11 < a < -4



=> -11 < a < -4 and -4 <= a < 3
=> -11 < a < 3
Or, 3 > a > -11

So, Answer will be C
Hope it helps!

Watch the following video to learn how to Solve Inequality + Absolute value Problems


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