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|x – 3| > 4

We will have to take two cases

Case 1: Whatever is inside the modulus is >= 0
=> x-3 >= 0 => x >= 3
=> |x-3| = x-3 (as |X| = X when X >= 0)
=> x-3 > 4
=> x > 4+3
=> x > 7
7 >= 3
=> x > 7 is a solution

Case 2: Whatever is inside the modulus is < 0
=> x-3 < 0 => x < 3
=> |x-3| = -(x-3) (as |X| = -X when X < 0)
=> -(x-3) > 4
=> -x + 3 > 4
=> -x > 4-3
=> -x > 1
=> x < -1 (multiplying both the sides with -1 will reverse the sign of the inequality)
=> The condition was x < 3
-1 < 3
=> x < -1 is a solution

So, solution is x < -1 and x > 7

So, Answer will be E
Hope it helps!

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

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