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So, this can be done by traditional algebra. However, solving such inequalities in most cases can be done by option elimination as well.

|3x – 19| > 5 – x

Check if the inequality holds true for x = 0
LHS => |3(0) - 19| = 19
RHS => 5 - 0 = 5

x = 0 satisfies the inequality, hence should be a part of the solution.

Hence, we can eliminate options - A, C, E

Now choose a value of x which is in option 4 but not in option 1.

Let's say x = 10

LHS => |3(10) - 19| = 11
RHS => 5 - 10 = -5
x = 10 satisfies the inequality, hence should be a part of the solution. Hence, option B is eliminated.

Answer = option D
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If |3x – 19| > 5 – x , which of the following represents the complete range of values of x ?

A. x > 6
B. x < 7
C. 6 < x < 7
D. All real values
E. No real value
D is the correct answer choice.

Video explanation:

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