Question

Which compound inequality is equivalent to lax- bl > C for all real numbers a, b, and c, where c greater than or equal to 0?

Answer 1

C is correct because when you get rid of absolute value you negate the right side. And solve as if there is no absolute value.
This means that one equation becomes 2.
Solve these and you get C.

Answer 2

Answer:

  none of the above

Step-by-step explanation:

When the inequality symbol points left, the absolute value inequality can be "unfolded" to a compound inequality as follows:

  |ax -b| < c   ⇔   -c < ax -b < c

That is, the constant can be negated and put on the left with an inequality symbol that points in the same direction.

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When the inequality symbol points to the right, as it does in this problem statement, unfolding it in the same fashion gives an expression that looks like the second answer choice:

  -c > ax -b > c . . . . . . nonsense

This is false on the face of it, because -c > c is a false statement (when c>0). Such an inequality presumes an AND conjunction:

  -c > ax -b  AND  ax -b > c

Since the two solution spaces do not overlap, the AND of them is the null set.

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The same inequality can be unfolded in the same way, if an OR conjunction is used:

  -c > ax -b  OR  ax -b > c

Rewriting this to match the form of the given answer choices, it would be ...

  ax -b < -c  OR  ax -b > c . . . . . . . matches no given choice

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Attached is an example of such an inequality, so you can ponder this answer with a concrete example. You may want to discuss this question with your teacher.