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How do you solve #abs(2x-5)> -1#?

Answer 1

Given #abs(2x-5) > -1#

We first note that by definition: #color(white)("XXXXX")##abs( "anything") >= 0# Therefore #color(white)("XXXXX")##abs(2x-5)> -1# is valid for any value of #x#

The "solution" is #color(white)("XXXXX")# All #x epsilon RR# (or, actually, all #x epsilon CC# if you want to expand to complex numbers).

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Answer 2

Eli Assouline

To solve the inequality ( |2x - 5| > -1 ), note that the absolute value of any real number is always non-negative. Therefore, the absolute value ( |2x - 5| ) is always greater than or equal to zero.

Since the inequality states that ( |2x - 5| > -1 ), which is always true for any real number ( x ), the solution is all real numbers ( x ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you solve #abs(2x-5)&gt; -1#?

How do you solve #abs(2x-5)> -1#?