Help with inequation question: 1/x-2<1 please?

Question

Why is the answer x<2 and x>3 and why isn't it 2<x<3?

Answer 1

hi,

1/x-2 &lt; 1

=&gt; 1/x &lt; 1+2

=&gt; 1/x &lt; 3

=&gt; x &gt; 1/3

this is your required answer. Hope this helps. Bye.

Answer 2

1/x-2&lt;1

+2      +2

1/x&lt;3

Therefore x&lt;3

Answer 3

I assume you mean 1/(x-2)&lt;1   right?

You have to solve a simpler equation first:

Replace

x - 2 --&gt; x

1 / x &lt; 1

The answer to this is, of course  x &gt; 1

Then you solve:

x - 2 &gt; 1

Which is x &gt; 3

If the question really is 1/x-2&lt;1

Tha simplifies to 1/x &lt; 3

which is

x &lt; (1/3)  or   x &lt; 0.3333

Answer 4

Assuming the inequation is 1/(x-2) &lt; 1.

When multiplying both sides of an inequation, you must reverse the inequation sign when the multiplying number is negative.

For example, multiplying the inequation &quot;a &lt; b&quot; with the number -1 gives you the inequation &quot;-a &gt; -b&quot;.

In the inequation 1/(x-2) &lt; 1 we multiply both side with (x-2). However it is tricky because we have to distinguish 2 cases.

1) When (x-2) &gt; 0 (which means x &gt; 2)

   the inequation becomes 1&lt;x-2, and then x&gt;3

   As a result, the inequation is true for any x &gt; 3

  

2)  When (x-2) &lt; 0 (which means x &lt; 2)

   the inequation becomes 1&gt;x-2, and then x&lt;3

   As a result, the inequation is true for any x &lt; 2

The final answer is &quot;x&lt;2 or x&gt;3&quot;

Answer 5

1/(x-2) &lt; 1

If x-2 &gt; 0

Cross multiply to get:

1 &lt; x - 2

3 &lt; x

x &gt; 3

Check this with the assumption:

3-2 &gt; 0 is true

This is one part of a valid answer.

If x-2 &lt; 0:

Cross multiply to get:

1 &gt; x - 2 ------ multiplying both sides by a negative value willl reverse the inquality.

3 &gt; x

x &lt; 3

But the assumption limits this to the values such that x&lt;2

x &lt; 2

x ∈ {(-∞,2) U (3,∞)}

Answer 6

Easy. Treat it as an equation and:-

1/(x-2) &lt; 1       multiply bs by (x - 2)  (bs = both sides)

1 &lt; x - 2      Add 2 to bs

3 &lt; x     ie   x &gt; 3