# Thread: Inequalities with a variable in the denominator

1. ## Inequalities with a variable in the denominator

I'm past this, but couldn't remember the rule....I have
$
\frac {5} {(x+1)^3} < N
$

Then there is a rule for solving this.... You can't multiply both sides by $(x+1)^3$,because you don't know if $x$ is negative or positive. The book says you should get
$
{\sqrt [3]{5/|N|}}
$

Thanks!

2. Originally Posted by Chokfull
I'm past this, but couldn't remember the rule....I have
$
\frac {5} {(x+1)^3} < N
$

Then there is a rule for solving this.... You can't multiply both sides by $(x+1)^3$,because you don't know if $x$ is negative or positive. The book says you should get
$
{\sqrt [3]{5/|N|}}
$

Thanks!
With these types of equalities you have a couple of chouces.

As you said you cannot multiply both sides by $(x+1)^3$, however you can multiply both by $(x+1)^6$, as you know this is a positive number.

You can also make both sides into fractions and add or subtract to rearrange.

For example, $\frac {5} {(x+1)^3} <\frac{N(x+1)^3}{(x+1)^3}$, leading to $\frac {5} {(x+1)^3} - \frac{N(x+1)^3}{(x+1)^3}<0$

Can you solve this from here?

3. Originally Posted by craig
With these types of equalities you have a couple of chouces.

As you said you cannot multiply both sides by $(x+1)^3$, however you can multiply both by $(x+1)^6$, as you know this is a positive number.

You can also make both sides into fractions and add or subtract to rearrange.

For example, $\frac {5} {(x+1)^3} <\frac{N(x+1)^3}{(x+1)^3}$, leading to $\frac {5} {(x+1)^3} - \frac{N(x+1)^3}{(x+1)^3}<0$

Can you solve this from here?
OK, but how does that get an absolute value? I'll fiddle with it for now but don't see how it gets ${\sqrt [3]{5/|N|}}
$