Inequality

**Stuck Man** · November 18th 2010, 04:36 AM

I know how to solve the inequality x > (2/(x-1)) but how can I know not to mutiply all by (x-1)^2? That is a method for solving some inequalities.

**Wilmer** · November 18th 2010, 05:13 AM

Originally Posted by Stuck Man

I know how to solve the inequality x > (2/(x-1)) but how can I know not to mutiply all by (x-1)^2? That is a method for solving some inequalities.

Well, the equation could have been "typed" this way: x(x - 1) > 2 ; get my drift?

**skeeter** · November 18th 2010, 05:16 AM

Originally Posted by Stuck Man

I know how to solve the inequality x > (2/(x-1)) but how can I know not to mutiply all by (x-1)^2? That is a method for solving some inequalities.

you may multiply by $(x-1)^2$ provided you understand $x \ne 1$ and you can deal algebraically with the resulting cubic expression.

In this case, I would prefer to use this method ...

$x > \frac{2}{x-1}$

$x - \frac{2}{x-1} > 0$

$\frac{x(x-1)}{x-1} - \frac{2}{x-1} > 0$

$\frac{x(x-1) - 2}{x-1} > 0$

$\frac{x^2 - x - 2}{x-1} > 0$

$\frac{(x-2)(x+1)}{x-1} > 0$

critical values are $x = 2$ , $x = -1$ , and $x = 0$

these three values are where the expression on the left side of the inequality equals 0 or is undefined.

last step is to check a value from each interval between the critical values to see if the values in that interval make the original inequality true or false.

**Stuck Man** · November 18th 2010, 05:28 AM

I had a mistake with my algebra. So both methods can be used. I don't understand what Wilmer's manipulation achieves.

**Wilmer** · November 18th 2010, 05:55 AM

All I was trying to say is if equation was initially shown as x(x - 1) > 2,
then you wouldn't even think of: "how can I know not to mutiply all by (x-1)^2?".
Anyhow, not important...

**Archie Meade** · November 18th 2010, 01:50 PM

Originally Posted by Stuck Man

I know how to solve the inequality x > (2/(x-1)) but how can I know not to mutiply all by (x-1)^2? That is a method for solving some inequalities.

Suppose $x<1$

Then if you multiply both sides of the inequality by $(x-1)$

you would have to reverse the inequality.

This is because $x-1$ is negative when $x<1$

$5>4$ but $-5<-4$

The reversal happens if you change the sign of both sides,
which happens when you multiply both sides by a negative value,
or divide both sides by a negative value.

When you are dealing with inequalities, expressions involving x may be positive for certain x
and negative for other x.

Hence, if you multiply both sides by a square, you avoid that scenario (since real squares are non-negative).

Adding and subtracting the same value to both sides does not introduce any sign reversal,
nor does multiplying by 1, as shown by skeeter.

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