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.

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- Nov 18th 2010, 04:36 AMStuck ManInequality
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.

- Nov 18th 2010, 05:13 AMWilmer
- Nov 18th 2010, 05:16 AMskeeter
you may multiply by $\displaystyle (x-1)^2$ provided you understand $\displaystyle x \ne 1$ and you can deal algebraically with the resulting cubic expression.

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

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

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

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

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

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

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

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

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. - Nov 18th 2010, 05:28 AMStuck Man
I had a mistake with my algebra. So both methods can be used. I don't understand what Wilmer's manipulation achieves.

- Nov 18th 2010, 05:55 AMWilmer
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... - Nov 18th 2010, 01:50 PMArchie Meade
Suppose $\displaystyle x<1$

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

you would have to reverse the inequality.

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

$\displaystyle 5>4$ but $\displaystyle -5<-4$

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.