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, 05:36 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, 06:13 AMWilmer
- Nov 18th 2010, 06:16 AMskeeter
you may multiply by provided you understand and you can deal algebraically with the resulting cubic expression.

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

critical values are , , and

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, 06:28 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, 06:55 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, 02:50 PMArchie Meade
Suppose

Then if you multiply both sides of the inequality by

you would have to reverse the inequality.

This is because is negative when

but

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.