# Precalc project: positive values in an inequality

• hpdwnsn95

## Homework Statement

Determine the set of positive values of x that satisfy the following inequality:
(1/x) - (1/(x-1)) > (1/(x-2))

a) (0, 1) union (2^1/2, 2) b) (0, 1/2) union (1, 2) c) (1/2, 1) union (2^1/2, 2(2^1/2))
d) (0, 2^1/2) union (3/2, 2) e) (1, 2^1/2) union (2, 2(2^1/2))

## Homework Equations

i'm not sure that there are any relevant equations, just started precalc a few weeks ago and I'm not sure how to solve the problem

## The Attempt at a Solution

I've tried graphing the equation and then tracing x to find y, but I haven't been able to find an answer that way. I've also tried solving for x which i get 1/2 > x for, but I'm not sure if that applies to this problem

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## Homework Statement

Determine the set of positive values of x that satisfy the following inequality:
(1/x) - (1/(x-1)) > (1/(x-2))

a) (0, 1) union (2^1/2, 2) b) (0, 1/2) union (1, 2) c) (1/2, 1) union (2^1/2, 2(2^1/2))
d) (0, 2^1/2) union (3/2, 2) e) (1, 2^1/2) union (2, 2(2^1/2))

## Homework Equations

i'm not sure that there are any relevant equations, just started precalc a few weeks ago and I'm not sure how to solve the problem

## The Attempt at a Solution

I've tried graphing the equation and then tracing x to find y, but I haven't been able to find an answer that way. I've also tried solving for x which i get 1/2 > x for, but I'm not sure if that applies to this problem
The first step would be to combine the two fractions on the left side.

## Homework Statement

Determine the set of positive values of x that satisfy the following inequality:
(1/x) - (1/(x-1)) > (1/(x-2))

a) (0, 1) union (2^1/2, 2) b) (0, 1/2) union (1, 2) c) (1/2, 1) union (2^1/2, 2(2^1/2))
d) (0, 2^1/2) union (3/2, 2) e) (1, 2^1/2) union (2, 2(2^1/2))

## Homework Equations

i'm not sure that there are any relevant equations, just started precalc a few weeks ago and I'm not sure how to solve the problem

## The Attempt at a Solution

I've tried graphing the equation and then tracing x to find y, but I haven't been able to find an answer that way. I've also tried solving for x which i get 1/2 > x for, but I'm not sure if that applies to this problem

Is this actually a multiple choice question? Is that what the a) b) c) etc refer to ?

That would be my guess.

## Homework Statement

Determine the set of positive values of x that satisfy the following inequality:
(1/x) - (1/(x-1)) > (1/(x-2))

a) (0, 1) union (2^1/2, 2) b) (0, 1/2) union (1, 2) c) (1/2, 1) union (2^1/2, 2(2^1/2))
d) (0, 2^1/2) union (3/2, 2) e) (1, 2^1/2) union (2, 2(2^1/2))

## Homework Equations

i'm not sure that there are any relevant equations, just started precalc a few weeks ago and I'm not sure how to solve the problem

## The Attempt at a Solution

I've tried graphing the equation and then tracing x to find y, but I haven't been able to find an answer that way. I've also tried solving for x which i get 1/2 > x for, but I'm not sure if that applies to this problem

Remember that if this is a multiple choice question, you do not have to find the answer, you only have to identify which of the 5 offerings is the answer.

While not a great method of solving a problem, substituting the offerings into the original inequality may be the fastest way for you to solve this.

## Homework Statement

Determine the set of positive values of x that satisfy the following inequality:
(1/x) - (1/(x-1)) > (1/(x-2))

a) (0, 1) union (2^1/2, 2) b) (0, 1/2) union (1, 2) c) (1/2, 1) union (2^1/2, 2(2^1/2))
d) (0, 2^1/2) union (3/2, 2) e) (1, 2^1/2) union (2, 2(2^1/2))

## Homework Equations

i'm not sure that there are any relevant equations, just started precalc a few weeks ago and I'm not sure how to solve the problem

## The Attempt at a Solution

I've tried graphing the equation and then tracing x to find y, but I haven't been able to find an answer that way. I've also tried solving for x which i get 1/2 > x for, but I'm not sure if that applies to this problem

If you do not know or see a clever trick to use, then solving the problem using basic principles is the best way. In the sample problem, start by adding the additive inverse of the righthand member to both sides; then perform the arithmetic steps for the rational expressions. You have some rational expression related to zero. Now just focus on the numerator.

Given that x is positive, I would separate this problem into three parts:
x between 0 and 1: both x- 1 and x- 2 are negative.

x between 1 and 2: x- 1 is positive but x- 2 is negative

x greater than 2: both x- 1 and x- 2 are positive.

For each of those, multiply the entire inequality by x(x-1)(x-2).

Is this actually a multiple choice question? Is that what the a) b) c) etc refer to ?

yes it is

Remember that if this is a multiple choice question, you do not have to find the answer, you only have to identify which of the 5 offerings is the answer.

While not a great method of solving a problem, substituting the offerings into the original inequality may be the fastest way for you to solve this.

I would do that but I don't understand the answer format, i"ve never seen union answers before, can you (or anyone else) expain them?

Do you know what the definition of the union of two sets is?

Do you know what the definition of the union of two sets is?

i think union stands for and right? so it's like one set & one set
but I'm not sure how would plug either set into the problem because there's no y in the problem

One piece (interval) of the x-axis plus (as in union) another piece. --- Actually, all the numbers in one interval together with all the numbers in another interval together with ...

I would do that but I don't understand the answer format, i"ve never seen union answers before, can you (or anyone else) expain them?

The first option

(0, 1) union (2^1/2, 2)

suggests it is true for any value between 0 and 1, as well as values between √2 and 2, but false for values between 1 and √2, and false for values above 2
it is clearly undefined for x = 0,1 or2 due to a fraction becoming 1/0

You could substitute 0.5 [1/2], and check it is true, 1.25 [5/4] and check it is false, 1.5 [3/2] and check it is true and 3 to check it is false.

Solving is better, but substitution could be your last resort.