# Inequality with fractions solve 2/(x^2−1)≤1/(x+1)

• MHB
• Yankel
In summary, if x^2-1 is negative, then x<-1. If x^2-1 is positive, then x>-1. If x^2 is positive, then x>3. If x^2 is negative, then x<-1.
Yankel
Dear all,

I am trying to solve this inequality:

$\frac{2}{x^{2}-1}\leq \frac{1}{x+1}$

I've tried several things, from multiplying both sides by

$(x^{2}-1)^{2}$

finding the common denominator, but didn't get the correct answer, which is:

$2<x<3$

or

$x<-1$How to you solve this one ?

Solve 2/(x^2-1) = 1/(x+1)

Solution: x = 3 or x = -1

Now throw the < in there!

Yankel said:
Dear all,

I am trying to solve this inequality:

$\frac{2}{x^{2}-1}\leq \frac{1}{x+1}$

I've tried several things, from multiplying both sides by

$(x^{2}-1)^{2}$

finding the common denominator, but didn't get the correct answer, which is:

$2<x<3$

or

$x<-1$How to you solve this one ?

You don't want to multiply an equality by any expression involving a variable, because you don't know the sign, and it is important to know the sign of a value when multiplying with an inequality. What I would do is move everything to one side, so you have zero on the other:

$$\displaystyle \frac{2}{x^2-1}-\frac{1}{x+1}\le0$$

Now, combine terms on the LHS:

$$\displaystyle \frac{2-(x-1)}{x^2-1}\le0$$

$$\displaystyle \frac{3-x}{(x+1)(x-1)}\le0$$

Can you proceed from here?

MarkFL,

I tried your way but got stuck exactly where you stopped.

I tried multiplying each side by

$(x^{2}-1)^{2}$

which is always positive or zero, can't be negative so I don't have to worry about the sign. Then I got expression with fifth power, couldn't go on.

Yankel said:
MarkFL,

I tried your way but got stuck exactly where you stopped.

I tried multiplying each side by

$(x^{2}-1)^{2}$

which is always positive or zero, can't be negative so I don't have to worry about the sign. Then I got expression with fifth power, couldn't go on.

$$\displaystyle x^2-1$$ is negative on $$(-1,1)$$, so you do need to worry about the sign. Let's go back to:

$$\displaystyle \frac{3-x}{(x+1)(x-1)}\le0$$

From this, we obtain 3 critical values (places where the expression may change sign)

$$\displaystyle x\in\{-1,1,3\}$$

These 3 critical values divide the real number line into 4 intervals...can you state these intervals?

MarkFL said:
$$\displaystyle x^2-1$$ is negative on $$(-1,1)$$, so you do need to worry about the sign.
Yes, but Yankel is multiplying by $\color{red}(\color{black}x^2-1\color{red})^2$.

Olinguito said:
Yes, but Yankel is multiplying by $\color{red}(\color{black}x^2-1\color{red})^2$.

Okay I did miss that, but that still leads to the same critical values and is a needless move.

$$\displaystyle \frac{3-x}{(x+1)(x-1)}\le0$$

By inspection of Mark's sexy equation:
x<>1 and x<>-1 (division by 0)

equality occurs only when x=3

0 is greater only if x>3

Wilmer said:
$$\displaystyle \frac{3-x}{(x+1)(x-1)}\le0$$

By inspection of Mark's sexy equation:
x<>1 and x<>-1 (division by 0)

equality occurs only when x=3

0 is greater only if x>3

What I would do is divide the real number line into the following intervals (based on the critical values, the strength of the inequality and division by zero):

$$\displaystyle (-\infty,-1)$$

$$\displaystyle (-1,1)$$

$$\displaystyle (1,3)$$

$$\displaystyle [3,\infty)$$

Then, since all critical values are of odd multiplicity, I would observe that the sign of the expression will alternate across all intervals, so I only need test one interval, and I would choose the one containing $$x=0$$, and we see the sign of the expression is negative, and so we conclude:

$$\displaystyle (-\infty,-1)$$ + not part of the solution

$$\displaystyle (-1,1)$$ - part of the solution

$$\displaystyle (1,3)$$ + not part of the solution

$$\displaystyle [3,\infty)$$ - part of the solution

And so the solution to the original inequality is:

$$\displaystyle (-1,1)\,\cup\,[3,\infty)$$

Alternatively, we can multiply both sides with $(x^2-1)$ and take cases:
$$\begin{array}{ccccccc}&&&\frac 2{x^2-1} \le \frac 1{x+1}\\ (x^2-1 > 0) &\land& (2 \le \frac{x^2-1}{x+1}) &\lor& (x^2-1 < 0) &\land& (2 \ge \frac{x^2-1}{x+1}) \\ (x^2 > 1) &\land& (2 \le x-1) &\lor& (x^2 < 1) &\land& (2 \ge x-1) \\ (x^2 > 1) &\land& (x \ge 3) &\lor& (x^2 < 1) &\land& (x\le 3) \\ &&(x \ge 3) &\lor&(-1<x<1) \end{array}$$

## What is inequality with fractions?

Inequality with fractions is an expression that compares two fractions using the symbols <, >, ≤, or ≥ to indicate which fraction is larger or smaller.

## How do I solve "2/(x^2−1)≤1/(x+1)" for x?

To solve this inequality, first rewrite it as 2/(x^2-1) - 1/(x+1) ≤ 0. Then, find the common denominator and simplify the fractions. Next, factor the denominator and find the critical values where the expression is undefined. Finally, plot the critical values on a number line and test a value within each interval to determine the solution set for x.

## Can I multiply or divide both sides of the inequality by a negative number?

Yes, you can multiply or divide both sides of the inequality by a negative number without changing the direction of the inequality. However, if you multiply or divide by a negative number, you must flip the direction of the inequality. For example, if you multiply both sides of the inequality -2/3 < x by -3, the resulting inequality is 2 > -3x.

## What happens if the inequality has an equals sign (≤ or ≥) instead of just < or >?

If the inequality has an equals sign, it means that the two fractions can be equal to each other. In other words, the solution set includes the values that make the inequality true when the fractions are equal. This is represented by a shaded region on the number line.

## Is it possible to have more than one solution to an inequality with fractions?

Yes, it is possible to have multiple solutions to an inequality with fractions. The solution set can include a range of values or multiple discrete values that make the inequality true. For example, in the inequality 2/(x^2-1) < 1/(x+1), the solution set is x < -1 or 0 < x < 1.

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