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

 

[Wilmer]

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

Solution: x = 3 or x = -1

Now throw the < in there!

 

[MarkFL]

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?

 

[Yankel]

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.

 

[MarkFL]

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?

 

[Olinguito]

\(\displaystyle x^2-1\) is negative on \((-1,1)\), so you do need to worry about the sign.

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

 

[MarkFL]

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.

 

[Wilmer]

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

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

 

[MarkFL]

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

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:

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

 

[I like Serena]

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