Rational Inequalities: Problems with Solutions

Solution:
$\frac{5}{x+8}<0$

This inequality is possible only when $x+8<0$
So $x<-8$

The solution interval is $x\in \left( -\infty ,-8\right)$

Solution:
$\frac{10}{2x+5}\geq 0$

This inequality is possible when $2x+5>0$

Then $x>-\frac{5}{2}$

So the solution interval is $x\in \left( -\frac{5}{2},\infty \right)$

Solution:
We must solve when $x^{2}-1<0$
So $\left( x+1\right) \left( x-1\right) <0$

Then we must analyse the following table of intervals

$\begin{array}{cccccc} & \left( -\infty ,-1\right) & -1 & \left( -1,1\right) & 1 & \left( 1,+\infty \right) \\ \left( x+1\right) & \left( -\right) & 0 & \left( +\right) & 2 & \left( +\right) \\ \left( x-1\right) & \left( -\right) & -2 & \left( -\right) & 0 & \left( +\right) \\ \left( x+1\right) \left( x-1\right) & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$

So the solution to $\frac{1}{x^{2}-1}<0$ is $x\in \left( -1,1\right) $

Solution:
$\frac{x-3}{x+2}<0$

The roots are $x=3\qquad x=-2$

The table of intervals

$\begin{array}{cccccc} & \left( -\infty ,-3\right) & -2 & \left( -2,3\right) & 3 & \left( 3,+\infty \right) \\ \left( x+2\right) & \left( -\right) & 0 & \left( +\right) & 5 & \left( +\right) \\ \left( x-3\right) & \left( -\right) & -5 & \left( -\right) & 0 & \left( +\right) \\ \frac{x-3}{x+2} & \left( +\right) & \notin & \left( -\right) & 0 & \left( +\right) \end{array}$

Тhe solution is $x\in \left( -2,3\right)$

Solution:
$\frac{x+5}{x}\geq 0$
The roots are $x=-5\qquad x=0$
and we analyse the following table

$\begin{array}{cccccc} & \left( -\infty ,-5\right) & -5 & \left( -5,0\right) & 0 & \left( 0,+\infty \right) \\ \left( x+5\right) & \left( -\right) & 0 & \left( +\right) & 5 & \left( +\right) \\ x & \left( -\right) & -5 & \left( -\right) & 0 & \left( +\right) \\ \frac{x+5}{x} & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) \end{array}$

The solution is $x\in (-\infty ,-5]\cup \left( 0,+\infty \right) $

Solution:
$\frac{2x+6}{x-3}\leq 0$
The roots are $x=-3\qquad x=3$

Let we investigate the table

$\begin{array}{cccccc} & \left( -\infty ,-3\right) & -3 & \left( -3,3\right) & 3 & \left( 3,+\infty \right) \\ 2x+6 & \left( -\right) & 0 & \left( +\right) & 12 & \left( +\right) \\ x-3 & \left( -\right) & -6 & \left( -\right) & 0 & \left( +\right) \\ \frac{2x+6}{x-3} & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) \end{array}$

The solution is $x\in \lbrack -3,3)$

Solution:
$\frac{3x-1}{x+2}>0$
The roots are $x=\frac{1}{3}\qquad x=-2$

and analyse the solution intervals

$\begin{array}{cccccc} & \left( -\infty ,-2\right) & -2 & \left( -2,\frac{1}{3}\right) & \frac{1}{3} & \left( \frac{1}{3},+\infty \right) \\ 3x-1 & \left( -\right) & -7 & \left( -\right) & 0 & \left( +\right) \\ x+2 & \left( -\right) & 0 & \left( +\right) & \frac{7}{3} & \left( +\right) \\ \frac{3x-1}{x+2} & \left( +\right) & \notin & \left( -\right) & 0 & \left( +\right) \end{array}$

Then the solution is $x\in \left( -\infty ,-2\right) \cup \left( \frac{1}{3},+\infty \right)$

Solution:
First thing to do is to get everything on the left into a single rational expression.
$\frac{x+1}{x-1}+2=\frac{x+1+2x-2}{x-1}>0\Longrightarrow \frac{3x-1}{x-1}>0$

The roots are $x=\frac{1}{3}\qquad x=1$

And now we need to analyse the solution intervals

$\begin{array}{cccccc} & \left( -\infty ,\frac{1}{3}\right) & \frac{1}{3} & \left( \frac{1}{3},1\right) & 1 & \left( 1,+\infty \right) \\ 3x-1 & \left( -\right) & 0 & \left( +\right) & 2 & \left( +\right) \\ x-1 & \left( -\right) & -\frac{2}{3} & \left( -\right) & 0 & \left( +\right) \\ \frac{3x-1}{x-1} & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) \end{array}$

The solution is $x\in \left( -\infty ,\frac{1}{3}\right) \cup \left(1,+\infty \right)$

Solution:
$\frac{x-2}{x+3}\leq 1$

First thing to do is to get a zero on the right side and then get everything on the left into a single rational expression.
$\frac{x-2}{x+3}-1\leq 0$

$\frac{x-2-2x-3}{x+3}\leq 0\Longrightarrow \frac{-x-5}{x+3}\leq 0$
Now we can determine the roots, which are $x=-5\qquad x=-3$
and analyse the solution intervals

$\begin{array}{cccccc} & \left( -\infty ,-5\right) & -5 & \left( -5,-3\right) & -3 & \left( -3,+\infty \right) \\ x+5 & \left( -\right) & 0 & \left( +\right) & 7 & \left( +\right) \\ x+3 & \left( -\right) & -2 & \left( -\right) & 0 & \left( +\right) \\ \frac{x+5}{x+3} & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) \end{array}$

Then the solution to $\frac{x+5}{x+3}\geq 0$ is $x\in (-\infty ,-5]\cup \left( -3,+\infty \right) $

Solution:
First thing to do is to get a zero on the right side and then get everything on the left into a single rational expression.
$\frac{2x-3}{5x+2}\geq -2$
$\frac{2x-3}{5x+2}+2\geq 0$

Then $\frac{2x-3+10x+4}{5x+2}\geq 0\Longrightarrow \frac{12x+1}{5x+2}\geq 0$
and we obtain the roots

$x=-\frac{1}{12}\qquad x=-\frac{2}{5}$

Let we analyse the table

$\begin{array}{cccccc} & \left( -\infty ,-\frac{2}{5}\right) & -\frac{2}{5} & \left( -\frac{2}{5},-\frac{1}{12}\right) & -\frac{1}{12} & \left( -\frac{1}{12},+\infty \right) \\ 12x+1 & \left( -\right) & -\frac{19}{5} & \left( -\right) & 0 & \left( +\right) \\ 5x+2 & \left( -\right) & 0 & \left( +\right) & \frac{19}{12} & \left( +\right) \\ \frac{12x+1}{5x+2} & \left( +\right) & \notin & \left( -\right) & 0 & \left( +\right) \end{array}$

Then the solution is $x\in \left( -\infty ,-\frac{2}{5}\right) \cup \lbrack -\frac{1}{12},+\infty )$

Solution:
$\frac{3x-1}{2x-1}<-4$

$\frac{3x-1}{2x-1}+4<0$

$\frac{3x-1+8x-4}{2x-1}<0$

$\frac{11x-5}{2x-1}<0$

$x=\frac{5}{11}\qquad x=\frac{1}{2}$

$\begin{array}{cccccc} & \left( -\infty ,\frac{5}{11}\right) & \frac{5}{11} & \left( \frac{5}{11},% \frac{1}{2}\right) & \frac{1}{2} & \left( \frac{1}{2},+\infty \right) \\ 11x-5 & \left( -\right) & 0 & \left( +\right) & \frac{1}{2} & \left( +\right) \\ 2x-1 & \left( -\right) & -\frac{1}{11} & \left( -\right) & 0 & \left( +\right) \\ \frac{11x-5}{2x-1} & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) \end{array}$

$x\in \left( \frac{5}{11},\frac{1}{2}\right) $

Solution:
First thing to do is to get a zero on the right side and then get everything on the left into a single rational expression.
$x\leq 3-\frac{6}{x+2}\Longrightarrow x-3+\frac{6}{x+2}\leq 0$

$\frac{\left( x-3\right) \left( x+2\right) +6}{x+2}\leq 0$
So $\frac{x^{2}-x}{x+2}\leq 0$ $\Longrightarrow \frac{x(x-1)}{x+2}\leq 0$

The roots are $x=0, x=1$ and $x=-2$

$\begin{array}{ccccccc} & \left( -\infty ,-2\right) & -2 & \left( -2,0\right) & 0 & \left( 0,1\right) & \left( 1,\infty \right) \\ x & \left( -\right) & -2 & \left( -\right) & 0 & \left( +\right) & \left( +\right) \\ \left( x-1\right) & \left( -\right) & -3 & \left( -\right) & -1 & \left( -\right) & \left( +\right) \\ \left( x+2\right) & \left( -\right) & 0 & (+) & 2 & \left( +\right) & \left( +\right) \\ \frac{x(x-1)}{x+2} & \left( -\right) & \notin & \left( +\right) & 0 & \left( -\right) & \left( +\right) \end{array}$

The solution is $x\in \left( -\infty ,-2\right) \cup \left[ 0,1\right]$

Solution:
The roots are $x=0, x=1$ and $x=-5$

We can analyse the following table of intervals

$\begin{array}{cccccccc} & \left( -\infty ,-5\right) & -5 & \left( -5,0\right) & 0 & \left( 0,1\right) & 1 & \left( 1,\infty \right) \\ x & \left( -\right) & -5 & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) \\ x-1 & \left( -\right) & -6 & \left( -\right) & -1 & \left( -\right) & 0 & \left( +\right) \\ x+5 & \left( -\right) & 0 & \left( +\right) & 5 & \left( +\right) & 6 & \left( +\right) \\ \frac{x\left( x-1\right) }{x+5} & \left( -\right) & \notin & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$

So $\frac{x\left( x-1\right) }{x+5}\geq 0$ when $x\in (-5,0]\cup \lbrack1,\infty)$

Solution:
The roots are $x=0, x=1$ and $ x=-1$ and we analyse

$\begin{array}{cccccccc} & \left( -\infty ,-1\right) & -1 & \left( -1,0\right) & 0 & \left( 0,1\right) & 1 & \left( 1,\infty \right) \\ x & \left( -\right) & -1 & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) \\ 1+x & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) & 2 & \left( +\right) \\ 1-x & \left( +\right) & -2 & \left( +\right) & 1 & \left( +\right) & 0 & \left( -\right) \\ \frac{\left( 1+x\right) \left( 1-x\right) }{x} & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) & 0 & \left( -\right) \end{array}$

The equation $\frac{\left( 1+x\right) \left( 1-x\right) }{x}<0$ is true when $x\in \left(-1,0\right) \cup \left( 1,\infty \right)$

Solution:
First thing to do is to get a zero on the right side and then get everything on the left into a single rational expression.
$\frac{x^{2}-2x+3}{x+1}\leq 1\Longrightarrow \frac{x^{2}-2x+3}{x+1}-1\leq 0$

So, $\frac{x^{2}-2x+3-x-1}{x+1}\leq 0\Longrightarrow \frac{x^{2}-3x+2}{x+1}\leq 0\Longrightarrow \frac{\left( x-1\right) \left( x-2\right) }{x+1}\leq 0$

The roots are $x=-1, x=1$ and $x=2$

$\begin{array}{cccccccc} & \left( -\infty ,-1\right) & -1 & \left( -1,1\right) & 1 & \left( 1,2\right) & 2 & \left( 2,\infty \right) \\ x-1 & \left( -\right) & -2 & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) \\ x-2 & \left( -\right) & -3 & \left( -\right) & -1 & \left( -\right) & 0 & \left( +\right) \\ x+1 & \left( -\right) & 0 & \left( +\right) & 2 & \left( +\right) & 3 & \left( +\right) \\ \frac{\left( x-1\right) \left( x-2\right) }{x+1} & \left( -\right) & \notin & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}$

The inequality $\frac{\left( x-1\right) \left( x-2\right) }{x+1}\leq 0$ is true when $x\in \left( -\infty ,-1\right) \cup \left[ 1,2\right]$

Solution:
$\frac{x}{x^{2}-1}\geq 0$

Тhen $\frac{x}{\left( x+1\right) \left(x-1\right) }\geq 0$

The roots are $x=0, x=-1$ and $x=1$

We need to analyse the following intervals

$\begin{array}{cccccccc} & \left( -\infty ,-1\right) & -1 & \left( -1,0\right) & 0 & \left( 0,1\right) & 1 & \left( 1,\infty \right) \\ x-1 & \left( -\right) & -2 & \left( -\right) & -1 & \left( -\right) & 0 & \left( +\right) \\ x & \left( -\right) & -1 & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) \\ x+1 & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) & 2 & \left( +\right) \\ \frac{x}{\left( x+1\right) \left( x-1\right) } & \left( -\right) & \notin & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) \end{array}$

Then $\frac{x}{\left( x+1\right) \left( x-1\right) }\geq 0$ when $x\in (-1,0]\cup \lbrack 1,\infty )$

Solution:
$\frac{\left( x^{2}-1\right) }{\left( x^{2}-4\right) }\geq 0$
We must find the roots of

$\left( x^{2}-1\right) =\left( x-1\right) \left(x+1\right)$
$\left( x^{2}-4\right) =\left( x-2\right) \left( x+2\right) \Longrightarrow x=-2,-1,1,2$ are the roots.

We need to analyse the table of possible intervals

$\begin{array}{cccccccccc} & \left( -\infty ,-2\right) & -2 & \left( -2,-1\right) & -1 & \left( -1,1\right) & 1 & \left( 1,2\right) & 2 & \left( 2,\infty \right) \\ x-2 & \left( -\right) & -4 & \left( -\right) & -3 & \left( -\right) & -1 & \left( -\right) & 0 & \left( +\right) \\ x-1 & \left( -\right) & -3 & \left( -\right) & -2 & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) \\ x+1 & \left( -\right) & -1 & \left( -\right) & 0 & \left( +\right) & 2 & \left( +\right) & 3 & \left( +\right) \\ x+2 & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) & 3 & \left( +\right) & 4 & \left( +\right) \\ \frac{\left( x^{2}-1\right) }{\left( x^{2}-4\right) } & \left( +\right) & \notin & \left( -\right) & 0 & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) \end{array}$

Then $\frac{\left( x^{2}-1\right) }{\left( x^{2}-4\right) }\geq 0$ when $x\in \left( -\infty ,-2\right) \cup \left[ -1,1\right] \cup \left( 2,\infty\right)$

Solution:
$\frac{2}{x-2}+\frac{3}{x-3}\leq 0$

First thing to do is to get everything on the left into a single rational expression.

$\frac{2}{x-2}+\frac{3}{x-3}=\frac{2\left( x-3\right) +3\left( x-2\right) }{\left( x-2\right) \left( x-3\right) }=\frac{5x-12}{\left( x-2\right) \left( x-3\right) }\leq 0$

The roots are $x=\frac{12}{5}, x=2$ and $x=3$ and now we can find the solution interval, analysing the following table

$\begin{array}{cccccccc} & \left( -\infty ,2\right) & 2 & \left( 2,\frac{12}{5}\right) & \frac{12}{5% } & \left( \frac{12}{5},3\right) & 3 & \left( 3,\infty \right) \\ 5x-12 & \left( -\right) & -2 & \left( -\right) & 0 & \left( +\right) & 3 & \left( +\right) \\ x-2 & \left( -\right) & 0 & \left( +\right) & \frac{2}{5} & \left( +\right) & 1 & \left( +\right) \\ x-3 & \left( -\right) & -1 & \left( -\right) & -\frac{3}{5} & \left( -\right) & 0 & \left( +\right) \\ \frac{5x-12}{\left( x-2\right) \left( x-3\right) } & \left( -\right) & \notin & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) \end{array}$

Then the solution is $x\in \left( -\infty ,2\right) \cup \lbrack \frac{12}{5},3)$

Solution:
After factoring $x^{2}+2x-3=\left( x+3\right) \left( x-1\right)$ and
$x^{2}-1=\left( x-1\right) \left( x+1\right)$

we get 3 roots $x=-3, x=-1$ and $x=1$
So

$\frac{x^{2}+2x-3}{x^{2}-1}=\frac{\left( x+3\right) \left( x-1\right) }{\left( x-1\right) \left( x+1\right) }=\frac{\left( x+3\right) }{\left(x+1\right) }<0$
Then

$\begin{array}{cccccc} & \left( -\infty ,-3\right) & -3 & \left( -3,-1\right) & -1 & \left( -1,+\infty \right) \\ \left( x+3\right) & \left( -\right) & 0 & \left( +\right) & 2 & \left( +\right) \\ \left( x+1\right) & \left( -\right) & -2 & \left( -\right) & 0 & \left( +\right) \\ \frac{\left( x+3\right) }{\left( x+1\right) } & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) \end{array}$

The solution is $x\in \left( -3,-1\right) $