Rational Inequalities: Problems with Solutions

Solution:
[tex]\frac{5}{x+8}<0[/tex]

This inequality is possible only when [tex]x+8<0[/tex]
So [tex]x<-8[/tex]

The solution interval is [tex]x\in \left( -\infty ,-8\right)[/tex]

Solution:
[tex]\frac{10}{2x+5}\geq 0[/tex]

This inequality is possible when [tex]2x+5>0[/tex]

Then [tex]x>-\frac{5}{2}[/tex]

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

Solution:
[tex]\frac{x-3}{x+2}<0[/tex]

The roots are [tex]x=3\qquad x=-2[/tex]

The table of intervals

[tex]\begin{array}{cccccc} & \left( -\infty ,-2\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}[/tex]

The solution is [tex]x\in \left( -2,3\right)[/tex]

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

[tex]\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}[/tex]

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

Solution:
[tex]\frac{3x-1}{x+2}>0[/tex]
The roots are [tex]x=\frac{1}{3}\qquad x=-2[/tex]

and analyse the solution intervals

[tex]\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}[/tex]

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

Solution:
[tex]\frac{2x+6}{x-3}\leq 0[/tex]
The roots are [tex]x=-3\qquad x=3[/tex]

Let us investigate the table

[tex]\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}[/tex]

The solution is [tex]x\in \lbrack -3,3)[/tex]

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

Then we must analyse the following table of intervals

[tex]\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}[/tex]

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

Solution:
Domain: [tex]x\neq 0[/tex].
For [tex]x\neq 0[/tex] we have [tex]x^2\gt 0[/tex], so we can multiply by the positive denominator: [tex]1\gt x^2 \Leftrightarrow -1\lt x\lt 1[/tex].
We must exclude [tex]x=0[/tex].
Answer: [tex]x\in(-1, 0)\cup(0, 1)[/tex].

Solution:
Domain: [tex]x\neq 1[/tex].
For [tex]x\neq 1[/tex]: [tex]\frac{x-1}{(x-1)^3}=\frac{1}{(x-1)^2}\gt 0[/tex].
Answer: [tex]x\in(-\infty, 1)\cup(1, +\infty)[/tex].

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

The roots are [tex]x=\frac{1}{3}\qquad x=1[/tex]

And now we need to analyse the solution intervals

[tex]\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}[/tex]

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

Solution:
[tex]\frac{x-2}{x+3}\leq 1[/tex]

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.
[tex]\frac{x-2}{x+3}-1\leq 0[/tex]

[tex]\frac{x-2-\left( x+3\right) }{x+3}\leq 0\Longrightarrow \frac{-5}{x+3}\leq 0[/tex]

The numerator is the constant [tex]-5[/tex], so the sign of the fraction depends only on the denominator, and [tex]x=-3[/tex] must be excluded.

[tex]\begin{array}{cccc} & \left( -\infty ,-3\right) & -3 & \left( -3,+\infty \right) \\ -5 & \left( -\right) & -5 & \left( -\right) \\ x+3 & \left( -\right) & 0 & \left( +\right) \\ \frac{-5}{x+3} & \left( +\right) & \notin & \left( -\right) \end{array}[/tex]

The fraction is negative only when [tex]x+3>0[/tex]

The solution is [tex]x\in \left( -3,+\infty \right)[/tex]

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.
[tex]\frac{2x-3}{5x+2}\geq -2[/tex]
[tex]\frac{2x-3}{5x+2}+2\geq 0[/tex]

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

[tex]x=-\frac{1}{12}\qquad x=-\frac{2}{5}[/tex]

Let us analyse the table

[tex]\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}[/tex]

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

Solution:
[tex]\frac{3x-1}{2x-1}<-4[/tex]

[tex]\frac{3x-1}{2x-1}+4<0[/tex]

[tex]\frac{3x-1+8x-4}{2x-1}<0[/tex]

[tex]\frac{11x-5}{2x-1}<0[/tex]

[tex]x=\frac{5}{11}\qquad x=\frac{1}{2}[/tex]

[tex]\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}[/tex]

[tex]x\in \left( \frac{5}{11},\frac{1}{2}\right) [/tex]

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.
[tex]x\leq 3-\frac{6}{x+2}\Longrightarrow x-3+\frac{6}{x+2}\leq 0[/tex]

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

The roots are [tex]x=0, x=1[/tex] and [tex]x=-2[/tex]

[tex]\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}[/tex]

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

Solution:
The roots are [tex]x=0, x=1[/tex] and [tex]x=-5[/tex]

We can analyse the following table of intervals

[tex]\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}[/tex]

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

Solution:
The roots are [tex]x=0, x=1[/tex] and [tex] x=-1[/tex] and we analyse

[tex]\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}[/tex]

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

Solution:
[tex]\frac{x}{3}-\frac{4}{x}=\frac{x^2-12}{3x}=\frac{\left(x-2\sqrt{3}\right)\left(x+2\sqrt{3}\right)}{3x}\gt 0[/tex]
Zeros of the numerator and the denominator: [tex]-2\sqrt{3}, 0, 2\sqrt{3}[/tex]. Signs from left to right: [tex]-, +, -, +[/tex].
Answer: [tex]x\in\left(-2\sqrt{3}, 0\right)\cup\left(2\sqrt{3}, +\infty\right)[/tex].
Do not multiply both sides by [tex]x[/tex]: its sign is unknown.

Solution:
Multiply by [tex]-1[/tex] to turn [tex]3-x[/tex] into [tex]x-3[/tex]; the inequality sign is reversed: [tex](x-2)(x-3)(x+1)\geq 0[/tex]
Zeros of the factors: [tex]-1, 2, 3[/tex]. Signs from left to right: [tex]-, +, -, +[/tex].
[tex]x\in[-1, 2]\cup[3, +\infty)[/tex]
Keeping only [tex]x\geq 0[/tex]: [tex]x\in[0, 2]\cup[3, +\infty)[/tex].

Solution:
[tex]x^3+1=(x+1)\left(x^2-x+1\right)[/tex]
[tex]x^2-x+1\gt 0[/tex] for every [tex]x[/tex] (its discriminant is negative and the leading coefficient is positive), so it does not affect the sign.
[tex]x+1\leq 0 \Rightarrow x\leq -1[/tex]
Answer: [tex]x\in(-\infty, -1][/tex]. The bracket at [tex]-1[/tex] is square because the inequality is non-strict and [tex]x=-1[/tex] is a solution.

Solution:
[tex]x^4+x^6=x^4\left(1+x^2\right)[/tex]
Both factors are non-negative: [tex]x^4\geq 0[/tex] and [tex]1+x^2\gt 0[/tex]. Their product is never negative, so it can only be equal to [tex]0[/tex].
Therefore [tex]x^4\left(1+x^2\right)\leq 0[/tex] only for [tex]x=0[/tex].

Solution:
[tex](x-1)^2(x-2)(x+2)\lt 0[/tex]
The factor [tex](x-1)^2[/tex] is positive for [tex]x\neq 1[/tex] and equals [tex]0[/tex] at [tex]x=1[/tex]. The inequality is strict, so [tex]x=1[/tex] is not a solution, and for [tex]x\neq 1[/tex] we need:
[tex](x-2)(x+2)\lt 0 \Rightarrow -2\lt x\lt 2,\ x\neq 1[/tex]
Answer: [tex]x\in(-2, 1)\cup(1, 2)[/tex].

Solution:
[tex]\frac{x}{x^{2}-1}\geq 0[/tex]

Then [tex]\frac{x}{\left( x+1\right) \left(x-1\right) }\geq 0[/tex]

The roots are [tex]x=0, x=-1[/tex] and [tex]x=1[/tex]

We need to analyse the following intervals:

[tex]\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}[/tex]

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

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

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

We need to analyse the table of possible intervals

[tex]\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}[/tex]

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

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

we get 3 roots [tex]x=-3, x=-1[/tex] and [tex]x=1[/tex]
So

[tex]\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[/tex]
Then

[tex]\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}[/tex]

The solution is [tex]x\in \left( -3,-1\right) [/tex]

Solution:
Domain: [tex]x\neq\pm 3[/tex].
[tex]\frac{x+3}{(3-x)(3+x)}=\frac{1}{3-x}[/tex] (for [tex]x\neq -3[/tex])
[tex]\frac{1}{3-x}\geq 0 \Leftrightarrow 3-x\gt 0 \Leftrightarrow x\lt 3[/tex]
After cancelling we must not forget that [tex]x=-3[/tex] is not in the domain.
Answer: [tex]x\in(-\infty, -3)\cup(-3, 3)[/tex].

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.
[tex]\frac{x^{2}-2x+3}{x+1}\leq 1\Longrightarrow \frac{x^{2}-2x+3}{x+1}-1\leq 0[/tex]

So, [tex]\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[/tex]

The roots are [tex]x=-1, x=1[/tex] and [tex]x=2[/tex]

[tex]\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}[/tex]

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

Solution:
[tex]\frac{2}{x-2}+\frac{3}{x-3}\leq 0[/tex]

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

[tex]\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[/tex]

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

[tex]\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}[/tex]

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

Solution:
[tex]\frac{x^2}{x+2}-(x-3)=\frac{x^2-(x-3)(x+2)}{x+2}=\frac{x+6}{x+2}\lt 0[/tex]
[tex]-6\lt x\lt -2[/tex]
The integer solutions are [tex]-5, -4, -3[/tex], i.e. there are [tex]3[/tex] of them.