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Practice
Rational Inequalities
Rational Inequalities: Problems with Solutions
By Prof. Hernando Guzman Jaimes (University of Zulia - Maracaibo, Venezuela)
Problem 1
Solve the following inequality:
$\frac{5}{x+8}<0$
$x\in \left( -\infty ,5\right)$
$x\in \left( -8,5\right)$
$x\in \left( -5,8\right)$
$x\in \left( -\infty ,-8\right)$
Solution:
$\frac{5}{x+8}<0$
This inequality is possible only when $x+8<0$
So $x<-8$
The solution interval is $x\in \left( -\infty ,-8\right)$
Problem 2
Solve the following inequality
$\frac{10}{2x+5}\geq 0$
$x\in \left( -\frac{5}{2},\infty \right)$
$x\in \left( \frac{5}{2},\infty \right)$
$x\in \left( -\infty ,-\frac{5}{2}\right)$
$x\in \left( -\infty ,\frac{5}{2}\right)$
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 $x\in \left( -\frac{5}{2},\infty \right)$
Problem 3
Solve for x
$\frac{1}{x^{2}-1}<0$
$x\in \left( 0,1\right)$
$x\in \left( -1,1\right)$
$x\in \left( -1,0\right) $
$x\in \left( -\infty ,\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) $
Problem 4
Solve:
$\frac{x-3}{x+2}<0$
$x\in \left( -3,2\right)$
$x\in \left[ -2,3\right]$
$x\in \left( -2,3\right)$
$x\in \left( -\infty ,3\right)$
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 $x\in \left( -2,3\right)$
Problem 5
$\frac{x+5}{x}\geq 0$
$x\in (-5,0)\cup \left( 5,+\infty \right)$
$x\in (-\infty ,-5]\cup \left( 0,+\infty \right) $
$x\in \left( 0,+\infty \right) $
$x\in (-\infty ,-5]$
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) $
Problem 6
Determine the intervals, in which the inequality is true:
$\frac{2x+6}{x-3}\leq 0$
$x\in \lbrack -3,\infty )$
$x\in \lbrack -\infty ,3)$
$x\in \lbrack -3,3)$
$x\in \lbrack -2,6)$
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 $x\in \lbrack -3,3)$
Problem 7
$\frac{3x-1}{x+2}>0$
$x\in \left( -\infty ,-2\right) \cup \left( \frac{1}{3},+\infty \right) $
$x\in \left( -\infty ,-2\right) $
$x\in \left( \frac{1}{3},+\infty \right) $
$x\in \left( -2,\frac{1}{3}\right) \cup \left( 2,\infty \right) $
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 $x\in \left( -\infty ,-2\right) \cup \left( \frac{1}{3},+\infty \right)$
Problem 8
$\frac{x+1}{x-1}+2>0$
$x\in \left( -\infty ,\frac{1}{3}\right) \cup \left( 1,+\infty \right) $
$x\in \left( -\infty ,1\right) \cup \left( 3,+\infty \right) $
$x\in \left( -\infty ,0\right) \cup \left( \frac{1}{3},1\right) $
None of these.
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 $x\in \left( -\infty ,\frac{1}{3}\right) \cup \left(1,+\infty \right)$
Problem 9
$\frac{x-2}{x+3}\leq 1$
$x\in \left( -3,+\infty \right) $
$x\in (-5,-3]$
$x\in (-\infty ,-5]$
None of these.
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) $
Problem 10
$\frac{2x-3}{5x+2}\geq -2$
$x\in \left( -\infty ,\frac{1}{12}\right) \cup \lbrack \frac{2}{5},+\infty )$
$x\in \left( -\infty ,-\frac{2}{5}\right) $
$x\in \left( -\infty ,-\frac{2}{5}\right) \cup \lbrack -\frac{1}{12},+\infty )$
$x\in \lbrack -\frac{1}{12},+\infty )$
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 $x\in \left( -\infty ,-\frac{2}{5}\right) \cup \lbrack -\frac{1}{12},+\infty )$
Problem 11
$\frac{3x-1}{2x-1}<-4$
$x\in \left( -\infty ,\frac{1}{2}\right) $
$x\in \left( \frac{5}{11},\infty \right) $
$x\in \left( -4,1\right) $
$x\in \left( \frac{5}{11},\frac{1}{2}\right) $
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) $
Problem 12
$x\leq 3-\frac{6}{x+2}$
$x\in \left( -\infty ,-1\right) \cup \left[ 0,2\right] $
$x\in \left( -\infty ,-2\right) \cup \left[ 0,1\right] $
$x\in \left( -\infty ,-2\right) \cup \left[ 0,6\right] $
None of these.
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 $x\in \left( -\infty ,-2\right) \cup \left[ 0,1\right]$
Problem 13
$\frac{x\left(x-1\right) }{x+5}\geq 0$
$x\in (-\infty ,0]\cup \lbrack 1,\infty )$
$x\in (-\infty ,0]\cup \lbrack 5,\infty )$
$x\in (-5,0]\cup \lbrack 1,\infty )$
$x\in (-5,0]\cup \lbrack 5,\infty)$
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 $x\in (-5,0]\cup \lbrack1,\infty)$
Problem 14
$\frac{\left( 1+x\right) \left( 1-x\right) }{x}<0$
$x\in \left( -1,0\right) \cup \left( 0,\infty \right)$
$x\in \left( -\infty ,0\right) \cup \left( 1,\infty \right)$
$x\in \left( -1,1\right)$
$x\in \left( -1,0\right) \cup \left( 1,\infty \right)$
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 $x\in \left(-1,0\right) \cup \left( 1,\infty \right)$
Problem 15
$\frac{x^{2}-2x+3}{x+1}\leq 1$
$x\in \left( -\infty ,-1\right) \cup \left[ 1,2\right] $
$x\in \left( -\infty ,-1\right) $
$x\in \left[ 1,2\right] $
None of these.
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 $x\in \left( -\infty ,-1\right) \cup \left[ 1,2\right]$
Problem 16
$\frac{x}{x^{2}-1}\geq 0$
$x\in (-1,0]$
$x\in (-1,0]\cup \lbrack 1,\infty )$
$x\in \lbrack 1,\infty )$
None of these.
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 $x\in (-1,0]\cup \lbrack 1,\infty )$
Problem 17
$\frac{\left(x^{2}-1\right) }{\left( x^{2}-4\right) }\geq 0$
$x\in \left( -\infty ,-4\right) \cup \left[ -1,1\right] \cup \left( 4,\infty \right) $
$x\in \left( -\infty ,-2\right) \cup \left[ -1,1\right] \cup \left( 2,\infty \right) $
$x\in \left( -\infty ,-1\right) \cup \left( 1,\infty \right) $
None of these.
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 $x\in \left( -\infty ,-2\right) \cup \left[ -1,1\right] \cup \left( 2,\infty\right)$
Problem 18
$\frac{2}{x-2}+\frac{3}{x-3}\leq 0$
$x\in \left( -\infty ,2\right) \cup \lbrack \frac{12}{5},3)$
$x\in \left( -\infty ,-3\right) \cup \lbrack -2,\infty )$
$x\in \left( -\infty ,-2\right) \cup \lbrack 0,3)$
$x\in \left( -\infty ,-2\right) \cup \lbrack 3,\infty )$
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 $x\in \left( -\infty ,2\right) \cup \lbrack \frac{12}{5},3)$
Problem 19
$\frac{x^{2}+2x-3}{x^{2}-1}<0$
$x\in \left( -1,3\right) $
$x\in \left( -3,0\right) \cup \left( 1,2\right) $
$x\in \left( -\infty ,-\frac{1}{3}\right) $
$x\in \left( -3,-1\right) $
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) $
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