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Practice
Inequalities, Intervals
Inequalities, Intervals: Problems with Solutions
By
Prof. Hernando Guzman Jaimes (University of Zulia - Maracaibo, Venezuela)
Problem 1
Which the following intervals is the solution to
$-5 < x \leq 3$
$x\in \lbrack -5,3]$
$x\in (-5,3)$
$x\in (-5,3]$
$x\in \lbrack -5,3)$
Solution:
Solution $x\in (-5,3]$
It means that -5 is not a solution but 3 is a solution to the inequality.
Graphical solution:
Problem 2
Which of the following set of numbers belongs to the interval
$\left[ 7,9\right]$
$\left\{ 0,\frac{25}{3},8,\frac{41}{5},10\right\} $
$\left\{ 7,\frac{25}{3},8,\frac{41}{5},9\right\} $
$\left\{ 0,\frac{25}{3},8,\frac{41}{5},9\right\} $
$\left\{ 7,\frac{25}{3},8,\frac{41}{5},10\right\} $
Solution:
Answer: $\left\{ 7,\frac{25}{3},8,\frac{41}{5},9\right\}$
Since
$7\in \left[ 7,9\right]$
$\frac{25}{3}=8.33\in \left[ 7,9\right]$
$8\in \left[ 7,9\right]$
$\frac{41}{5}=8.2\in \left[ 7,9\right]$
$9\in \left[ 7,9\right] $
Problem 3
Solve the inequality
$2x+3>-2$
$x\in \left( \frac{-5}{2},+\infty \right)$
$x\in \left( -2,+\infty \right)$
$x\in \left( 3,+\infty \right)$
$x\in \left( \frac{3}{2},+\infty \right)$
Solution:
Answer: $x\in \left( \frac{-5}{2},+\infty \right)$
$2x+3>-2$
Now we use inequality properties to isolate the variable $x$ on the left side.
$2x+3-3>-2-3$
$2x>-5$
$\left( \frac{1}{2}\right)2x>-5\left( \frac{1}{2}\right)$
so
$x>\frac{-5}{2}$, then the solution interval is $x\in \left( \frac{-5}{2},+\infty \right)$
Problem 4
What is the solution to the inequality?
$3x-9<6$
$x\in \left( -\infty ,6\right)$
$x\in \left( -\infty ,-9\right)$
$x\in \left( -\infty ,5\right)$
$x\in \left( -\infty ,\frac{6}{9}\right)$
Solution:
Answer: $x\in \left( -\infty ,5\right)$
$3x-9<6$
We have to isolate the variable $x$ on the left side.
$3x-9+9<6+9$
$3x<15$
$\left( \frac{1}{3}\right) 3x<\left( \frac{1}{3}\right) 15$
$x<5$
In interval notation, the solution is $x\in \left(-\infty ,5\right)$
Problem 5
Write the solution in interval notation and graph it on the real number line.
$\frac{3}{2}x+4\leq 10$
$x\in (-\infty ,10]$
$x\in (-\infty ,4]$
$x\in (-\infty ,\frac{3}{2}]$
$x\in (4,+\infty)$
Solution:
$\frac{3}{2}x+4\leq 10$
Using inequality properties we have to isolate the variable $x$ on the left side.
$\frac{3}{2}x+4-4\leq 10-4$
$\frac{3}{2}x\leq 6$
$\left( \frac{2}{3}\right) \frac{3}{2}x\leq \left( \frac{2}{3}\right)6$
$x\leq 4$
So the solution is $x\in (-\infty ,4]$
Problem 6
Write the solution in interval notation and graph it on the real number line.
$5-\frac{5}{2}x\geq -4$
$x\in (-\infty ,-4]$
$x\in (-\infty ,\frac{18}{5}]$
$x\in (-\infty ,5]$
$x\in (-\infty ,\frac{5}{2}]$
Solution:
$5-\frac{5}{2}x\geq -4$
We have to isolate the variable $x$ on the left side.
$5-\frac{5}{2}x\geq -4$
$5-\frac{5}{2}x-5\geq -4-5$
$-\frac{5}{2}x\geq -9$ so
$\left( -\frac{2}{5}\right) \left( -\frac{5}{2}\right) x\leq -9\left( -\frac{2}{5}\right) \Longrightarrow x\leq \frac{18}{5}$
The solution interval is
$x\in (-\infty ,\frac{18}{5}]$
Problem 7
Write the solution in interval notation and graph it on the real number line.
$\frac{5}{2}-x>x$
$x\in (-\infty ,-1)$
$x\in (-\infty ,\frac{5}{4})$
$x\in (-\infty ,1)$
$x\in (-\infty ,2)$
Solution:
$\frac{5}{2}-x>x$
We have to isolate the variable $x$ on the left side
$\frac{5}{2}>x+x$
$\frac{5}{2}>2x$
$\left( \frac{1}{2}\right) \frac{5}{2}>\left( \frac{1}{2}\right) 2x$
$\frac{5}{4}>x$
$x<\frac{5}{4}$
So the solution interval is $x\in (-\infty ,\frac{5}{4})$
Problem 8
Select the solution in interval notation and graph it on the real number line.
$-\left( 1-x\right) \geq 2x-1$
$x\in (-\infty ,0]$
$x\in (-\infty ,-1]$
$x\in (-\infty ,2]$
$x\in (-\infty ,1]$
Solution:
$-\left( 1-x\right) \geq 2x-1$
We have to isolate the variable $x$ on the left side
$-1+x\geq 2x-1$
$1-1+x\geq 2x-1+1$
$x\geq 2x$
then
$0\geq 2x-x\Longrightarrow 0\geq x\Longrightarrow x\leq 0$
The solution interval is $x\in (-\infty ,0]$
Problem 9
$2+x\geq 3\left( x-1\right)$
$x\in (-\infty ,3]$
$x\in (-\infty ,2]$
$x\in (-\infty ,\frac{3}{2}]$
$x\in (-\infty ,\frac{5}{2}]$
Solution:
$2+x\geq 3\left( x-1\right) $
We have to isolate the variable $x$ on the left side.
$2+x\geq 3x-3$
$3+2+x\geq 3x-3+3$
$5+x\geq 3x$
So $5\geq 3x-x$
$5\geq 2x$
$\left( \frac{1}{2}\right) 5\geq \left( \frac{1}{2}\right) 2x$
$\frac{5}{2}\geq x$
So $x\leq \frac{5}{2}$
The solution is $x\in (-\infty ,\frac{5}{2}]$
Problem 10
Solve the inequality:
$-7x+3\leq 4-x$
$x\in \lbrack 3,+\infty )$
$x\in \lbrack -7,+\infty )$
$x\in \lbrack -4,+\infty )$
$x\in \lbrack -\frac{1}{6},+\infty )$
Solution:
$-7x+3\leq 4-x$
$-7x+3-3\leq 4-x-3$
$-7x\leq 1-x$
$-7x+x\leq 1$
$-6x\leq 1$
$\left( -\frac{1}{6}\right) \left( -6x\right) \geq 1\left( -\frac{1}{6}\right) $
so $x\geq -\frac{1}{6}$
or $x\in \lbrack -\frac{1}{6},+\infty )$
Problem 11
$-7\leq -x<2$
$x\in (-2,7]$
$x\in (-7,2]$
$x\in \lbrack -2,7]$
$x\in (2,7)$
Solution:
$-7\leq -x<2$
Using inequality properties we have to isolate the variable $x$ on the left side.
$-7(-1)\geq -x(-1)>2(-1)\Longrightarrow 7\geq x>-2$ so
$-2< x\leq 7$
Ths solution is $x\in (-2,7]$
Problem 12
Solve and graph the compound inequality.
$-\frac{20}{3}< \frac{2}{3}x < x$
$x\in \left( \frac{2}{3},+\infty \right) $
$x\in \left( -10,+\infty \right) $
$x\in \left( 0,+\infty \right) $
$x\in \left( -\frac{20}{3},\frac{2}{3}\right) $
Solution:
$-\frac{20}{3}<\frac{2}{3}x < x$
We have to solve two inequalities and make the intersection of their solution intervals.
$\left\{ \begin{array}{c} -\frac{20}{3}< \frac{2}{3}x \\ \frac{2}{3}x < x \end{array} \right\} \Longrightarrow \left\{ \begin{array}{c} \frac{2}{3}x > -\frac{20}{3} \\ \frac{2}{3}x-x < 0 \end{array} \right\} \Longrightarrow \left\{ \begin{array}{c} \left( \frac{3}{2}\right) \frac{2}{3}x > -\frac{20}{3}\left( \frac{3}{2}\right) \\ -\frac{1}{3}x < 0 \end{array} \right\} $
Then $\left\{ \begin{array}{c} x>-10 \\ \left( -3\right) \left( -\frac{1}{3}\right) x>0 \end{array} \right\} \Longrightarrow \left\{ \begin{array}{c} x>-10 \\ x>0 \end{array} \right\} $
The solution is
$x\in \left( -10,+\infty \right) \cap \left( 0,+\infty \right) \Longrightarrow x\in \left( 0,+\infty \right) $
Problem 13
Solve and graph the compound inequality.
$-7 < x-2 < 1$
$x\in \left( -5,3\right) $
$x\in \left( -3,5\right) $
$x\in \left[ -5,3\right] $
$x\in \left[ -3,5\right] $
Solution:
$-7 < x-2 < 1 \Longrightarrow -7+2 < x-2+2 < 1+2$
So
$-5 < x < 3$
The solution is $x\in \left( -5,3\right) $
Problem 14
Solve and graph the compound inequality.
$3 < x-4\leq 10$
$x\in (4,10]$
$x\in (7,14]$
$x\in (-4,14]$
$x\in (7,10]$
Solution:
$3 < x-4\leq 10$
$3+4 < x-4+4\leq 10+4\Longrightarrow 7 < x\leq 14$
So the solution interval is $x\in (7,14]$
Problem 15
Solve and graph the compound inequality.
$-1 < \frac{x-4}{4}\leq \frac{1}{2}$
$x\in (0,6]$
$x\in (\frac{1}{4},\frac{1}{2}]$
$x\in (-1,6]$
$x\in (\frac{1}{4},6]$
Solution:
$-1<\frac{x-4}{4}\leq \frac{1}{2}$
$-1\left( 4\right) < \left( 4\right) \frac{x-4}{4}\leq \frac{1}{2}\left(4\right)$
$\Longrightarrow -4< x-4\leq 2$
$\Longrightarrow -4+4< x-4+4\leq 2+4$
So $0< x\leq 6$
In interval notation, the solution is $x\in (0,6]$
Problem 16
Solve and graph the compound inequality.
$2\leq \frac{4x+2}{3}\leq 10$
$x\in \lbrack 2,7]$
$x\in \lbrack 1,10]$
$x\in \lbrack 2,10]$
$x\in \lbrack 1,7]$
Solution:
$2\leq \frac{4x+2}{3}\leq 10$
$2\left( 3\right) \leq \left( 3\right) \frac{4x+2}{3}\leq 10\left( 3\right)$
$\Longrightarrow 6\leq 4x+2\leq 30$ so
$6-2\leq 4x+2-2\leq 30-2\Longrightarrow 4\leq 4x\leq 28$
$\Longrightarrow \left( \frac{1}{4}\right) 4\leq \left( \frac{1}{4}\right) 4x\leq 28\left( \frac{1}{4}\right)$
Answer: $1\leq x\leq 7$
In interval notation, the solution is $x\in \lbrack 1,7]$
Problem 17
Solve the absolute-value inequality:
$\left\vert x-4\right\vert \leq 9$
$x\in \lbrack -5,9]$
$x\in \lbrack -5,13]$
$x\in \lbrack 9,13]$
$x\in \lbrack -4,9]$
Solution:
Here we apply the property:
$\left\vert A\right\vert \leq B\Longrightarrow -B\leq A\leq B$
$\left\vert x-4\right\vert \leq 9\Longrightarrow -9\leq x-4\leq 9$
$-9+4\leq x-4+4\leq 9+4\Longrightarrow -5\leq x\leq 13$
The solution interval is $x\in \lbrack -5,13]$
Problem 18
Solve the absolute-value inequality:
$\left\vert 2x-7\right\vert \leq 1$
$x\in \lbrack 3,4]$
$x\in \lbrack 1,4]$
$x\in \lbrack 3,7]$
$x\in \lbrack 2,7]$
Solution:
Here we apply the property:
$\left\vert A\right\vert \leq B\Longrightarrow -B\leq A\leq B$
$\left\vert 2x-7\right\vert \leq 1\Longrightarrow -1\leq 2x-7\leq 1$
$-1+7\leq 2x-7+7\leq 1+7\Longrightarrow 6\leq 2x\leq 8$
Then $\left( \frac{1}{2}\right) 6\leq \left( \frac{1}{2}\right) 2x\leq 8\left( \frac{1}{2}\right) \Longrightarrow 3\leq x\leq 4$
The solution interval is $x\in \lbrack 3,4]$
Problem 19
Solve the absolute-value inequality:
$\left\vert x+\sqrt{2}\right\vert \geq 1$
$x\in (-\infty ,-\sqrt{2}] \cup \lbrack \sqrt{2},+\infty )$
$x\in \lbrack -\sqrt{2},\sqrt{2})$
$x\in (-\infty ,-1-\sqrt{2}] \cup \lbrack 1-\sqrt{2},+\infty )$
$x\in \lbrack -1-\sqrt{2},1-\sqrt{2})$
Solution:
We can appy the following property:
$\left\vert A\right\vert \geq B\Longrightarrow \left\{ \begin{array}{c} A\geq B \\ \text{or} \\ -A\geq B \end{array} \right\}$
We have to find the union of the intervals.
$\left\vert x+\sqrt{2}\right\vert \geq 1\Longrightarrow \left\{ \begin{array}{c} x+\sqrt{2}\geq 1 \\ \text{or} \\ -\left( x+\sqrt{2}\right) \geq 1 \end{array} \right\} \Longrightarrow \left\{ \begin{array}{c} x\geq 1-\sqrt{2} \\ \text{or} \\ -x-\sqrt{2}\geq 1 \end{array} \right\} $
then $\left\{ \begin{array}{c} x\geq 1-\sqrt{2} \\ \text{or} \\ -1-\sqrt{2}\geq x \end{array} \right\} \Longrightarrow \left\{ \begin{array}{c} x\geq 1-\sqrt{2} \\ \text{or} \\ x\leq -1-\sqrt{2} \end{array} \right\} $
Answer: $x\in \lbrack 1-\sqrt{2},+\infty )\cup (-\infty ,-1-\sqrt{2}]$
Problem 20
Solve the absolute-value inequality:
$\left\vert \frac{3x-1}{-4}\right\vert <2$
$x\in \lbrack -4,3]$
$x\in \lbrack -2,3]$
$x\in \lbrack -1,\frac{3}{2}]$
$x\in \lbrack -\frac{7}{3},3]$
Solution:
Here we have to apply the property
$\left\vert A\right\vert \leq B\Longrightarrow -B\leq A\leq B$
$\left\vert \frac{3x-1}{-4}\right\vert <2\Longrightarrow -2\leq \frac{3x-1}{-4}\leq 2$
$-2(-4)\geq (-4)\frac{3x-1}{-4}\geq 2(-4)$
$\Longrightarrow 8\geq 3x-1\geq -8$
$\Longrightarrow -8\leq 3x-1\leq 8$
$\Longrightarrow -8+1\leq 3x-1+1\leq 8+1$
$\Longrightarrow -7\leq 3x\leq 9$
So $-7\left( \frac{1}{3}\right) \leq \left( \frac{1}{3}\right) 3x\leq 9\left( \frac{1}{3}\right)$
$\Longrightarrow -\frac{7}{3}\leq x\leq 3$
The solution $x\in \lbrack -\frac{7}{3},3]$
Problem 21
Solve the absolute-value inequality:
$\left\vert \frac{3-5x}{3}\right\vert \geq 5$
$x\in (-\infty ,-\frac{12}{5}]\cup \lbrack \frac{18}{5},+\infty )$
$x\in (-\infty ,3]\cup \lbrack 5,+\infty )$
$x\in (-\frac{12}{5},\frac{18}{5}]$
$x\in (-\infty ,-\infty )$
Solution:
We have to apply the following property:
$\left\vert A\right\vert \geq B\Longrightarrow \left\{ \begin{array}{c} A\geq B \\ \text{or} \\ -A\geq B \end{array} \right\} $
and find the union of both intervals.
$\left\vert \frac{3-5x}{3}\right\vert \geq 5\Longrightarrow \left\{ \begin{array}{c} \frac{3-5x}{3}\geq 5 \\ \text{or} \\ -\left( \frac{3-5x}{3}\right) \geq 5 \end{array} \right\}$
$ \left\{ \begin{array}{c} 3-5x\geq 15 \\ \text{or} \\ -3+5x\geq 15 \end{array} \right\} $
$\left\{ \begin{array}{c} 3-5x\geq 15 \\ \text{or} \\ -3+5x\geq 15 \end{array} \right\}$
$\left\{ \begin{array}{c} -3+3-5x\geq 15-3 \\ \text{or} \\ 3-3+5x\geq 15+3 \end{array} \right\} $
$\left\{ \begin{array}{c} -5x\geq 12 \\ \text{or} \\ 5x\geq 18 \end{array} \right\}$
$\left\{ \begin{array}{c} x\leq -\frac{12}{5} \\ \text{or} \\ x\geq \frac{18}{5} \end{array} \right\} $
So $x\in (-\infty ,-\frac{12}{5}]\cup \lbrack \frac{18}{5},+\infty )$
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