Inequalities, Intervals: Problems with Solutions

Solution:
Solution $x\in (-5,3]$

It means that -5 is not a solution but 3 is a solution to the inequality.
Graphical solution:

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] $

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 $x\in \left( \frac{-5}{2},+\infty \right)$

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 $x\in \left(-\infty ,5\right)$

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 $x\in (-\infty ,4]$

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
$x\in (-\infty ,\frac{18}{5}]$

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 $x\in (-\infty ,\frac{5}{4})$

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 $x\in (-\infty ,0]$

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 $x\in (-\infty ,\frac{5}{2}]$

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 $x\in \lbrack -\frac{1}{6},+\infty )$

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 $x\in (-2,7]$

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) $

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) $

Solution:
$3 < x-4\leq 10$

$3+4 < x-4+4\leq 10+4\Longrightarrow 7 < x\leq 14$

So the solution interval is $x\in (7,14]$

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 $x\in (0,6]$

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 $x\in \lbrack 1,7]$

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 $x\in \lbrack -5,13]$

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 $x\in \lbrack 3,4]$

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: $x\in \lbrack 1-\sqrt{2},+\infty )\cup (-\infty ,-1-\sqrt{2}]$

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 $x\in \lbrack -\frac{7}{3},3]$

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 $x\in (-\infty ,-\frac{12}{5}]\cup \lbrack \frac{18}{5},+\infty )$