Solve the following inequalities.

$color{#4257b2}text{(a)} 2x-1le13$

Add both of side by $(1)$ as follows:

$$
2xle13+1 2xle14
$$

Divide both of sides by $(2)$ as follows:

$$
xle7
$$

$color{#4257b2}text{(b)} -2x-1>-1$

Add both of side by $(1)$ as follows:

$$
-2x>-1+1 -2x>0
$$

Divide both of sides by $(-2)$ as follows:

$$
x<0
$$

$color{#4257b2}text{(c)} 2x-8>4x+12$

Isolate the variables on the left side as follows:

$$
2x-4x>12+8 -2x>20
$$

Divide both of side by $(-2)$ as follows:

$$
x<-10
$$

$color{#4257b2}text{(d)} 5(x-3)ge2x$

Use distributive property as follows:

$$
5x-15ge2x
$$

Isolate the variables on the left side as follows:

$$
5x-2xge15 3xge15
$$

Divide both of side by $(3)$ as follows:

$$
xge5
$$

$color{#4257b2}text{(e)} -4(5-3x)<2(3x+8)$

Use distributive property as follows:

$$
-20+12x<6x+16
$$

Isolate the variables on the left side as follows:

$$
12x-6x<16+20 6x<36
$$

Divide both of side by $(6)$ as follows:

$$
x<6
$$

$color{#4257b2}text{(f)} dfrac{x-2}{3}le2x-3$

Use distributive property as follows:

$$
x-2le3(2x-3) x-2le6x-9
$$

Isolate the variables on the left side as follows:

$$
x-6xle-9+2 -5xle-7
$$

Divide both of side by $(-5)$ as follows:

$$
xgedfrac{7}{5} xge1.4
$$

$$
text{color{Brown}(a) $ xleq 7$ (b) $x< 0$ (c) $x<-10$ \

(d) $ xgeq 5$ (e) $x < 6$ (f) $x geq dfrac{7}{5}$}
$$

To determine if $0$ is in the solution set of the given inequalities, we can substitute $x=0$ in each example and check if the inequality still holds.

*(a)* Substituting $x=0$ in this inequality, it yields:
$$
begin{align*}
3cdot0&le 4cdot0+1\
0&le 1.
end{align*}$$
Since $0$ is less than $1$, the inequality holds and $0$ is in the solution set of this inequality.

*(b)* Substituting $x=0$ in this inequality, it yields:
$$
begin{align*}
-6cdot0&<0+ 4<12\
0&< 4<12.
end{align*}$$
Since $0$ is less than $4$ and $4$ is less than 12, the inequality holds and $0$ is in the solution set of this inequality.

*(c)* Substituting $x=0$ in this inequality, it yields:
$$
begin{align*}
-0+1&>0+12\
0&>12.
end{align*}$$
Since $0$ is not greater than 12, the inequality does not hold and $0$ is not in the solution set of this inequality.

*(d)* Substituting $x=0$ in this inequality, it yields:
$$
begin{align*}
3cdot0&le0+1le0-1\
0&le 1le -1.
end{align*}$$
Since $0$ is less than $1$, but $1$ is not less than $-1$, the inequality does not hold and $0$ is not in the solution set of this inequality.

*(e)* Substituting $x=0$ in this inequality, it yields:
$$
begin{align*}
0cdot(2cdot0-1)&le07\
0&le 7.
end{align*}$$
Since $0$ is less than $7$, the inequality holds and $0$ is in the solution set of this inequality.

*(f)* Substituting $x=0$ in this inequality, it yields:
$$
begin{align*}
0+6&<(0+2)(5cdot0+3)\
6&< 6.
end{align*}$$
Since $6$ is equal, but not less than $6$, the inequality does not hold and $0$ is not in the solution set of this inequality.

(a) We would like to solve this inequality algebraically.

$$
-5 < 2x+7 < 11
$$

$diamond$ Note that: The subtraction and addtion properties of inequalities state,

$$
color{#4257b2} text{If} a leq b, text{then} a pm c leq b pm c
$$

Now, we can subtract $(7)$ from each side as follows:

$$
-5-7 < 2x+7-7 < 11-7
$$

$$
-12 < 2x < 4
$$

Then we will divide each side by $(2)$ as follows:

$$
-dfrac{12}{2} < dfrac{2x}{2} < dfrac{4}{2}
$$

$$
-6 < x < 2
$$

$$
text{color{#4257b2}$therefore$ The solution set is $left(-6, 2 right)$}
$$

(b) We would like to solve this inequality algebraically.

$$
11 < 3x-1 < 23
$$

$diamond$ Note that: The subtraction and addtion properties of inequalities state,

$$
color{#4257b2} text{If} a leq b, text{then} a pm c leq b pm c
$$

Now, we will add $(1)$ to each side as follows:

$$
11+1 < 3x-1+1 < 23+1
$$

$$
12 < 3x < 24
$$

Then we will divide each side by $(3)$ follows:

$$
dfrac{12}{3} < dfrac{3x}{3} < dfrac{24}{3}
$$

$$
4 < x < 8
$$

$$
text{color{#4257b2}$therefore$ The solution set is $left(4, 8 right)$}
$$

(c) We would like to solve this inequality algebraically.

$$
-1 leq -x+9 leq 13
$$

$diamond$ Note that: The subtraction and addtion properties of inequalities state,

$$
color{#4257b2} text{If} a leq b, text{then} a pm c leq b pm c
$$

Now, we will subtract $(9)$ from each side as follows:

$$
-1-9 leq -x+9-9 leq 13-9
$$

$$
-10 leq -x leq 4
$$

Then we will divide each side by $(-1)$ as follows:

$$
dfrac{-10}{-1} geq dfrac{-x}{-1} geq dfrac{4}{-1}
$$

$$
10 geq x geq -4
$$

$color{#4257b2}diamond Note that: Dividing by a negative number reverses the inequality sign.$

$$
text{color{#4257b2}$therefore$ The solution set is $left[-4, 10 right]$}
$$

(d) We would like to solve this inequality algebraically.

$$
0 leq -2(x+4) leq 6
$$

$diamond$ Note that: The subtraction and addtion properties of inequalities state,

$$
color{#4257b2} text{If} a leq b, text{then} a pm c leq b pm c
$$

First, remove the brackets

$$
0 leq -2x-8 leq 6
$$

Then, we will add $(8)$ to each side as follows:

$$
0+8 leq -2x-8+8 leq 6+8
$$

$$
8 leq -2x leq 14
$$

Then we will divide each side by $(-2)$ as follows:

$$
dfrac{8}{-2} geq dfrac{-2x}{-2} geq dfrac{14}{-2}
$$

$$
-4 geq x geq -7
$$

$color{#4257b2}diamond Note that: Dividing by a negative number reverses the inequality sign.$

$$
text{color{#4257b2}$therefore$ The solution set is $left[-7, -4 right]$}
$$

(e) We would like to solve this inequality algebraically.

$$
59 < 7x+10 < 73
$$

$diamond$ Note that: The subtraction and addtion properties of inequalities state,

$$
color{#4257b2} text{If} a leq b, text{then} a pm c leq b pm c
$$

Now, we can subtract $(10)$ from each side as follows:

$$
59-10 < 7x+10-10 < 73-10
$$

$$
49 < 7x < 63
$$

Then we will divide each side by $(7)$ as follows:

$$
dfrac{49}{7} < dfrac{7x}{7} < dfrac{63}{7}
$$

$$
7 < x < 9
$$

$$
text{color{#4257b2}$therefore$ The solution set is $left(7, 9 right)$}
$$

(f) We would like to solve this inequality algebraically.

$$
18 leq -12(x-1) leq 48
$$

$diamond$ Note that: The subtraction and addtion properties of inequalities state,

$$
color{#4257b2} text{If} a leq b, text{then} a pm c leq b pm c
$$

First, remove the brackets

$$
18 leq -12x+12 leq 48
$$

Then, we will subtract $(12)$ from each side as follows:

$$
18-12 leq -12x+12-12 leq 48-12
$$

$$
6 leq -12x leq 36
$$

Then we will divide each side by $(-12)$ as follows:

$$
dfrac{6}{-12} geq dfrac{-12x}{-12} geq dfrac{36}{-12}
$$

$$
-dfrac{1}{2} geq x geq -3
$$

$color{#4257b2}diamond Note that: Dividing by a negative number reverses the inequality sign.$

$$
text{color{#4257b2}$therefore$ The solution set is $left[-3, -dfrac{1}{2} right]$}
$$

$$
text{color{Brown} (a) $left(-6, 2 right)$ (b) $left(4, 8 right)$ (c) $left[-4, 10 right]$ \ \

(d) $left[-7, -4 right]$ (e) $left(7, 9 right)$ (f) $left[-3, -dfrac{1}{2} right]$}
$$

$$
text{color{#4257b2}(a) Create a linear inequality for which the solution set is $x>4$}
$$

$$
-2x+5>9-3x
$$

Isolate the variables on the left side as follows:

$$
-2x+3x>9-5 x>4
$$

$$
text{color{#4257b2}(b) Create a linear inequality for which the solution set is $xledfrac{3}{2}$}
$$

$$
2(2x+3)ge3(2x+1)
$$

Use distributive property as follows:

$$
4x+6ge6x+3
$$

Isolate the variables on the left side as follows:

$$
4x-6xge3-6 -2xge-3
$$

Divide both of sides by $(-2)$ as follows:

$$
xledfrac{3}{2}
$$

$$
text{color{Brown}(a) $-2x+5>9-3x$
\ \
(b) $2(2x+3)ge3(2x+1)$}
$$

(a) We would like to write the solution using set notation.

$text{color{#4257b2} The solution set is $left{xmid -6 leq x leq 4 right}$}$

(a) We would like to create a double inequality for which this is the solution set.

$$
text{color{#4257b2} The double inequality is $( -6 leq x leq 4 )$}
$$

$$
text{color{Brown} (a) The solution set is $left{xmid -6 leq x leq 4 right}$ \

(b) The double inequality is $( -6 leq x leq 4 )$}
$$

Which of the following inequalities has a solution.

$$
text{color{#4257b2}For $x-3<3-x<x-5$}
$$

Add $(-x)$ for entire the inequality as follows:

$$
x-x-3<3-x-x<x-x-5 -3<3-2x<-5
$$

Add $(-3)$ for entire the inequality as follows:

$$
-3-3<3+3-2x<-5-3 -6<-2xx>4
$$

This inequality has a solution

$$
text{color{#4257b2}For $-3>3-x>x-5$}
$$

Add $(x)$ for entire the inequality as follows:

$$
-3+x>3-x+x>-5+x
$$

This inequality has no solution, due the variables on the tow sides of inequality.

$$
text{color{Brown}The inequality which has a solution is $x-3<3-x<x-5$}
$$

Consider the graph in the textbook, then answer for the following questions.

$$
text{color{#4257b2}(a) Write the inequality that modelled by the graph.}
$$

$$
f(x)=3 xin R f(x)=dfrac{1}{2} x+1 x in R
$$

$$
text{color{#4257b2}(b) Find the solution by examining the graph.}
$$

From the graph, for $(y=3)$

Value of $(x=0)$ and $(x)$, values available for all real number. because the function is constant at $(y=3)$.

From the graph, for $left(y=dfrac{1}{2} x+1right)$

Value of $(x=-2)$

$$
text{color{#4257b2}(c) Confirm your solution by solve the inequality algabrically.}
$$

For equation of, $(y=3)$

Value of $(x=0)$ and $(x)$, values available for all real number. because the function is constant at $(y=3)$.

For equation of, $left(y=dfrac{1}{2} x+1right)$

Use zero property as follows:

$$
dfrac{1}{2} x+1=0 dfrac{1}{2} x=-1
$$

Multiply both of sides by $(2)$ as follows:

$$
x=-1cdot2 x=-2
$$

$$
text{color{Brown}
(a) $f(x)=3 xin R f(x)=dfrac{1}{2} x+1 x in R$
\ \
(b) From the graph $x=0 x=-2$
\ \
(c) Solve algebraically $x=0 x=-2$}
$$

The relation between Celsius and Fahrenheit represented by $C=dfrac{5}{9}(F-32)$

The temperature house between $18text{textdegree}$ and $22text{textdegree}$ Celsius

$$
text{color{#4257b2}(a) Write the double linear inequality}
$$

$$
18 text{textdegree} Cle xle22 text{textdegree} C
$$

$$
text{color{#4257b2}(b) Find the range of temperature in Fahrenheit degree.}
$$

For$18text{textdegree}$C

$$
C=dfrac{5}{9}(F-32) 18=dfrac{5}{9}(F-32)
$$

Multiply both of sides by $left(dfrac{9}{5}right)$ as follows:

$$
18cdotdfrac{9}{5}=F-32
$$

Isolate the variables on the only side as follows:

$$
F=dfrac{162}{5}+32 F=dfrac{160+162}{5}
$$

$$
F=dfrac{322}{5} F=64.4text{textdegree}text{F}
$$

For $22text{textdegree}$C

$$
C=dfrac{5}{9}(F-32) 22=dfrac{5}{9}(F-32)
$$

Multiply both of sides by $left(dfrac{9}{5}right)$ as follows:

$$
22cdotdfrac{9}{5}=F-32
$$

Isolate the variables on the only side as follows:

$$
F=dfrac{198}{5}+32 F=dfrac{160+198}{5}
$$

$$
F=dfrac{358}{5} F=71.6 text{textdegree}text{F}
$$

The range of temperature in degree of Fahrenheit is $64.4 text{textdegree} Fle xle71.6 text{textdegree} F$

$$
text{color{#c34632}(a) $18 text{textdegree} Cle xle22 text{textdegree} C$
\ \
(b) $64.4 text{textdegree} Fle xle71.6 text{textdegree} F$}
$$

$$
text{color{#4257b2}How long can each volunteer talk to prospective dollars.}
$$

The calls has a billed of $(0.5)$ dollar for the first three minute then billed $(0.1)$ dollar for each minute and the maximum billed per call is not more than $(2)$ dollar.

$because$ First three minute is billed of $(0.5)$ dollar

$therefore$ The price remaining for call is $(2-0.5=1.5)$ dollars

The remaining minute after first three minutes are $=dfrac{1.5}{0.1}=15$ minutes

The minutes needed to talk to prospective dollars are $(15+3=18$ minutes)

$$
text{color{#c34632} $18$ minutes}
$$

$$
text{color{#4257b2}(a) Find a equation that allow conversion from Celsius to Fahrenheit temperature with using question $(12)$}
$$

$$
C=dfrac{5}{9}(F-32)
$$

We need to get the value of $(F)$ in the left side, so follow below steps.

Multiply both of sides by $left(dfrac{9}{5}right)$ as follows:

$$
dfrac{9}{5} C=F-32
$$

Add both of sides by $(32)$ as follows:

$$
F=dfrac{9}{5} C+32 F=1.8C+32
$$

$$
text{color{Brown} $F=1.8C+32$}
$$

For the inequality $|2x-1|<7$

For ${color{#4257b2}2x-1<7} 2x<7+1 2x<8$

Divide both of sides by $(2)$ as follows:

$$
x<dfrac{8}{2} x<4
$$

For ${color{#4257b2}-2x+1<7} -2x<7-1 -2x-dfrac{6}{2} x>-3
$$

$$
text{color{#4257b2}(a) Graph the inequality}
$$

$$
text{color{#4257b2}(b) Solve the inequality from the graph }
$$

From the graph, the solution for the inequality is $-3<x<4$

$$
text{color{#c34632} (a) See the graph (b) $-3<x<4$}
$$

$$
text{color{#4257b2}Will the solution to the double inequality is has a upper and lower limit?}
$$

For all double inequality has the upper and lower limit.

Its including but not limited to:

$$
0<x<4, -3<x<2, 1ge xle3
$$

$$
text{color{Brown}$0<x<4, -3<x<2, 1ge xle3$}
$$

*(a)* Note that the given inequality is not linear, but also has $x$ as an exponent in $2^x$. In this inequality, it would be very hard to isolate $x$ and find a set of solutions algebraically.

*(b)* Inequalities of this kind we can solve graphicly. In a graphing calculator input both sides of inequality as separate functions $y_1=2^x-3$ (red in *Figure 1*) and $y_2=x+1$ (blue in *Figure 1*). Note the region where the graph of the first function is below the graph of the second function. Furthermore, we can press TRACE and use arrows to move the cursor to the points where graphs meet. There we can read out the values of $x$. The graphs of these functions are depicted in *Figure 1*.

*Figure 1.* The graphs of the two functions

Note that red curve is under the blue curve for
$$
-3.94<x<2.76$$

$-3.94<x<2.76$

Determine if the inequality directions should be maintained , switched or not for the following terms.

Assume the inequality is $(2<4)$

$$
text{color{#4257b2}(a) Cubing both sides}
$$

$$
(2)^3<(4)^3 8<64
$$

The inequality still true

$$
text{color{#4257b2}(b) Square both sides}
$$

$$
(2)^2<(4)^2 4<16
$$

The inequality still true

$$
text{color{#4257b2}(c) Make both of sides the exponent with base of $(2)$}
$$

$$
(2)^2<(2)^4 4<16
$$

The inequality still true

$$
text{color{#4257b2}(d) Make both of sides the exponent with base of $(0.5)$}
$$

$$
(0.5)^2<(0.5)^4 0.250.0625$

$$
text{color{#4257b2}(e) Taking a reciprocating for both sides }
$$

$$
dfrac{1}{2}dfrac{1}{4}$

$$
text{color{#4257b2}(f) Tacking square root for both sides}
$$

$$
sqrt{2}<sqrt{4} 1.4<2
$$

The inequality still true

$$
text{color{Brown}(a) The inequality still true (b) The inequality still true
\ \
(c) The inequality still true (d) The inequality is not true
\ \
(e) The inequality is not true (f) The inequality still true}
$$

Solve the following inequalities and graph it on number line.

$$
color{#4257b2}text{(a)} x^2<4
$$

Use square root property as follows:

$$
sqrt{x^2}<sqrt{4} x<pm2
$$

$$
x<2 x<-2
$$

$$
color{#4257b2}text{(b)} 4x^2+5ge41
$$

Add both of sides by $(-5)$ as follows:

$$
4x^2ge41-5 4x^2ge36
$$

Divide both of sides by $(4)$ as follows:

$$
x^2ge9
$$

Use square root property as follows:

$$
sqrt{x^2}gesqrt{9} xgepm3
$$

$$
xge3 xge-3
$$

$$
color{#4257b2}text{(c)} |2x+2|<8
$$

For $2x+2<8 2x<8-2 2x<6$

Divide both of sides by $(2)$ as follows:

$$
x<3
$$

For $-2x-2<8 -2x<8+2 -2x-5
$$

$$
-5<x<3
$$

$$
color{#4257b2}text{(d)} -3x^3ge81
$$

Divide both of sides by $(-3)$ as follows:

$$
x^3le-27
$$

Use cubic root property as follows:

$$
sqrt[3]{x^3}le-sqrt[3]{27}
$$

$$
xle-3
$$

$$
text{color{Brown}(a) $x=(-2, 2)$ (b) $x=(-3, 3)$
\ \
(c) $-5<x<3$ (d) $xle-3$}
$$