Step 1: Establishing the Initial Inequality

The given linear inequality is:

$\frac{x}{2} \ge \frac{5x-2}{3} - \frac{7x-3}{5}$

Step 2: Clearing the Denominators

To simplify, identify the Least Common Multiple (LCM) of the denominators $2$, $3$, and $5$, which is $30$ [By the property of arithmetic operations on rational expressions]. Multiply both sides of the inequality by $30$:

$30 \cdot \left( \frac{x}{2} \right) \ge 30 \cdot \left( \frac{5x-2}{3} - \frac{7x-3}{5} \right)$

$15x \ge 10(5x - 2) - 6(7x - 3)$

Step 3: Algebraic Expansion and Simplification

Distribute the constants across the terms within the parentheses:

$15x \ge 50x - 20 - 42x + 18$ [By Distributive Property of Multiplication]

Combine like terms on the right-hand side:

$15x \ge (50x - 42x) + (-20 + 18)$

$15x \ge 8x - 2$

Step 4: Isolating the Variable $x$

Subtract $8x$ from both sides of the inequality [Per the Addition Property of Inequality]:

$15x - 8x \ge -2$

$7x \ge -2$

Divide both sides by $7$:

$x \ge -\frac{2}{7}$

Step 5: Representing the Solution Set

The solution in interval notation is $[-\frac{2}{7}, \infty)$. This indicates that all real numbers greater than or equal to $-\frac{2}{7}$ satisfy the original inequality.

Step 6: Graphical Representation on the Number Line

Final Solution: $x \in \left[ -\frac{2}{7}, \infty \right)$

Step 1: Setting up the Inequality

Given the linear inequality: $x + \frac{x}{2} + \frac{x}{3} < 11$ [Per the property of algebraic summation of fractions]

Step 2: Finding a Common Denominator

To add the terms involving $x$, we identify the Least Common Multiple (LCM) of the denominators $\{1, 2, 3\}$, which is $6$. Rewriting the expression: $\frac{6x}{6} + \frac{3x}{6} + \frac{2x}{6} < 11$ [By the property of equivalent fractions]

Step 3: Combining Like Terms

Summing the numerators: $\frac{6x + 3x + 2x}{6} < 11$ $\frac{11x}{6} < 11$ [By the distributive property of multiplication $a(b+c) = ab + ac$]

Step 4: Isolating the Variable $x$

To isolate $x$, we multiply both sides of the inequality by $6$: $11x < 11 \times 6$ $11x < 66$ [Per the Multiplication Property of Inequality: if $c > 0$, then $a < b \implies ac < bc$]

Divide both sides by $11$: $x < \frac{66}{11}$ $x < 6$ [Per the Division Property of Inequality: if $c > 0$, then $a < b \implies \frac{a}{c} < \frac{b}{c}$]

Visual Representation of the Solution Set

Step 5: Expressing the Final Solution in Interval Notation

Since the inequality is strict ($<$), the value $6$ is excluded from the solution set. For all real $x$ such that $x < 6$: $x \in (-\infty, 6)$ [By the definition of open intervals in real numbers]

Final Solution: $x < 6$ or $x \in (-\infty, 6)$

Step 1: Analyzing the Inequality

The given linear inequality is: $24x < 100$ [Per the property of inequalities, we can divide both sides by a positive number without changing the direction of the inequality sign.]

Step 2: Isolating the Variable

Divide both sides of the inequality by $24$: $x < \frac{100}{24}$
Reducing the fraction by dividing the numerator and denominator by their greatest common divisor ($4$): $x < \frac{25}{6}$
Converting to a decimal representation for clarity: $x < 4.166...$ [By the definition of real number division and inequality properties.]

Step 3: Applying the Domain Constraint

The problem specifies that $x$ must be a natural number ($x \in \mathbb{N}$). Natural numbers are defined as the set of positive integers: $\{1, 2, 3, 4, 5, ...\}$.
We seek all natural numbers $x$ such that $x < 4.166...$.
Comparing the set of natural numbers against the upper bound:

• $1 < 4.166$ (True)
• $2 < 4.166$ (True)
• $3 < 4.166$ (True)
• $4 < 4.166$ (True)
• $5 < 4.166$ (False)

Visual Representation of the Solution Space

Final Solution: The set of natural numbers satisfying the inequality is $\{1, 2, 3, 4\}$.

Step 1: Isolation of the Variable $x$

Given the linear inequality: $3x + 8 > 2$

To isolate the term containing $x$, we subtract $8$ from both sides of the inequality [Per the Addition/Subtraction Property of Inequalities]: $3x + 8 - 8 > 2 - 8$ $3x > -6$

Next, we divide both sides by $3$. Since $3 > 0$, the direction of the inequality sign remains unchanged [Per the Multiplication/Division Property of Inequalities]: $\frac{3x}{3} > \frac{-6}{3}$ $x > -2$

Step 2: Defining the Solution Set for Integers

We are tasked to find the solution set where $x$ must be an integer ($\mathbb{Z}$). The inequality $x > -2$ represents all real numbers strictly greater than $-2$. Within the domain of integers, this set includes every integer starting from the next consecutive integer after $-2$.

The set of integers $x$ such that $x > -2$ is given by: $S = \{ -1, 0, 1, 2, 3, 4, \dots \}$

Visual Representation of the Solution on the Number Line

Conclusion

By solving the algebraic inequality and restricting the domain to the set of integers, we identify that any integer greater than $-2$ satisfies the condition.

Final Solution: The solution set for $x$ is $\{x \in \mathbb{Z} : x > -2\}$, which is $\{-1, 0, 1, 2, 3, \dots \}$.

Step 1: Establishing the Mathematical Framework

Let the marks obtained by Ravi in the third unit test be represented by the variable $x$. The objective is to determine the minimum value of $x$ such that the average of the three test scores is at least 60 marks.

The given marks for the first two tests are $70$ and $75$. The arithmetic mean of three values is calculated as the sum of the values divided by the count of the values [Per the definition of Arithmetic Mean].

Step 2: Formulating the Linear Inequality

The condition "at least 60" is mathematically translated as $\geq 60$. Constructing the inequality based on the average formula:

$$\frac{70 + 75 + x}{3} \geq 60$$

Step 3: Algebraic Manipulation and Solution

To isolate $x$, we apply the properties of inequalities [Per the Multiplication Property of Inequality, where multiplying by a positive constant preserves the inequality sign]:

1. Multiply both sides by 3:
   $70 + 75 + x \geq 60 \times 3$
   $145 + x \geq 180$
2. Subtract 145 from both sides [Per the Addition/Subtraction Property of Inequality]:
   $x \geq 180 - 145$
   $x \geq 35$

Step 4: Verification and Graphical Representation

The solution $x \geq 35$ indicates that Ravi must score a minimum of 35 marks to achieve an average of 60.

Step 5: Data Summary

Final Solution: Ravi must obtain a minimum of 35 marks in the third test to secure an average of at least 60 marks.

Step 1: Establishing the Algebraic Inequality

The given linear inequality is:

$5x - 3 < 7$

To isolate the variable $x$, we apply the addition property of inequalities [Per the properties of inequalities, adding the same real number to both sides of an inequality does not change its direction].

Add $3$ to both sides:

$5x - 3 + 3 < 7 + 3$

$5x < 10$

Step 2: Isolating the Variable $x$

To solve for $x$, divide both sides of the inequality by $5$. Since $5$ is a positive real number, the inequality sign remains unchanged [Per the multiplication/division property of inequalities, if $c > 0$, then $a < b \implies \frac{a}{c} < \frac{b}{c}$].

$\frac{5x}{5} < \frac{10}{5}$

$x < 2$

Step 3: Determining the Solution Set for Real Numbers

When $x$ is defined as a real number ($x \in \mathbb{R}$), the solution is the set of all real values strictly less than $2$. This corresponds to the interval $(-\infty, 2)$.

Step 4: Graphical Representation

The solution set is represented on the real number line by an open circle at $2$ (indicating $2$ is excluded) and shading the region to the left.

Final Solution: $x \in (-\infty, 2)$

Step 1: Expanding the Terms

Given the inequality:

$37 - (3x + 5) > 9x - 8(x - 3)$

Apply the distributive property to remove the parentheses [Per the Distributive Property of Multiplication over Addition/Subtraction]:

$37 - 3x - 5 > 9x - 8x + 24$

Step 2: Combining Like Terms

Combine constant terms and variable terms on each side of the inequality separately [By Arithmetic Simplification]:

Left Side: $(37 - 5) - 3x = 32 - 3x$

Right Side: $(9x - 8x) + 24 = x + 24$

The inequality becomes:

$32 - 3x > x + 24$

Step 3: Isolating Variable Terms

Add $3x$ to both sides of the inequality [Per the Addition Property of Inequalities]:

$32 > x + 3x + 24$

$32 > 4x + 24$

Subtract $24$ from both sides [Per the Subtraction Property of Inequalities]:

$32 - 24 > 4x$

$8 > 4x$

Step 4: Solving for x

Divide both sides by the positive coefficient $4$ [Per the Division Property of Inequalities; since $4 > 0$, the inequality sign remains unchanged]:

$\frac{8}{4} > x$

$2 > x$

This is equivalent to stating $x < 2$.

Step 5: Visual Representation

Step 6: Final Solution Set Interval

Since the inequality is $x < 2$, the set of all real numbers $x$ satisfying this inequality is all values less than $2$.

Interval notation: $(-\infty, 2)$

Final Solution: $x \in (-\infty, 2)$

Step 1: Establishing the Initial Inequality

Given the linear inequality in one variable:

$3x - 7 > 5x - 1$

Step 2: Isolating the Variable Term

To collect the variable $x$ on one side, subtract $3x$ from both sides of the inequality [Per the Addition/Subtraction Property of Inequalities, adding or subtracting the same value does not change the inequality sign].

$3x - 3x - 7 > 5x - 3x - 1$

$-7 > 2x - 1$

Step 3: Isolating the Constant Term

Add $1$ to both sides of the inequality to isolate the term containing $x$:

$-7 + 1 > 2x - 1 + 1$

$-6 > 2x$

Step 4: Solving for $x$

Divide both sides by $2$ [Since $2 > 0$, the direction of the inequality remains unchanged per the Multiplication/Division Property of Inequalities].

$\frac{-6}{2} > \frac{2x}{2}$

$-3 > x$

Step 5: Expressing the Solution Set

The inequality $-3 > x$ can be rewritten as $x < -3$. In set-builder notation, the solution is $\{x : x \in \mathbb{R}, x < -3\}$. In interval notation, the solution set is $(-\infty, -3)$.

Visual Representation of the Solution

Final Solution: The solution set for the inequality $3x - 7 > 5x - 1$ is $x < -3$, represented in interval notation as $(-\infty, -3)$.

Step 1: Given Variables & Initial Setup

We are provided with the following linear inequality in one variable $x$:

$2(2x + 3) - 10 < 6(x - 2)$

Step 2: Distribution and Algebraic Expansion

Distribute the constants on both sides of the inequality [Per the Distributive Property of Multiplication over Addition/Subtraction]:

$2(2x) + 2(3) - 10 < 6(x) - 6(2)$

$4x + 6 - 10 < 6x - 12$

Step 3: Combining Like Terms

Simplify the numerical terms on the left-hand side:

$4x - 4 < 6x - 12$

Step 4: Transposition of Terms

To isolate the variable $x$, subtract $4x$ from both sides [Per the Addition/Subtraction Property of Inequalities]:

$4x - 4x - 4 < 6x - 4x - 12$

$-4 < 2x - 12$

Add $12$ to both sides to isolate the term containing $x$:

$-4 + 12 < 2x$

$8 < 2x$

Step 5: Solving for x

Divide both sides by $2$. Since $2 > 0$, the direction of the inequality remains unchanged [Per the Multiplication/Division Property of Inequalities]:

$\frac{8}{2} < \frac{2x}{2}$

$4 < x$

This is equivalent to $x > 4$.

Step 6: Representation of the Solution Set

The solution for the real value of $x$ is the interval $(4, \infty)$.

Final Solution: The solution set for the inequality $2 (2x + 3) – 10 < 6 (x – 2)$ is $x \in (4, \infty)$ or $x > 4$.

Step 1: Defining Variables and Constraints

Let the length of the shortest side of the triangle be $x$ cm, where $x > 0$. Based on the problem statement, we define the other two sides as follows:

• Shortest side $= x$
• Longest side $= 3x$
• Third side $= 3x - 2$

[Per the definitions provided by the problem constraints]

Step 2: Establishing the Perimeter Inequality

The perimeter of a triangle is the sum of its three sides. The problem states that the perimeter is at least $61$ cm. Therefore:

$x + 3x + (3x - 2) \geq 61$

Combining like terms:

$7x - 2 \geq 61$

$7x \geq 63$

$x \geq 9$ [By the property of linear inequalities, dividing both sides by a positive number preserves the inequality sign]

Step 3: Verifying the Triangle Inequality Theorem

For any triangle with sides $a, b, c$, the sum of any two sides must be strictly greater than the third side ($a + b > c$). Given sides are $x$, $3x$, and $3x-2$.

[By the Triangle Inequality Theorem, all sides must satisfy these conditions simultaneously]

Step 4: Geometric Representation of the Triangle Constraints

Step 5: Synthesizing the Minimum Value

From Step 2, we found $x \geq 9$. From Step 3, the condition $x > 2$ must be satisfied for the triangle to exist. Since $9 > 2$, the value $x = 9$ satisfies all conditions of the problem.

Final Solution: The minimum length of the shortest side is 9 cm.

Step 1: Distribution of Scalar Multipliers

Given the inequality:

$3(2 - x) \ge 2(1 - x)$

Apply the Distributive Property of Multiplication over Subtraction, $a(b - c) = ab - ac$ [Per the Distributive Law of Real Numbers]:

$3 \cdot 2 - 3 \cdot x \ge 2 \cdot 1 - 2 \cdot x$

$6 - 3x \ge 2 - 2x$

Step 2: Isolation of Variable Terms

To collect variable terms on one side and constant terms on the other, add $3x$ to both sides of the inequality [Per the Addition Property of Inequalities]:

$6 - 3x + 3x \ge 2 - 2x + 3x$

$6 \ge 2 + x$

Step 3: Isolation of the Variable $x$

Subtract $2$ from both sides of the inequality to isolate $x$ [Per the Subtraction Property of Inequalities]:

$6 - 2 \ge 2 + x - 2$

$4 \ge x$

Rewriting the inequality for standard form: $x \le 4$.

Step 4: Graphical Representation of the Solution Set

Step 5: Defining the Interval Notation

Since the inequality is $x \le 4$, the solution includes all real numbers less than or equal to $4$. In interval notation, this is expressed as $(-\infty, 4]$.

Final Solution: The solution set for $x$ is $x \le 4$, or in interval notation, $(-\infty, 4]$.

Step 1: Establishing the Algebraic Inequality

Given the inequality: $5x - 3 > 3x - 5$

Our objective is to isolate the variable $x$ on one side of the inequality. [Per the properties of inequalities, we may add or subtract the same real number from both sides without changing the sign of the inequality].

Step 2: Performing Algebraic Transposition

Subtract $3x$ from both sides:

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

$2x - 3 > -5$

Add $3$ to both sides:

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

$2x > -2$

Step 3: Solving for $x$

Divide both sides by $2$. [Per the Multiplication/Division Property of Inequalities, if a positive number is used to divide both sides, the inequality sign remains unchanged].

$\frac{2x}{2} > \frac{-2}{2}$

$x > -1$

Step 4: Graphical Representation on the Number Line

The solution set $x > -1$ implies all real numbers strictly greater than $-1$. On a number line, this is represented by an open circle at $-1$ (to signify that $-1$ is not included) and a ray extending to the right towards positive infinity.

The interval notation for this solution is $x \in (-1, \infty)$.

Final Solution: $x > -1$ or $x \in (-1, \infty)$

Step 1: Expansion of Algebraic Expressions

To solve the inequality $3(x - 1) \le 2(x - 3)$, we first apply the Distributive Property of Multiplication over subtraction, denoted as $a(b - c) = ab - ac$ [Per the Distributive Law of Real Numbers].

$3x - 3 \le 2x - 6$

Step 2: Isolation of Variable Terms

To consolidate the variable $x$ on one side of the inequality, we subtract $2x$ from both sides of the inequality [Per the Addition Property of Inequalities: if $a \le b$, then $a - c \le b - c$].

$3x - 2x - 3 \le 2x - 2x - 6$

$x - 3 \le -6$

Step 3: Isolation of the Constant

To solve for $x$, we add $3$ to both sides of the inequality [Per the Addition Property of Inequalities].

$x - 3 + 3 \le -6 + 3$

$x \le -3$

Step 4: Graphical Representation

The solution set represents all real numbers less than or equal to $-3$. This is expressed in interval notation as $(-\infty, -3]$.

Step 5: Theoretical Justification of the Solution Set

Given the domain is all real numbers ($x \in \mathbb{R}$), the inequality $x \le -3$ encompasses every point on the real number line starting from $-\infty$ up to and including $-3$.

Final Solution: The solution to the inequality $3(x - 1) \le 2(x - 3)$ is $x \le -3$, or in interval notation, $(-\infty, -3]$.

Step 1: Expanding the Inequality

Given the linear inequality: $3(1 - x) < 2(x + 4)$

Apply the distributive property of multiplication over subtraction and addition [$a(b \pm c) = ab \pm ac$]:

$3 - 3x < 2x + 8$

Step 2: Isolating the Variable Terms

To collect all $x$ terms on one side, subtract $2x$ from both sides [Per the Addition Property of Inequality]:

$3 - 3x - 2x < 8$

$3 - 5x < 8$

Subtract $3$ from both sides [Per the Addition Property of Inequality]:

$-5x < 8 - 3$

$-5x < 5$

Step 3: Solving for x

Divide both sides by $-5$. Note that dividing or multiplying an inequality by a negative number reverses the direction of the inequality sign [Per the Multiplicative Inverse Property of Inequality]:

$x > \frac{5}{-5}$

$x > -1$

Step 4: Graphic Representation on the Number Line

The solution $x > -1$ indicates all real numbers greater than $-1$, excluding $-1$ itself. We represent this with a hollow circle at $-1$ and a shaded ray extending to the right.

Final Solution: The solution to the inequality is $x \in (-1, \infty)$, representing all real values strictly greater than $-1$.

Step 1: Initial Algebraic Setup

Given the linear inequality:
$\frac{3(x - 2)}{5} \le \frac{5(2 - x)}{3}$

To eliminate the denominators, calculate the Least Common Multiple (LCM) of $5$ and $3$, which is $15$. Multiply both sides of the inequality by $15$. [Per the Multiplication Property of Inequality, multiplying by a positive constant does not reverse the inequality sign].

$15 \times \left( \frac{3(x - 2)}{5} \right) \le 15 \times \left( \frac{5(2 - x)}{3} \right)$

Step 2: Simplification of the Expression

Simplify the fractions by canceling common factors:
$3 \times 3(x - 2) \le 5 \times 5(2 - x)$
$9(x - 2) \le 25(2 - x)$

Step 3: Distributive Expansion

Apply the distributive law of multiplication over subtraction [$a(b - c) = ab - ac$]:
$9x - 18 \le 50 - 25x$

Step 4: Grouping Variable and Constant Terms

Collect all terms containing $x$ on the left side by adding $25x$ to both sides, and all constant terms on the right side by adding $18$ to both sides [Per the Addition Property of Inequality].
$9x + 25x \le 50 + 18$
$34x \le 68$

Step 5: Solving for x

Divide both sides by $34$. Since $34 > 0$, the inequality sign remains unchanged [Per the Division Property of Inequality].
$x \le \frac{68}{34}$
$x \le 2$

Visual Representation of the Solution Set

Step 6: Defining the Interval Notation

The inequality $x \le 2$ includes all real numbers less than or equal to $2$. In interval notation, this is expressed as:
$x \in (-\infty, 2]$

Final Solution: x \in (-\infty, 2]

Step 1: Given Variables & Initial Setup

The objective is to solve the linear inequality for real values of $x$:

$$\frac{x}{3} > \frac{x}{2} + 1$$

Step 2: Isolating Variable Terms

To solve for $x$, we collect all terms containing $x$ on the left-hand side of the inequality. We subtract $\frac{x}{2}$ from both sides [Per the Addition/Subtraction Property of Inequalities]:

$$\frac{x}{3} - \frac{x}{2} > 1$$

Step 3: Calculating Common Denominator

Find the Least Common Multiple (LCM) of the denominators $3$ and $2$, which is $6$. Convert the fractions to equivalent fractions with a denominator of $6$:

$$\frac{2x}{6} - \frac{3x}{6} > 1$$

Combine the fractions:

$$\frac{2x - 3x}{6} > 1$$

$$\frac{-x}{6} > 1$$

Step 4: Isolating $x$

Multiply both sides of the inequality by $6$ [Per the Multiplication Property of Inequalities, since $6 > 0$, the inequality sign remains unchanged]:

$$-x > 6$$

Multiply both sides by $-1$. [Per the Multiplication Property of Inequalities, since we are multiplying by a negative number, the inequality sign must be reversed]:

$$x < -6$$

Step 5: Visual Representation

The solution set represents all real numbers strictly less than $-6$.

Step 6: Defining the Solution Set

In interval notation, the set of all real numbers satisfying $x < -6$ is expressed as:

$$x \in (-\infty, -6)$$

Final Solution: The solution to the inequality is $x < -6$, or in interval notation, $x \in (-\infty, -6)$.

Step 1: Establishing the Mathematical Model

Let the marks obtained by Sunita in the fifth examination be denoted by the variable $x$. The condition for receiving Grade 'A' is defined by the requirement that the arithmetic mean of the five examinations must be greater than or equal to 90.

The average is calculated as the sum of all marks divided by the total number of examinations ($n = 5$):

$\frac{87 + 92 + 94 + 95 + x}{5} \geq 90$ [Definition of Arithmetic Mean]

Step 2: Simplifying the Inequality

First, sum the known values in the numerator:

$87 + 92 + 94 + 95 = 368$

Substitute this back into the inequality:

$\frac{368 + x}{5} \geq 90$

Step 3: Algebraic Isolation of $x$

To solve for $x$, multiply both sides of the inequality by 5. Since 5 is a positive constant, the direction of the inequality sign remains unchanged [Per the Multiplication Property of Inequalities]:

$368 + x \geq 90 \times 5$

$368 + x \geq 450$

Now, subtract 368 from both sides of the inequality [Per the Addition/Subtraction Property of Inequalities]:

$x \geq 450 - 368$

$x \geq 82$

Step 4: Visualizing the Required Threshold

Step 5: Interpretation of Results

The inequality $x \geq 82$ implies that for Sunita to maintain an average of 90 or more, her marks in the fifth examination must be at least 82. Since the maximum marks possible in the examination is 100, the solution set for $x$ is the interval $[82, 100]$.

Final Solution: The minimum marks that Sunita must obtain in the fifth examination to secure Grade ‘A’ is 82.

Step 1: Algebraic Manipulation of the Inequality

Given the linear inequality: $5x - 3 < 7$.

To isolate the term involving $x$, we apply the addition property of inequality by adding $3$ to both sides of the expression [Per the Addition Property of Inequalities]:

$5x - 3 + 3 < 7 + 3$

$5x < 10$

Step 2: Solving for $x$

To solve for $x$, we divide both sides of the inequality by $5$. Since $5 > 0$, the direction of the inequality sign remains unchanged [Per the Division Property of Inequalities for positive constants]:

$\frac{5x}{5} < \frac{10}{5}$

$x < 2$

Step 3: Determining the Solution Set for $x \in \mathbb{Z}$

The condition specifies that $x$ must be an integer ($\mathbb{Z}$). The set of all integers strictly less than $2$ includes all negative integers, zero, and positive integers up to $1$.

This can be represented as the set: $\{\dots, -3, -2, -1, 0, 1\}$.

Visual Representation of the Solution Set

Final Solution: The solution set for $x$ where $x$ is an integer is $\{\dots, -3, -2, -1, 0, 1\}$.

Step 1: Establishing the Mathematical Conditions

Let the two consecutive odd positive integers be represented as $x$ and $x+2$.

• The integers must be positive: $x \in \{1, 3, 5, 7, 9, ...\}$ [Definition of positive odd integers].
• Both integers must be smaller than 10: $x < 10$ and $x + 2 < 10$.
• Their sum must be more than 11: $x + (x + 2) > 11$.

Step 2: Solving the Inequality for the Sum

We solve the inequality representing the sum constraint:

$x + x + 2 > 11$

$2x + 2 > 11$ [Algebraic Addition Property]

$2x > 9$ [Subtraction Property of Inequality]

$x > 4.5$ [Division Property of Inequality]

Step 3: Solving the Inequality for the Upper Bound

We analyze the constraint that both integers are smaller than 10:

$x + 2 < 10$

$x < 8$ [Subtraction Property of Inequality]

Step 4: Determining the Intersection of Conditions

Combining the results from Step 2 and Step 3, we obtain the range for $x$:

$4.5 < x < 8$

Given that $x$ must be an odd positive integer [Given condition], we examine the integers within the interval $(4.5, 8)$:

• $x = 5$
• $x = 7$

Step 5: Tabulating the Valid Pairs

Visual Representation of Solution Set

Final Solution: The pairs of consecutive odd positive integers are (5, 7) and (7, 9).

Step 1: Analyzing the Linear Inequality

The given inequality is:

$-12x > 30$

To isolate the variable $x$, we divide both sides of the inequality by $-12$. [By the Property of Inequality: Dividing or multiplying both sides of an inequality by a negative number reverses the direction of the inequality sign].

$\frac{-12x}{-12} < \frac{30}{-12}$

Step 2: Simplifying the Expression

Perform the division and reduce the fraction to its lowest terms:

$x < -\frac{30}{12}$

$x < -\frac{5}{2}$

$x < -2.5$

Step 3: Applying Constraints for Natural Numbers

The problem specifies that $x$ must be a natural number ($\mathbb{N}$). The set of natural numbers is defined as $\mathbb{N} = \{1, 2, 3, 4, ...\}$.

We evaluate the condition $x < -2.5$ against the set of natural numbers:

• The set of natural numbers contains only positive integers ($x \geq 1$).
• The solution set $x < -2.5$ contains only negative real numbers less than $-2.5$ (e.g., $-3, -4, -5, ...$).
• There is no intersection between the set of natural numbers ($\mathbb{N}$) and the interval $(-\infty, -2.5)$.

Step 4: Visualizing the Number Line

As illustrated, the red region representing the solution $x < -2.5$ does not overlap with the natural numbers, which begin at $1$ on the positive side of the origin ($0$).

Final Solution: Since there are no natural numbers $x$ such that $x < -2.5$, the solution set is the empty set, denoted by $\emptyset$ or $\{\}$.