Chapter 5 — Linear Inequalities

Theory

A linear inequality in one variable can be solved similar to linear equations, with one key rule:

• If we multiply or divide both sides by a positive number, the inequality sign remains unchanged.
• If we multiply or divide by a negative number, the inequality sign reverses.

The solution of an inequality represents a range of values, not a single value.

Solution Roadmap

1. Isolate the variable \(x\)
2. Convert inequality into simplified form
3. Apply domain restriction (natural numbers / integers)
4. List valid values

Solution

$$\begin{aligned} 24x < 100 \ x < \dfrac{100}{24} \\ x < \dfrac{25}{6} \end{aligned}$$

(i) When \(x\) is a natural number:

Since \(x < \dfrac{25}{6} \approx 4.16\), the natural numbers satisfying this are: \[\{1,2,3,4\}\]

(ii) When \(x\) is an integer:

All integers less than \( \dfrac{25}{6} \) are included: \[\{\dots,-2,-1,0,1,2,3,4\}\]

Graphical Representation

Final Answer

(i) \( \{1,2,3,4\} \)
(ii) \( \{\dots,-2,-1,0,1,2,3,4\} \)

Significance

• Board Exams: Tests clarity of inequality solving and understanding of number systems.
• JEE / NEET: Forms the base for solving interval-based problems, modulus inequalities, and domain restrictions.
• Common Mistake: Students often forget to apply domain constraints after solving inequality.

Theory

In linear inequalities, when both sides are multiplied or divided by a negative number, the inequality sign reverses.

Also, solutions must satisfy both:

• The inequality condition
• The given domain (natural numbers or integers)

Solution Roadmap

1. Isolate \(x\)
2. Divide by coefficient of \(x\)
3. Reverse inequality sign (since coefficient is negative)
4. Apply domain restriction

Solution

$$\begin{aligned} -12x &\gt 30 \ x < \dfrac{30}{-12} \\ x < -\dfrac{5}{2} \end{aligned}$$

(i) When \(x\) is a natural number:

Natural numbers are \(1,2,3,\dots\). Since all natural numbers are positive and the inequality requires \(x < -\dfrac{5}{2}\), no natural number satisfies this condition.

(ii) When \(x\) is an integer:

Integers less than \(-\dfrac{5}{2}\) are: \[\{\dots,-5,-4,-3\}\]

Graphical Representation

Final Answer

(i) No solution in natural numbers
(ii) \( \{\dots,-5,-4,-3\} \)

Significance

• Board Exams: Very important for testing sign reversal while dividing by negative numbers.
• JEE / NEET: Frequently used in compound inequalities and interval-based problems.
• Common Error: Forgetting to reverse inequality sign after dividing by negative number.

Theory

A linear inequality can be solved using algebraic operations similar to equations. When adding or subtracting the same number from both sides, the inequality sign remains unchanged.

The solution represents a set of values, which may be:

• Discrete (integers)
• Continuous interval (real numbers)

Solution Roadmap

1. Move constant terms to one side
2. Simplify inequality
3. Divide by coefficient of \(x\)
4. Apply domain (integer / real)

Solution

$$\begin{aligned} 5x - 3 &\lt 7 \\ 5x &\lt 10 \\ x &\lt 2 \end{aligned}$$

(i) When \(x\) is an integer:

All integers less than 2 are:

$$\{\dots,-2,-1,0,1\}$$

(ii) When \(x\) is a real number:

The solution includes all real numbers less than 2: \[x \in (-\infty, 2)\]

Graphical Representation

Final Answer

(i) \( \{\dots,-2,-1,0,1\} \)
(ii) \( x \in (-\infty, 2) \)

Significance

• Board Exams: Tests algebraic manipulation and correct interval notation.
• JEE / NEET: Fundamental for solving inequalities involving modulus, quadratic expressions, and functions.
• Key Insight: Same inequality gives different answers depending on domain (discrete vs continuous).

Theory

Linear inequalities are solved using standard algebraic operations. When adding or subtracting the same number from both sides, the inequality sign remains unchanged.

The solution set depends on the domain of \(x\):

• Integers → discrete values
• Real numbers → continuous interval

Solution Roadmap

1. Shift constant term to RHS
2. Simplify inequality
3. Divide by coefficient of \(x\)
4. Apply domain restriction

Solution

$$\begin{aligned} 3x + 8 &\gt 2 \\ 3x &\gt -6 \\ x &\gt -2 \end{aligned}$$

(i) When \(x\) is an integer:

All integers greater than \(-2\) are:

$$\{-1,0,1,2,\dots\}$$

(ii) When \(x\) is a real number:

The solution includes all real numbers greater than \(-2\): \[x \in (-2,\infty)\]

Graphical Representation

Final Answer

(i) \( \{-1,0,1,2,\dots\} \)
(ii) \( x \in (-2,\infty) \)

Significance

• Board Exams: Tests correct algebraic transformation and interval notation.
• JEE / NEET: Fundamental in solving compound inequalities and domain-based problems.
• Key Insight: Same inequality gives discrete vs continuous answers depending on domain.

Theory

In linear inequalities, variables can be shifted to one side and constants to the other. The goal is to isolate \(x\). Since we are only adding/subtracting terms here, the inequality sign remains unchanged.

The solution will be expressed as an interval on the real number line.

Solution Roadmap

1. Bring all variable terms to one side
2. Shift constants to the other side
3. Simplify
4. Write solution in interval form

Solution

$$\begin{aligned} x + 3 &\lt 5x + 7 \\ x - 5x &\lt 7 - 3 \\ -4x &\lt 4 \\ x &\gt -1 \end{aligned}$$

Therefore, the solution set is: \[x \in (-1,\infty)\]

Graphical Representation

Final Answer

\(x \in (-1,\infty)\)

Significance

• Board Exams: Tests algebraic manipulation and correct handling of inequality signs.
• JEE / NEET: Builds foundation for solving inequalities involving polynomials and rational expressions.
• Common Error: Mistakes in simplifying \(x - 5x\); always combine like terms carefully.

Theory

In linear inequalities, we isolate the variable by shifting terms. When dividing both sides by a negative number, the inequality sign must be reversed.

Final answers for real numbers are expressed in interval notation.

Solution Roadmap

1. Bring variable terms to one side
2. Shift constants to the other side
3. Factor and simplify
4. Reverse inequality if dividing by negative
5. Write interval form

Solution

$$\begin{aligned} 3x - 7 &\gt 5x - 1 \\ 3x - 5x &\gt -1 + 7 \\ -2x &\gt 6 \\ x &\lt -3 \end{aligned}$$

Therefore, the solution set is: \[x \in (-\infty, -3)\]

Graphical Representation

Final Answer

\(x \in (-\infty, -3)\)

Significance

• Board Exams: Tests correct handling of negative coefficients and sign reversal.
• JEE / NEET: Core concept used in solving inequalities involving quadratic and rational expressions.
• Common Error: Forgetting to reverse inequality after dividing by \(-2\).

Theory

When inequalities contain brackets, first apply the distributive law to remove them. Then solve like a standard linear inequality.

Since no multiplication/division by a negative number is involved, the inequality sign remains unchanged.

Solution Roadmap

1. Expand both sides
2. Bring variable terms together
3. Shift constants
4. Write solution in interval form

Solution

$$\begin{aligned} 3(x-1) &\leq 2(x-3) \\ 3x - 3 &\leq 2x - 6 \\ 3x - 2x &\leq -6 + 3 \\ x &\leq -3 \end{aligned}$$

Therefore, the solution set is: \[x \in (-\infty,-3]\]

Graphical Representation

Final Answer

\(x \in (-\infty,-3]\)

Significance

• Board Exams: Tests correct expansion of brackets and simplification.
• JEE / NEET: Important for inequalities involving expressions and functions.
• Common Error: Mistakes while expanding \(3(x-1)\) or sign handling in constants.

Theory

When solving inequalities involving brackets, first apply the distributive property. After simplification, isolate the variable.

If the coefficient of \(x\) becomes negative, remember to reverse the inequality sign.

Solution Roadmap

1. Expand both sides
2. Bring like terms together
3. Simplify inequality
4. Reverse sign if dividing by negative

Solution

$$\begin{aligned} 3(2-x) &\geq 2(1-x) \\ 6 - 3x &\geq 2 - 2x \\ -3x + 2x &\geq 2 - 6 \\ -x &\geq -4 \\ x &\leq 4 \end{aligned}$$

Therefore, the solution set is: \[x \in (-\infty, 4]\]

Graphical Representation

Final Answer

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

Significance

• Board Exams: Tests expansion of brackets and inequality simplification.
• JEE / NEET: Important foundation for inequalities involving expressions and functions.
• Common Error: Forgetting to reverse inequality when dividing by negative (critical concept).

Theory

When an inequality involves fractions, it is often simplified by taking the LCM of denominators. This helps convert the expression into a single fraction.

After simplification, solve like a standard linear inequality.

Solution Roadmap

1. Take LCM of denominators
2. Combine terms into a single fraction
3. Simplify inequality
4. Isolate \(x\)

Solution

$$\begin{aligned} x + \dfrac{x}{2} + \dfrac{x}{3} &\lt 11 \\ \dfrac{6x + 3x + 2x}{6} &\lt 11 \\ \dfrac{11x}{6} &\lt 11 \\ 11x &\lt 66 \\ x &\lt 6 \end{aligned}$$

Therefore, the solution set is: \[x \in (-\infty, 6)\]

Graphical Representation

Final Answer

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

Significance

• Board Exams: Tests ability to handle fractional expressions using LCM.
• JEE / NEET: Important for inequalities involving rational expressions.
• Key Insight: Converting multiple fractions into a single expression simplifies solving significantly.

Theory

For inequalities involving fractions, eliminate denominators using the LCM. Multiplying both sides by a positive number keeps the inequality sign unchanged.

After simplification, solve like a standard linear inequality.

Solution Roadmap

1. Take LCM of denominators
2. Multiply both sides to remove fractions
3. Simplify and isolate \(x\)
4. Write interval form

Solution

$$\begin{aligned} \dfrac{x}{3} &> \dfrac{x}{2} + 1 \\\\ \text{Multiply }&\text{both sides by 6:} \\ 2x &> 3x + 6 \\ 2x - 3x &> 6 \\ -x &> 6 \\ x &< -6 \end{aligned}$$

Therefore, the solution set is: \[x \in (-\infty, -6)\]

Graphical Representation

Final Answer

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

Significance

• Board Exams: Tests correct handling of fractional inequalities using LCM.
• JEE / NEET: Core technique used in rational and algebraic inequalities.
• Common Error: Incorrectly combining fractions like \(\dfrac{x}{2} + 1 = \dfrac{x+2}{2}\) (this step is invalid).

Theory

For inequalities involving fractions, eliminate denominators using the LCM. Since the LCM here is positive, the inequality sign remains unchanged.

Also note that \(2 - x = -(x - 2)\), which can simplify expressions conceptually.

Solution Roadmap

1. Remove denominators using LCM
2. Expand both sides
3. Collect like terms
4. Solve for \(x\)

Solution

$$\begin{aligned} \dfrac{3(x-2)}{5} &\leq \dfrac{5(2-x)}{3} \\\\ \text{Multiply bo}&\text{th sides by 15:} \\ 9(x-2) &\leq 25(2-x) \\ 9x - 18 &\leq 50 - 25x \\ 9x + 25x &\leq 50 + 18 \\ 34x &\leq 68 \\ x &\leq 2 \end{aligned}$$

Therefore, the solution set is: \[x \in (-\infty, 2]\]

Graphical Representation

Final Answer

\(x \in (-\infty, 2]\)

Significance

• Board Exams: Tests fraction handling and correct expansion.
• JEE / NEET: Important for rational inequalities and sign-based transformations.
• Key Insight: Recognizing \(2 - x = -(x - 2)\) can simplify mental calculations.

Theory

For inequalities involving nested fractions, first simplify the inner expression, then remove denominators using the LCM. Since the LCM is positive, the inequality sign remains unchanged.

Solution Roadmap

1. Simplify bracket expressions
2. Convert to a single fraction
3. Eliminate denominators using LCM
4. Solve the resulting linear inequality

Solution

$$\begin{aligned} \dfrac{1}{2}\left( \dfrac{3x}{5}+4\right) &\geq \dfrac{1}{3}(x-6) \\ \dfrac{1}{2}\left( \dfrac{3x+20}{5} \right) &\geq \dfrac{1}{3}(x-6) \\ \dfrac{3x+20}{10} &\geq \dfrac{x-6}{3} \\\\ \text{Multiply bo}&\text{th sides by 30:} \\ 3(3x+20) &\geq 10(x-6) \\ 9x + 60 &\geq 10x - 60 \\ 9x - 10x &\geq -60 - 60 \\ -x &\geq -120 \\ x &\leq 120 \end{aligned}$$

Therefore, the solution set is: \[x \in (-\infty, 120]\]

Graphical Representation

Final Answer

\(x \in (-\infty, 120]\)

Significance

• Board Exams: Tests handling of nested fractions and algebraic simplification.
• JEE / NEET: Important for multi-step inequalities and rational expressions.
• Common Error: Mistyping denominator (like \(10\) written incorrectly) or skipping fraction simplification.

Theory

For inequalities with brackets, first expand using the distributive property. Then simplify by collecting like terms and isolate the variable.

If division by a negative number occurs, the inequality sign must be reversed.

Solution Roadmap

1. Expand both sides
2. Simplify expressions
3. Bring variables to one side
4. Solve and write interval form

Solution

$$\begin{aligned} 2(2x+3) - 10 &\lt 6(x-2) \\ 4x + 6 - 10 &\lt 6x - 12 \\ 4x - 4 &\lt 6x - 12 \\ 4x - 6x &\lt -12 + 4 \\ -2x &\lt -8 \\ x &\gt 4 \end{aligned}$$

Therefore, the solution set is: \[x \in (4,\infty)\]

Graphical Representation

Final Answer

\(x \in (4,\infty)\)

Significance

• Board Exams: Tests bracket expansion and correct simplification.
• JEE / NEET: Foundation for inequalities involving linear expressions and transformations.
• Common Error: Mistakes in combining constants (like \(6 - 10\)).

Theory

When expressions involve brackets with subtraction, carefully distribute the negative sign: \(a - (b+c) = a - b - c\).

After expansion, simplify both sides and isolate the variable.

Solution Roadmap

1. Expand both sides (handle negative sign carefully)
2. Simplify expressions
3. Bring variable terms together
4. Solve and write interval form

Solution

$$\begin{aligned} 37-(3x+5) &\geq 9x - 8(x-3) \\ 37 - 3x - 5 &\geq 9x - 8x + 24 \\ 32 - 3x &\geq x + 24 \\ -3x - x &\geq 24 - 32 \\ -4x &\geq -8 \\ x &\leq 2 \end{aligned}$$

Therefore, the solution set is: \[x \in (-\infty, 2]\]

Graphical Representation

Final Answer

\(x \in (-\infty, 2]\)

Significance

• Board Exams: Tests handling of negative sign before brackets.
• JEE / NEET: Important for multi-step simplification in inequalities.
• Common Error: Missing sign change in \(37 - (3x+5)\).

Theory

For inequalities involving multiple fractions, first take the LCM of denominators and combine terms into a single expression.

Then eliminate denominators and solve the resulting linear inequality.

Solution Roadmap

1. Take LCM of 3 and 5 to combine RHS
2. Simplify numerator
3. Multiply both sides to remove denominators
4. Solve for \(x\)

Solution

$$\begin{aligned} \dfrac{x}{4} &< \dfrac{5x-2}{3} - \dfrac{7x-3}{5} \\ \dfrac{x}{4} &< \dfrac{5(5x-2) - 3(7x-3)}{15} \\ \dfrac{x}{4} &< \dfrac{25x - 10 - 21x + 9}{15} \\ \dfrac{x}{4} &< \dfrac{4x - 1}{15} \\\\ \text{Multiply}&\text{ both sides by 60:} \\ 15x &< 4(4x - 1) \\ 15x &< 16x - 4 \\ -x &< -4 \\ x &> 4 \end{aligned}$$

Therefore, the solution set is: \[x \in (4,\infty)\]

Graphical Representation

Final Answer

\(x \in (4,\infty)\)

Significance

• Board Exams: Tests multi-step fraction simplification and LCM handling.
• JEE / NEET: Important for rational inequalities and algebraic transformations.
• Common Error: Skipping steps while combining fractions or incorrectly multiplying both sides.

Theory

For inequalities involving multiple fractions, first combine terms using the LCM, then eliminate denominators by multiplying both sides with a positive number.

Care must be taken while handling expressions like \(2 - x\), as sign errors are common.

Solution Roadmap

1. Take LCM of 4 and 5 to simplify RHS
2. Simplify numerator carefully
3. Multiply both sides by LCM (60)
4. Solve resulting linear inequality

Solution

$$\begin{aligned} \dfrac{2x-1}{3} &\geq \dfrac{3x-2}{4} - \dfrac{2-x}{5} \\ \dfrac{2x-1}{3} &\geq \dfrac{5(3x-2) - 4(2-x)}{20} \\ \dfrac{2x-1}{3} &\geq \dfrac{15x - 10 - 8 + 4x}{20} \\ \dfrac{2x-1}{3} &\geq \dfrac{19x - 18}{20} \\\\ \text{Multiply b}&\text{oth sides by 60:} \\ 20(2x - 1) &\geq 3(19x - 18) \\ 40x - 20 &\geq 57x - 54 \\ 40x - 57x &\geq -54 + 20 \\ -17x &\geq -34 \\ x &\leq 2 \end{aligned}$$

Therefore, the solution set is: \[x \in (-\infty, 2]\]

Graphical Representation

Final Answer

\(x \in (-\infty, 2]\)

Significance

• Board Exams: Tests multi-step fraction handling and algebraic accuracy.
• JEE / NEET: Important for rational inequalities and sign-sensitive simplifications.
• Common Error: Mistakes in expanding \(2 - x\) or skipping LCM step.

Theory

A linear inequality is solved by isolating the variable. When adding or subtracting terms, the inequality sign remains unchanged.

The solution represents all values of \(x\) satisfying the inequality, typically written in interval form.

Solution Roadmap

1. Bring variable terms to one side
2. Shift constants to the other side
3. Simplify
4. Write final answer in interval form

Solution

$$\begin{aligned} 3x - 2 &\lt 2x + 1 \\ 3x - 2x &\lt 1 + 2 \\ x &\lt 3 \end{aligned}$$

Therefore, the solution set is: \[x \in (-\infty, 3)\]

Graphical Representation

Final Answer

\(x \in (-\infty, 3)\)

Significance

• Board Exams: Direct application of basic linear inequality solving.
• JEE / NEET: Forms the foundation for solving compound and higher-degree inequalities.
• Key Insight: Simple inequalities like this build accuracy and speed for complex problems.

Theory

Linear inequalities are solved by isolating the variable. Addition or subtraction of the same quantity on both sides does not change the inequality sign.

The final answer is expressed as an interval representing all valid values of \(x\).

Solution Roadmap

1. Bring variable terms to one side
2. Shift constants to the other side
3. Simplify and isolate \(x\)
4. Write in interval notation

Solution

$$\begin{aligned} 5x - 3 &\geq 3x - 5 \\ 5x - 3x &\geq -5 + 3 \\ 2x &\geq -2 \\ x &\geq -1 \end{aligned}$$

Therefore, the solution set is: \[x \in [-1,\infty)\]

Graphical Representation

Final Answer

\(x \in [-1,\infty)\)

Significance

• Board Exams: Tests correct handling of inequality and interval notation.
• JEE / NEET: Fundamental for solving compound inequalities and interval intersections.
• Common Error: Forgetting that equality (\(\geq\)) includes the boundary point.

Theory

When solving inequalities with brackets, first expand using the distributive law. Then collect like terms and isolate the variable.

If division by a negative number occurs, the inequality sign must be reversed.

Solution Roadmap

1. Expand both sides
2. Bring variable terms together
3. Simplify inequality
4. Write solution in interval form

Solution

$$\begin{aligned} 3(1-x) &\lt 2(x+4) \\ 3 - 3x &\lt 2x + 8 \\ -3x - 2x &\lt 8 - 3 \\ -5x &\lt 5 \\ x &\gt -1 \end{aligned}$$

Therefore, the solution set is:

\[x \in (-1,\infty)\]

Graphical Representation

Final Answer

\(x \in (-1,\infty)\)

Significance

• Board Exams: Tests bracket expansion and inequality simplification.
• JEE / NEET: Builds foundation for solving inequalities with expressions and transformations.
• Common Error: Forgetting to reverse inequality after dividing by \(-5\).

Theory

For inequalities involving multiple fractions, first combine terms using the LCM. Then eliminate denominators by multiplying both sides with a positive number.

Since the LCM is positive, the inequality sign remains unchanged.

Solution Roadmap

1. Take LCM of 3 and 5 to combine RHS
2. Simplify numerator
3. Multiply both sides by LCM (30)
4. Solve the resulting inequality

Solution

$$\begin{aligned} \dfrac{x}{2} &\geq \dfrac{5x-2}{3} - \dfrac{7x-3}{5} \\ \dfrac{x}{2} &\geq \dfrac{5(5x-2) - 3(7x-3)}{15} \\ \dfrac{x}{2} &\geq \dfrac{25x - 10 - 21x + 9}{15} \\ \dfrac{x}{2} &\geq \dfrac{4x - 1}{15} \\\\ \text{Multiply }&\text{both sides by 30:} \\ 15x &\geq 2(4x - 1) \\ 15x &\geq 8x - 2 \\ 15x - 8x &\geq -2 \\ 7x &\geq -2 \\ x &\geq -\dfrac{2}{7} \end{aligned}$$

Therefore, the solution set is: \[x \in \left[-\dfrac{2}{7}, \infty\right)\]

Graphical Representation

Final Answer

\(x \in \left[-\dfrac{2}{7}, \infty\right)\)

Significance

• Board Exams: Tests LCM-based simplification and fraction handling.
• JEE / NEET: Important for rational inequalities and algebraic manipulation.
• Common Error: Incorrectly multiplying only one side (like writing \(\frac{15x}{2}\)).

Theory

Word problems involving averages can be converted into inequalities using the formula:

Average = (Sum of observations) / (Number of observations)

The phrase "at least" indicates a \(\geq\) inequality.

Solution Roadmap

1. Let unknown marks be \(x\)
2. Form average inequality
3. Solve inequality
4. Interpret result in context

Solution

Let the marks obtained in the third test be \(x\).

Total marks in three tests: \[70 + 75 + x = 145 + x\]

Given that the average should be at least 60:

$$\begin{aligned} \frac{145 + x}{3} &\geq 60 \\ 145 + x &\geq 180 \\ x &\geq 35 \end{aligned}$$

Therefore, Ravi must score at least 35 marks.

Final Answer

Minimum marks required \(= 35\)

Significance

• Board Exams: Tests ability to convert real-life situations into inequalities.
• JEE / NEET: Important for algebraic modeling and inequality-based word problems.
• Key Insight: Words like “at least”, “minimum” always imply \(\geq\).

Theory

Problems involving averages can be modeled using:

Average = (Total marks) / (Number of examinations)

The phrase "at least" implies a \(\geq\) inequality.

Solution Roadmap

1. Let required marks be \(x\)
2. Compute sum of known marks
3. Form inequality using average condition
4. Solve and interpret result

Solution

Let the marks obtained in the fifth examination be \(x\).

Sum of marks in first four examinations: \[87 + 92 + 94 + 95 = 368\]

Total marks in five examinations: \[368 + x\]

Given that the average must be at least 90:

$$\begin{aligned} \frac{368 + x}{5} &\geq 90 \\ 368 + x &\geq 450 \\ x &\geq 82 \end{aligned}$$

Therefore, Sunita must score at least 82 marks.

Final Answer

Minimum marks required \(= 82\)

Significance

• Board Exams: Tests conversion of real-life conditions into inequalities.
• JEE / NEET: Strengthens algebraic modeling and inequality formulation.
• Key Insight: “At least” → \(\geq\), and always convert average into total-based inequality.

Theory

Consecutive odd integers differ by 2. If the first odd integer is \(x\), then the next is \(x + 2\).

Such problems are solved by forming simultaneous inequalities based on given conditions.

Solution Roadmap

1. Assume consecutive odd integers
2. Apply upper bound condition
3. Apply sum condition
4. Find common solution satisfying both

Solution

Let the smaller odd integer be \(x\). Then the next consecutive odd integer is \(x + 2\).

Since both integers are less than 10: \[x + 2 < 10 \Rightarrow x < 8\]

Given that their sum is greater than 11:

$$\begin{aligned} x + (x + 2) &> 11 \\ 2x + 2 &> 11 \\ 2x &> 9 \\ x &> \dfrac{9}{2} \end{aligned}$$

Combining both conditions: \[\dfrac{9}{2} < x < 8\]

Since \(x\) must be a positive odd integer, possible values are: \[x = 5, \; 7\]

Corresponding pairs: \[ (5,7), \; (7,9) \]

Final Answer

Required pairs are \( (5,7) \) and \( (7,9) \)

Significance

• Board Exams: Tests formation of inequalities from word conditions.
• JEE / NEET: Strengthens handling of domain restrictions and discrete solutions.
• Key Insight: Always apply both inequality constraints before selecting valid integer values.

Theory

Consecutive even integers differ by 2. If the first even integer is \(x\), the next is \(x + 2\).

Such problems require forming inequalities and applying domain restrictions (even, positive, greater than 5).

Solution Roadmap

1. Assume consecutive even integers
2. Apply lower bound condition
3. Apply sum condition
4. Find values satisfying both conditions

Solution

Let the smaller even integer be \(x\). Then the next consecutive even integer is \(x + 2\).

Since both integers are greater than 5: \[x > 5\]

Given that their sum is less than 23:

$$\begin{aligned} x + (x + 2) &< 23 \\ 2x + 2 &< 23 \\ 2x &< 21 \\ x &< \dfrac{21}{2} \end{aligned}$$

Combining both conditions: \[5 < x < \dfrac{21}{2}\]

Since \(x\) must be an even positive integer, possible values are: \[x = 6,\; 8,\; 10\]

Corresponding pairs: \[ (6,8),\; (8,10),\; (10,12) \]

Final Answer

Required pairs are \( (6,8), (8,10), (10,12) \)

Significance

• Board Exams: Tests inequality formation with domain restrictions.
• JEE / NEET: Important for discrete solution filtering and integer constraints.
• Key Insight: Always apply both inequality and domain (even numbers) before final selection.

Theory

Let the shortest side be \(x\). Then:

• Longest side \(= 3x\)
• Third side \(= 3x - 2\)

The perimeter is the sum of all three sides. The phrase "at least" implies a \(\geq\) inequality.

Also, for a valid triangle, the sum of any two sides must be greater than the third side.

Solution Roadmap

1. Express all sides in terms of \(x\)
2. Form perimeter inequality
3. Solve inequality
4. Verify triangle validity

Solution

Let the shortest side be \(x\).

Then the sides are: \[x,\; 3x,\; 3x - 2\]

Perimeter condition:

$$\begin{aligned} x + 3x + (3x - 2) &\geq 61 \\ 7x - 2 &\geq 61 \\ 7x &\geq 63 \\ x &\geq 9 \end{aligned}$$

Triangle Validity Check

Check triangle inequality: \[\begin{aligned}x + (3x - 2) &> 3x \\ \Rightarrow 4x - 2 &> 3x \\ \Rightarrow x &> 2\end{aligned}\]

This condition is satisfied for \(x \geq 9\), hence the triangle is valid.

Final Answer

Minimum length of the shortest side \(= 9\) cm

Significance

• Board Exams: Tests modeling of real-life geometry into inequalities.
• JEE / NEET: Combines algebra with triangle inequality concepts.
• Key Insight: Always verify triangle inequality after forming algebraic expressions.

Theory

Let the shortest length be \(x\). Then:

• Second length \(= x + 3\)
• Third length \(= 2x\)

The total length cannot exceed the board length, giving a \(\leq\) inequality.

The phrase "at least 5 cm longer" implies a \(\geq\) inequality.

Solution Roadmap

1. Express all lengths in terms of \(x\)
2. Apply total length constraint
3. Apply comparison condition
4. Find intersection of inequalities

Solution

Let the shortest length be \(x\).

Total length condition:

$$\begin{aligned} x + (x + 3) + 2x &\leq 91 \\ 4x + 3 &\leq 91 \\ 4x &\leq 88 \\ x &\leq 22 \end{aligned}$$

Given that the third piece is at least 5 cm longer than the second:

$$\begin{aligned} 2x &\geq (x + 3) + 5 \\ 2x &\geq x + 8 \\ x &\geq 8 \end{aligned}$$

Combining both conditions: \[8 \leq x \leq 22\]

Final Answer

Possible lengths of the shortest piece: \(x \in [8, 22]\)

Significance

• Board Exams: Tests multi-condition inequality modeling.
• JEE / NEET: Strengthens skills in forming and intersecting inequalities.
• Key Insight: Always combine multiple constraints to get the final feasible range.