LINEAR INEQUALITIES

Theory Insight:
A compound inequality of the form \(a \leq f(x) \leq b\) represents two simultaneous inequalities: \[ a \leq f(x) \quad \text{and} \quad f(x) \leq b \] The solution is the intersection of both conditions. This is a foundational concept for interval notation and graphical interpretation on the number line.

Solution Roadmap:

• Break the compound inequality into two separate inequalities
• Solve each inequality step-by-step
• Find the common solution (intersection)
• Express answer in interval form and visualize on number line

Solution

$$\begin{aligned} 2 &\leq 3x - 4 \leq 5 \\\ 2 &\leq 3x - 4 \\ 3x - 4 &\geq 2 \\ 3x &\geq 6 \\ x &\geq 2 \\\ 3x - 4 &\leq 5 \\ 3x &\leq 9 \\ x &\leq 3 \\\ \therefore \; x &\in [2,3] \end{aligned}$$

Why this matters (Boards + Competitive Exams):

• Direct application of compound inequality solving — frequently asked in CBSE exams
• Builds foundation for interval notation used in calculus and coordinate geometry
• Essential for JEE/NEET when dealing with modulus, quadratic inequalities, and solution regions
• Graphical interpretation strengthens conceptual clarity for feasible regions in Linear Programming

Theory Insight:
When solving compound inequalities involving multiplication or division by a negative number, the inequality sign reverses. This is a critical rule: \[ \text{If } a < b \Rightarrow -a > -b \] The given inequality represents two simultaneous conditions, and the final solution is their intersection.

Solution Roadmap:

• Split the compound inequality into two parts
• Simplify each inequality carefully
• Reverse inequality sign when dividing by negative
• Find the common solution set
• Represent using interval notation and number line

Solution

$$\begin{aligned} 6 \leq -3(2x - 4) &< 12 \\\ -3(2x - 4) &\geq 6 \\ -6x + 12 &\geq 6 \\ -6x &\geq -6 \\ x &\leq 1 \\\ -3(2x - 4) &< 12 \\ -6x + 12 &< 12 \\ -6x &< 0 \\ x &> 0 \\\ \therefore \; x &\in (0,1] \end{aligned}$$

Why this matters (Boards + Competitive Exams):

• Tests the critical concept of inequality reversal — a common CBSE mistake area
• Frequently appears in JEE/NEET in disguised forms (modulus, rational inequalities)
• Builds precision in handling sign-sensitive algebraic transformations
• Strengthens understanding of open vs closed intervals in number line representation

Theory Insight:
When inequalities involve fractions, it is often efficient to eliminate denominators first (by multiplying throughout by the LCM). Also, remember:

• Multiplying by a positive number → inequality sign remains same
• Multiplying by a negative number → inequality sign reverses

This is a compound inequality, so we solve two parts and take their intersection.

Solution Roadmap:

• Split the compound inequality into two inequalities
• Remove fraction by multiplying with 2
• Solve each inequality carefully
• Handle sign reversal correctly when dividing by negative
• Intersect both solutions and express in interval form

Solution

$$\begin{aligned} -3 \leq 4 - \frac{7x}{2} &\leq 18 \\\ 4 - \frac{7x}{2} &\geq -3 \\ 8 - 7x &\geq -6 \quad (\text{multiplying by } 2) \\ -7x &\geq -14 \\ x &\leq 2 \\\ 4 - \frac{7x}{2} &\leq 18 \\ 8 - 7x &\leq 36 \\ -7x &\leq 28 \\ x &\geq -4 \\\ \therefore \; x &\in [-4, 2] \end{aligned}$$

Why this matters (Boards + Competitive Exams):

• Tests handling of fractional inequalities — very common in CBSE exams
• Builds speed by encouraging denominator elimination strategy
• Critical for JEE/NEET when solving rational and modulus inequalities
• Improves accuracy in sign handling during transformations

Theory Insight:
In compound inequalities involving fractions, multiply throughout by the LCM to simplify. Since we multiply by a positive number (5), the inequality signs remain unchanged. Always treat strict inequalities (<, >) carefully while writing interval notation.

Solution Roadmap:

• Split the compound inequality into two parts
• Multiply throughout by 5 to remove denominator
• Solve each inequality step-by-step
• Combine results and determine correct interval type

Solution

$$\begin{aligned} -15 < \frac{3(x-2)}{5} &\leq 0 \\\ \frac{3(x-2)}{5} &> -15 \\ 3(x-2) &> -75 \\ 3x - 6 &> -75 \\ 3x &> -69 \\ x &> -23 \\\ \frac{3(x-2)}{5} &\leq 0 \\ 3(x-2) &\leq 0 \\ 3x - 6 &\leq 0 \\ 3x &\leq 6 \\ x &\leq 2 \\\ \therefore \; x &\in (-23, 2] \end{aligned}$$

Why this matters (Boards + Competitive Exams):

• Tests proper handling of strict vs non-strict inequalities (very common CBSE error)
• Strengthens interval notation accuracy — crucial for board marking schemes
• Frequently used in JEE/NEET in rational and parameter-based inequalities
• Improves algebraic discipline in multi-step inequality transformations

Theory Insight:
First simplify the expression carefully. Note: \[ \frac{3x}{-5} = -\frac{3x}{5} \Rightarrow 4 - \left(-\frac{3x}{5}\right) = 4 + \frac{3x}{5} \] Reducing sign complexity early avoids major algebra mistakes. Then proceed as a standard compound inequality.

Solution Roadmap:

• Simplify the expression (handle negative denominator)
• Split into two inequalities
• Multiply by 5 to eliminate fraction
• Solve both and take intersection
• Write correct interval using strict/non-strict bounds

Solution

$$\begin{aligned} -12 < 4 + \frac{3x}{5} &\leq 2 \\\ 4 + \frac{3x}{5} &> -12 \\ \frac{3x}{5} &> -16 \\ 3x &> -80 \\ x &> -\frac{80}{3} \\\ 4 + \frac{3x}{5} &\leq 2 \\ \frac{3x}{5} &\leq -2 \\ 3x &\leq -10 \\ x &\leq -\frac{10}{3} \\\ \therefore \; x &\in \left(-\frac{80}{3}, -\frac{10}{3}\right] \end{aligned}$$

Why this matters (Boards + Competitive Exams):

• Tests handling of negative denominators — a frequent CBSE error trap
• Encourages simplification before solving (important for speed in exams)
• Common pattern in JEE/NEET rational inequalities
• Improves precision in fractional interval representation

Theory Insight:
For compound inequalities involving a fraction, multiply the entire inequality by the denominator (if positive) to simplify in one step. Since 2 is positive, inequality signs remain unchanged. This “single-step method” is faster and reduces algebraic errors.

Solution Roadmap:

• Multiply the entire inequality by 2
• Solve resulting linear compound inequality
• Isolate \(x\)
• Write final answer in interval form

Solution

$$\begin{aligned} 7 \leq \frac{3x+11}{2} &\leq 11 \\\ 14 \leq 3x+11 &\leq 22 \\ 14 - 11 \leq 3x &\leq 22 - 11 \\ 3 \leq 3x &\leq 11 \\ 1 \leq x &\leq \frac{11}{3} \\\ \therefore \; x \in \left[1, \frac{11}{3}\right] \end{aligned}$$

Why this matters (Boards + Competitive Exams):

• Demonstrates the fastest solving technique (multiply whole inequality at once)
• Reduces steps → saves time in CBSE board exams
• Important for JEE/NEET where multi-step inequalities appear under time pressure
• Builds foundation for solving chained inequalities in calculus and coordinate geometry

Theory Insight:
When two inequalities are given together, we solve them separately and take their intersection. Since both are strict inequalities, the final interval will be open at both ends.

Solution Roadmap:

• Solve each inequality independently
• Isolate \(x\) in both cases
• Take intersection of both results
• Represent solution on number line using open circles

Solution

$$\begin{aligned} 5x + 1 &> -24 \\ 5x &> -25 \\ x &> -5 \\\ 5x - 1 &< 24 \\ 5x &< 25 \\ x &< 5 \\\ \therefore \; x \in (-5, 5) \end{aligned}$$

Why this matters (Boards + Competitive Exams):

• Basic system of inequalities — very common CBSE question type
• Introduces concept of intersection of solution sets
• Frequently used in JEE/NEET for range-based constraints
• Strengthens graphical interpretation for Linear Programming problems

Theory Insight:
This is a system of linear inequalities. Each inequality gives a range of values of \(x\). The final solution is the intersection of both ranges. Expansion and simplification must be done carefully.

Solution Roadmap:

• Expand both inequalities
• Simplify and isolate \(x\)
• Find solution of each inequality
• Take intersection and express as interval

Solution

$$\begin{aligned} 2(x - 1) &< x + 5 \\ 2x - 2 &< x + 5 \\ x &< 7 \\\ 3(x + 2) &> 2 - x \\ 3x + 6 &> 2 - x \\ 4x &> -4 \\ x &> -1 \\\ \therefore \; x \in (-1, 7) \end{aligned}$$

Why this matters (Boards + Competitive Exams):

• Standard system of inequalities — very frequent in CBSE exams
• Builds algebraic manipulation speed (expansion + simplification)
• Core concept for JEE/NEET in constraint-based problems
• Foundation for graphical solution regions in Linear Programming

Theory Insight:
In a system of inequalities, the final solution is the intersection of all individual solutions. Sometimes one inequality gives a stronger restriction than the other — this is called the dominant constraint.

Solution Roadmap:

• Solve each inequality separately
• Reduce to simplest form
• Compare results and identify the stricter condition
• Write final solution as intersection

Solution

$$\begin{aligned} 3x - 7 &> 2(x - 6) \\ 3x - 7 &> 2x - 12 \\ x &> -5 \\\ 6 - x &> 11 - 2x \\ 6 + x &> 11 \\ x &> 5 \\\ \therefore \; x > 5 \end{aligned}$$

Why this matters (Boards + Competitive Exams):

• Illustrates concept of dominant inequality (important for faster solving)
• Common CBSE pattern where one inequality becomes redundant
• Very useful in JEE/NEET for constraint tightening problems
• Builds intuition for feasible regions extending to infinity

Theory Insight:
In systems of inequalities, expand expressions carefully and reduce to linear form. Pay special attention when dividing by negative numbers — inequality signs must reverse. Final solution is the intersection of both inequalities.

Solution Roadmap:

• Expand and simplify both inequalities
• Reduce to standard linear form
• Handle sign reversal carefully (if needed)
• Find intersection of both results

Solution

$$\begin{aligned} 5(2x - 7) - 3(2x + 3) &\leq 0 \\ 10x - 35 - 6x - 9 &\leq 0 \\ 4x - 44 &\leq 0 \\ 4x &\leq 44 \\ x &\leq 11 \\\ 2x + 19 &\leq 6x + 47 \\ 2x - 6x &\leq 47 - 19 \\ -4x &\leq 28 \\ x &\geq -7 \\\ \therefore \; x \in [-7, 11] \end{aligned}$$

Why this matters (Boards + Competitive Exams):

• Combines expansion + inequality solving — common CBSE exam pattern
• Tests sign reversal accuracy (high mistake probability area)
• Frequently appears in JEE/NEET as part of multi-constraint problems
• Builds strong base for solving feasible regions in Linear Programming

Theory Insight:
Temperature conversion between Fahrenheit and Celsius follows a linear relation. When a range is given in one unit, convert it into the other by forming a compound inequality. Since the transformation is linear and increasing, the order of inequality remains unchanged.

Solution Roadmap:

• Form compound inequality: \(68 \leq F \leq 77\)
• Substitute \(F = \frac{9}{5}C + 32\)
• Solve inequality step-by-step
• Express final answer in interval form

Solution

$$\begin{aligned} 68 \leq \frac{9}{5}C + 32 \leq 77 \\\ 68 - 32 \leq \frac{9}{5}C \leq 77 - 32 \\ 36 \leq \frac{9}{5}C \leq 45 \\\ 36 \cdot \frac{5}{9} \leq C \leq 45 \cdot \frac{5}{9} \\ 20 \leq C \leq 25 \\\ \therefore \; C \in [20, 25] \end{aligned}$$

Why this matters (Boards + Competitive Exams):

• Classic application-based inequality problem in CBSE exams
• Tests modelling skill: converting real-life statement → mathematical inequality
• Important for JEE/NEET in unit conversions and constraint problems
• Strengthens understanding of linear transformations preserving order

Theory Insight:
In mixture problems, concentration is defined as: \[ \text{Concentration} = \frac{\text{amount of solute}}{\text{total volume}} \] The final concentration must lie between two values, forming a compound inequality. This is a classic weighted-average inequality problem.

Solution Roadmap:

• Let added solution = \(x\) litres
• Compute total solute and total volume
• Form compound inequality for concentration
• Solve both inequalities and take intersection

Solution

$$\begin{aligned} \text{Let added solution} = x \text{ litres} \\\ \text{Solute in 8% solution} &= 0.08 \times 640 = 51.2 \\ \text{Solute in 2% solution} &= 0.02x \\\ \text{Total solute} &= 51.2 + 0.02x \\ \text{Total volume} &= 640 + x \\\ 0.04 < \frac{51.2 + 0.02x}{640 + x} &< 0.06 \\\ 0.04(640 + x) < 51.2 + 0.02x &< 0.06(640 + x) \end{aligned}$$ $$\begin{aligned} 25.6 + 0.04x &< 51.2 + 0.02x \\ 0.02x &< 25.6 \\ x &< 1280 \\\ 51.2 + 0.02x &< 38.4 + 0.06x \\ 12.8 &< 0.04x \\ x &> 320 \\\ \therefore \; x \in (320, 1280) \end{aligned}$$

Why this matters (Boards + Competitive Exams):

• High-weight application problem in CBSE board exams
• Tests mixture concept + inequality modelling simultaneously
• Very common in JEE/NEET (alligation and weighted averages)
• Builds real-life mathematical modelling skills

Theory Insight:
When water is added, the amount of solute (acid) remains constant, while total volume increases. Hence, concentration decreases. Such problems are modelled using: \[ \text{Concentration} = \frac{\text{fixed solute}}{\text{new volume}} \] This leads to a compound inequality.

Solution Roadmap:

• Let water added = \(x\) litres
• Compute fixed amount of acid
• Form inequality using new volume
• Solve both inequalities and intersect

Solution

$$\begin{aligned} \text{Let water added} = x \text{ litres} \\\ \text{Acid present} &= 0.45 \times 1125 = 506.25 \\\ \text{New volume} &= 1125 + x \\\ 0.25 < \frac{506.25}{1125 + x} &< 0.30 \end{aligned}$$ $$\begin{aligned} 0.25(1125 + x) &< 506.25 \\ 281.25 + 0.25x &< 506.25 \\ 0.25x &< 225 \\ x &< 900 \\\ 506.25 &< 0.30(1125 + x) \\ 506.25 &< 337.5 + 0.30x \\ 168.75 &< 0.30x \\ x &> 562.5 \\\ \therefore \; x \in (562.5, 900) \end{aligned}$$

Why this matters (Boards + Competitive Exams):

• Classic dilution problem — very common in CBSE board exams
• Tests understanding of constant solute principle
• Important for JEE/NEET mixture and concentration problems
• Builds modelling skills for real-life quantitative scenarios

Theory Insight:
This is a direct application of linear transformation in inequalities. Since IQ is proportional to mental age (MA), the inequality can be solved by simple scaling. As all multipliers are positive, inequality signs remain unchanged.

Solution Roadmap:

• Substitute \(CA = 12\) into the formula
• Form compound inequality for IQ
• Convert into inequality in terms of MA
• Simplify and express final interval

Solution

$$\begin{aligned} IQ &= \frac{MA}{12} \times 100 \\\ 80 \leq \frac{MA}{12} \times 100 &\leq 140 \\\ 0.8 \leq \frac{MA}{12} &\leq 1.4 \\ 9.6 \leq MA &\leq 16.8 \\\ \therefore \; MA \in [9.6,\;16.8] \end{aligned}$$

Why this matters (Boards + Competitive Exams):

• Tests ability to convert formula-based statements into inequalities
• Common CBSE application question (high scoring if steps are clear)
• Builds understanding of proportional relationships
• Useful in JEE/NEET for interpreting formula-based constraints