Linear Inequalities – Complete Guide with Solutions
Understanding Inequalities
What is an Inequality?
An inequality is a mathematical statement that compares two values or expressions to show that one is less than, greater than, less than or equal to, or greater than or equal to the other.
Unlike an equation (which says two things are equal), an inequality allows a range of possible values.
Common Inequality Symbols
Symbol
Meaning
Example
\( < \)
Less than
\( x < 10 \)
\( > \)
Greater than
\( y > 3 \)
\( \leq \)
Less than or equal to
\( x \leq 7 \)
\( \geq \)
Greater than or equal to
\( y \geq 2 \)
Why are Inequalities Useful?
Inequalities help us describe situations where values can vary but have limits or ranges, such as:
Budget constraints
Speed limits
Age or height restrictions
Temperature thresholds
Real-Life Example: Budget Constraint for Buying Groceries
Imagine you have ₹3000 to spend on fruits and vegetables this month.
You want to make sure your total spending does not exceed your budget.
Let:
\( x \) = amount spent on fruits (in ₹)
\( y \) = amount spent on vegetables (in ₹)
The inequality to represent your budget is:
\( x + y \leq 3000 \)
This means the combined spending on fruits and vegetables must be less than or equal to ₹3000.
Solution of Inequalities are Intervals
The solution to an inequality like \( x + y \leq 3000 \) is the set of all possible values of \( x \) and \( y \) that satisfy the inequality.
These solutions are often represented as intervals on the real number line because inequalities describe ranges or sets of values, not just single points.
For example, if we consider a simpler inequality like \( x \leq 5 \), the solution is all real numbers less than or equal to 5. This corresponds to the interval:
\[ (-\infty, 5] \]
Here, the round bracket ( at \(-\infty\) means the set extends indefinitely to the left, and the square bracket ] at 5 means 5 is included in the solution set.
Similarly, inequalities with two variables, like the grocery budget example, define a region of possible values that can be represented using intervals or more complex shapes.
Real-Life Example: Speed Limit on a Highway
Suppose the speed limit on a highway is 60 km/h.
You want to make sure your speed \( s \) (in km/h) does not exceed this limit.
The inequality representing this is:
\( s \leq 60 \)
This means your speed can be any value less than or equal to 60 km/h.
The solution to this inequality is the interval:
\[ (-\infty, 60] \]
which includes all speeds up to and including 60 km/h. (keep in mind speed can't be nagative so it will be \([0,60]\), for the shake of explantion we write \( (-\infty, 60] \) )
🔢 Types of Intervals
Interval Type
Symbolic Notation
Description
Open Interval
\((a, b)\)
All real numbers between \(a\) and \(b\), excluding both endpoints.
Closed Interval
\([a, b]\)
Includes both \(a\) and \(b\).
Half-Open (Left)
\([a, b)\)
Includes \(a\), excludes \(b\).
Half-Open (Right)
\((a, b]\)
Excludes \(a\), includes \(b\).
Infinite Interval
\[(-\infty, b]\]
,\[(a, \infty)\]
Extends to infinity; uses open brackets with \(\infty\).
Open vs Solid Circles
Endpoint Type
Symbol
Meaning
Included (Closed)
🔴 Solid Circle
The endpoint is part of the interval.
Excluded (Open)
⚪ Open Circle
The endpoint is not part of the interval.
📏 Steps to Draw an Interval
Draw a horizontal line with a center mark labeled 0.
Identify the type of interval and its endpoints.
Mark the endpoints:
Use 🔴 for included (closed) points.
Use ⚪ for excluded (open) points.
Draw a bold segment or arrow connecting the points based on interval type.
🧠 Examples
Example 1: \( (2, 5) \)
⚪──────⚪ (open on both ends)
Example 2: \( [2, 5] \)
🔴──────🔴 (closed on both ends)
Example 3: \( [2, 5) \)
🔴──────⚪ (left closed, right open)
Example 4: \( (-\infty, 3] \)
<──────🔴 (arrow to left, closed at 3)
Example 5: \( (0, \infty) \)
⚪──────> (open at 0, continues right)
Visual Representation of Intervals on the Real Number Line
Types of Tips We'll Include:
🧠 Concept Tips
Inequality sign flips when multiplying/dividing by a negative number.
Never include a value that makes the denominator zero — it's always excluded.
📏 Graphing Tips
Use open circles for strict inequalities like < and >, closed circles for ≤, ≥.
Infinity is always open — never include ∞ or -∞ in your solution.
🚨 Exam Tips
Always test one value from each interval — never assume signs without checking.
Watch out for trap options in MCQs: endpoints included/excluded.
🎯 Shortcut Tips
If both numerator and denominator are linear, test sign changes around roots quickly.
You can directly use sign charts for rational inequalities to avoid long calculations.
Tips and Cautions: What NOT to Do While Solving Inequalities
💡 Do not forget to reverse the inequality sign when multiplying or dividing both sides by a negative number.
💡 Do not add or subtract incorrectly—always perform the same operation on both sides to keep the inequality balanced.
💡 Do not treat inequalities like equations; inequalities represent ranges, not just single values.
💡 Do not ignore absolute value rules—split absolute value inequalities into two cases properly.
💡 Do not forget to express the solution set clearly using interval notation or set-builder notation.
💡 Do not forget that dividing by zero is undefined—avoid dividing by variables unless you know their sign.
💡 Do not skip steps—write each algebraic manipulation carefully to avoid mistakes.
💡 Do not confuse "less than" (<) with "less than or equal to" (≤), and similarly with "greater than".
💡 Do not forget to check your solution by plugging values back into the original inequality when possible.
10 Simple Linear Inequality Problems
Example 1: Solve \( x + 4 > 7 \)
Problem: Solve for \( x \).
To isolate \( x \), subtract 4 from both sides of the inequality. This keeps the inequality balanced and maintains the direction since we are subtracting a positive number.
\[ x + 4 > 7 \] \[ x > 7 - 4 \]\[\Rightarrow x > 3 \]
Answer: \( x \in (3, \infty) \)
Example 2: Solve \( 2x - 5 \leq 9 \)
Problem: Solve for \( x \).
Add 5 to both sides to move constants to the right side, maintaining the inequality direction. Then divide both sides by 2 to solve for \( x \). Dividing by a positive number does not change the inequality.
Subtract 5 from both sides first. Then multiply both sides by -1 to get \( x \) alone. Remember when multiplying or dividing by a negative number, the inequality sign flips direction.
The absolute value inequality \( |x| < 6 \) means that the distance of \( x \) from 0 is less than 6. This can be rewritten as a compound inequality without absolute value: \( -6 < x < 6 \).
\[ -6 < x < 6 \]
Answer: \( x \in (-6, 6) \)
Example 6: Solve \( |x - 2| \geq 5 \)
Problem: Solve the absolute value inequality.
Absolute value inequality \( |x - 2| \geq 5 \) means the distance between \( x \) and 2 is at least 5. This splits into two cases:
1) \( x - 2 \geq 5 \) which gives \( x \geq 7 \)
2) \( x - 2 \leq -5 \) which gives \( x \leq -3 \)
\[x - 2 \geq 5\]\[ \Rightarrow x \geq 7\]
\[ x - 2 \leq -5\]\[ \Rightarrow x \leq -3\]
Answer: \( x \in (-\infty, -3] \cup [7, \infty) \)
Example 7: Solve \( -3x > 12 \)
Problem: Solve for \( x \).
Divide both sides by -3 to isolate \( x \). Because we are dividing by a negative number, the inequality direction reverses.
Move all terms containing \( x \) to one side and constants to the other:
Subtract \( 2x \) from both sides and subtract 1 from both sides, then divide by the positive number 2.
Now we see how to solve inequalities of the following types
1. Less than (<)
\[
\displaystyle \frac{ax+b}{cx+d} < k
\]
2. Greater than (>)
\[
\displaystyle \frac{ax+b}{cx+d} > k
\]
3. Less than or equal to (≤)
\[
\displaystyle \frac{ax+b}{cx+d} \leq k
\]
4. Greater than or equal to (≥)
\[
\displaystyle \frac{ax+b}{cx+d} \geq k
\]
Algorith to solve the inequaties of above type
Step 1: Rewrite the Inequality
Bring all terms to one side so the inequality compares to zero:
Step 2: Simplify the Expression
Combine into a single rational expression. Make the coefficient of \(x\) positive in both numerator and denominator if needed.
Step 3: Find Critical Points
Solve for the values where the numerator and denominator are zero. These critical points divide the real line into intervals:
\[
\text{Numerator} = 0,\quad \text{Denominator} = 0
\]
Step 4: Determine Sign on Intervals
In the rightmost interval, the expression is positive. Alternate signs in each region from right to left. Mark them accordingly on the number line.
Step 5: Write the Solution
Depending on the sign in the inequality (>, ≥, <, ≤), choose intervals that satisfy it. Exclude or include boundary points appropriately:
Exclude points where the denominator is zero (undefined)
Include points where the expression equals zero if the inequality is ≤ or ≥
Negative in intervals:
\[
(-\infty, 2.5)\quad \text{and} \quad (4.125, \infty)
\]
Note: We exclude \(x = 2.5\) since denominator becomes 0 and \(x = 4.125\) since inequality is strict (<).
Example 1 : Solve \(2x - 7 > 5 - x \), \(11 - 5x \leq 1\)
Solution: lets first solve 1. First Inequality: \(2x - 7 > 5 - x\)
\[
2x - 7 > 5 - x \Rightarrow 3x > 12 \Rightarrow x > 4
\]
So, the solution is:
\(x > 4\)
On Number Line:
2. Second Inequality: \(11 - 5x \leq 1\)
\[
11 - 5x \leq 1 \Rightarrow -5x \leq -10 \Rightarrow x \geq 2
\]
So, the solution is:
\(x \geq 2\)
On Number Line:
3. Final Step: Take the Intersection
We need to find the values of \(x\) that satisfy both inequalities. So will take the intersection of both the interval ie
\[(4, \infty) \cap [2, \infty)\]
in above we can see that \(2, \infty)\) lies inside \((4, \infty)\), thus solution will be that interval which is common i.e
\[(4, \infty)\]
Final Solution on Real Line:
✅ Final Answer: The solution to the system is\(\;x \in (4, \infty)\)
(Students will think why we not included \(4\) in answer, while 4 is already in \([2, \infty)\), this is beacsue 4 is absent in \((4, \infty)\), and we take the intersection, in intersection we just write the elements which are common in both sets.)
Sign of expression \(\displaystyle \frac{33x - 6}{7x - 1}\):
\(x < \displaystyle \frac{1}{7}\): both negative → positive ✅
\(\displaystyle \frac{1}{7} < x < \displaystyle \frac{2}{11}\): num neg, den pos → negative ❌
\(x > \displaystyle \frac{2}{11}\): both negative → positive ✅
required interval:
\[
\frac{1}{7} < x < \displaystyle \frac{2}{11}
\]
(student may get confuse why we take \(\displaystyle \frac{1}{7} < x < \displaystyle \frac{2}{11}\) as solution in this inequality, see inequality \(\displaystyle \frac{33x - 6}{7x - 1}<0\) should be negative, that is why we chosse this interval)
\(x < 8\): num negative, den negative → positive ✅
\(8 < x < 23\): num negative, den positive → negative ❌
\(x > 23\): both positive → positive ✅
Solution:
\[
8 < x < 23
\]
Final Answer:
From 1st inequality: \(\displaystyle \frac{1}{7}x < \displaystyle \frac{2}{11} \)
From 2nd inequality: \(8 < x < 23\)
No intersection between these intervals.
✅ Solution Set:No solution
Word problem
Example 4: A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it.
The resulting mixture should have a concentration of more than 4% but less than 6%.
If there are 640 liters of the 8% solution, how many liters of the 2% solution must be added?
Step 1: Define the variable
Let \( x \) be the number of liters of the 2% boric acid solution added.
Step 2: Set up the inequality
Total acid from 8% solution: \( 0.08 \times 640 = 51.2 \) liters
Acid from 2% solution: \( 0.02x \)
Total acid = \( 51.2 + 0.02x \)
Total volume = \( 640 + x \)
Desired concentration: between 4% and 6%
✅ Final Answer:
The number of liters of 2% solution to be added must be between:
\[
\boxed{320 < x < 1280}
\]
So, more than 320 liters but less than 1280 liters must be added.
Exmaple 5: A company manufacture bikes and it costs and revenue function for a month are C =3000+3x/2 and R = 5x respectively, where x is number of bikes produced and sold in a week. How many bikes must be sold for the earn profit
SolutionA company manufactures bikes and has the following:
Cost function: \( C = 3000 + \frac{3x}{2} \)
Revenue function: \( R = 5x \)
\(x\) = number of bikes produced and sold per week.
Question: How many bikes must be sold to earn a profit?
Step 1: Set up the inequality for profit
Profit means:
\[
R > C \Rightarrow 5x > 3000 + \frac{3x}{2}
\]
Word Problem: A number increased by 5 is less than 12. What is the number?
Answer: \( x + 5 < 12 \Rightarrow x < 7 \)
Word Problem: The length of a rectangle is more than twice its width. If the width is \( w \), express the inequality.
Answer: \( \text{Length} > 2w \)
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