Q.Solve for x: x+14≤3≤x+16, (x>0).
Concept understanding — Linear Inequality Solutions
Linear Inequality Solutions – A First Look
Imagine you're standing on a number line. You know exactly where the number 5 is. But what if I asked you to stand on "all numbers greater than 5"? You can't stand on all of them at once — they stretch infinitely to the right. That's the core idea of an inequality: instead of one exact point, you get a whole region of possible values.
A linear inequality is just like a linear equation (ax+b=0), but instead of an equals sign, you have one of these: <, >, ≤, or ≥. The solution is not a single number — it's an interval (or a union of intervals) on the number line.
From Equation to Inequality
Start with a simple equation:
2x+3=7
Solve it: 2x=4⟹x=2. One point.
Now change it to an inequality:
2x+3>7
Solve it the same way — but the meaning changes. Subtract 3: 2x>4. Divide by 2: x>2.
The solution is all numbers greater than 2. On a number line, you draw an open circle at 2 (because 2 itself is not included) and shade everything to the right.
If you multiply or divide both sides of an inequality by a negative number, the inequality sign reverses. For example: −x<5 becomes x>−5. This is the single most common mistake students make.
The Four Types of Solutions
| Inequality | Meaning | Number line representation |
|---|---|---|
| x>a | All numbers strictly greater than a | Open circle at a, shade right |
| x≥a | All numbers greater than or equal to a | Closed (filled) circle at a, shade right |
| x<a | All numbers strictly less than a | Open circle at a, shade left |
| x≤a | All numbers less than or equal to a | Closed circle at a, shade left |
The solution set is usually written in interval notation:
- x>2 → (2,∞)
- x≤−3 → (−∞,−3]
Parentheses ( or ) mean the endpoint is not included. Brackets [ or ] mean it is included.
Solving a Linear Inequality: Step by Step
Solve 3x−5≤7x+3.
-
Bring variable terms to one side:
3x−5−7x≤3
−4x−5≤3
-
Isolate the variable term:
−4x≤8
-
Divide by the coefficient (here it's −4, so reverse the sign):
x≥−2
The solution is x≥−2, or in interval notation: [−2,∞).
Always check your answer by testing a number from the solution set. For x≥−2, test x=0: 3(0)−5=−5 and 7(0)+3=3. Is −5≤3? Yes. Now test a number outside, say x=−3: 3(−3)−5=−14 and 7(−3)+3=−18. Is −14≤−18? No — so the inequality fails, confirming our solution is correct.
Why This Matters
Linear inequalities are the foundation for:
- Compound inequalities (like −2<x≤5)
- Absolute value inequalities (like ∣x∣<3)
- Systems of inequalities (used in linear programming)
- Quadratic and rational inequalities (where sign charts become essential)
The key takeaway: an inequality solution is a range of values, not a single point. The algebra is nearly identical to solving equations — except for that one critical rule about multiplying/dividing by negatives.
The solution of a linear inequality is an interval (or union of intervals) on the real number line. Always represent it with a number line sketch and interval notation in exams — both are often required for full marks.
Representing the solution set of a linear inequality using interval notation and number-line diagrams is a key expected skill in the NCERT Class 11 Mathematics chapter on Linear Inequalities, and "linear inequality solution set interval notation" is a commonly searched topic for CBSE board revision. This representation skill is frequently assessed alongside the solving steps in "linear inequalities important questions" for board and competitive-exam practice.
Linear Inequality Solutions
We have a compound inequality with rational expressions. The key is to split it into two parts and solve each while respecting the constraint x>0.
Step 1: From x+16≥3:
6≥3(x+1)⟹6≥3x+3⟹3≥3x⟹x≤1
Step 2: From x+14≤3:
4≤3(x+1)⟹4≤3x+3⟹1≤3x⟹x≥31
Since x>0, we have x+1>1>0, so multiplying by (x+1) preserves inequality directions.
Step 3: Combine both conditions with x>0:
31≤x≤1
This already satisfies x>0.
The solution is x∈[31,1].
Splitting the double inequality (with x+1>0 since x>0) gives 31≤x≤1.
Since x>0, we have x+1>0, so multiplying by x+1 preserves the inequality directions. Split the compound inequality into two parts.
Part 1: x+14≤3
4≤3(x+1)⟹4≤3x+3⟹1≤3x⟹x≥31.
Part 2: 3≤x+16
3(x+1)≤6⟹x+1≤2⟹x≤1.
Taking the intersection of x≥31 and x≤1 (both consistent with x>0):
31≤x≤1.
The solution set is [31, 1], i.e. 31≤x≤1.
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