Q.Solve 4−x5x+8<2,x∈R.
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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 …
Rearranging gives 4−x7x<0, so x and (4−x) must have opposite signs, giving x<0 or x>4. …
Bring everything to one side over a common denominator, then run a sign analysis.
4−x5x+8<2
4−x5x+8−2<0
4−x(5x+8)−2(4−x)<0
4−x5x+8−8+2x<0⟹4−x7x<0
Since 7>0, this requires 4−xx<0, i.e. x and (4−x) have opposite signs. Critical points: x=0, x=4.
- x<0: x<0, (4−x)>0 ⇒ ratio <0 — valid.
- 0<x<4: both positive ⇒ ratio >0 — invalid.
- x>4: x>0, (4−x)<0 ⇒ ratio <0 — valid. …
Showing the 12 most recent of 35 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.Solution of 4x+3<6x+7, x∈R is(a) (−∞,−2)(b) (−2,∞)(c) (2,∞)(d) (−∞,2)
›Reveal solutionSolution
Collect x-terms on one side and constants on the other, keeping the inequality direction (no division by a negative here).
4x+3<6x+7
Subtract 4x from both sides:
3<2x+7
Subtract 7:
−4<2x …
- CBSE 2026Set ANNUAL1 markMCQQ.Solution of 3x>2x+1, x∈R is(a) (−6,∞)(b) (−∞,−6)(c) (−∞,6)(d) None of these
›Reveal solutionSolution
Clear the fractions by multiplying by the LCM (6), then isolate x.
3x>2x+1
Multiply every term by 6 (positive, direction unchanged):
2x>3x+6
Subtract 3x:
−x>6 …
- CBSE 2026Set ANNUAL1 markMCQQ.Solution of −8≤5x−3<7, x∈R is(a) [−1,2)(b) (−1,2)(c) (−1,2](d) [−1,2]
›Reveal solutionSolution
Add 3 to all three parts of the compound inequality, then divide by 5, preserving direction and endpoint inclusivity throughout.
−8≤5x−3<7
Add 3 to all parts:
−5≤5x<10
Divide by 5 (positive):
−1≤x<2 …
- CBSE 2026Set ANNUAL1 markMCQQ.If x>−4 then the solution of the inequality will be —(a) (−4,∞)(b) (2,2)(c) (1,4)(d) (∞,−4)
›Reveal solutionSolution
The solution set of x>−4 is the open interval (−4,∞), option (a).
The inequality x>−4 describes every real number strictly greater than −4. In interval notation, a round bracket is used at −4 because −4 itself is NOT included (strict inequality), and the interval extends w …
- CBSE 2026Set ANNUAL1 markMCQQ.From the following, which value of x satisfies the inequality x+4>7?(a) −3(b) 1(c) 4(d) 3
›Reveal solutionSolution
Only x=4 satisfies x+4>7, option (c).
Solve the inequality: x+4>7⇒x>3.
Now test each option:
- x=−3: −3<3 — fails
- x=1: 1<3 — fails
- x=4: 4>3 — satisfies (4+4=8>7 ✓) …
- CBSE 2025Set ANNUAL1 markMCQQ.The solution set of the inequation 24x<100; x is a natural number, is(a) {0,1,2,3,4}(b) {1,2,3,4}(c) {0,1,2,3}(d) {4}
›Reveal solutionSolution
The solution set is {1,2,3,4}.
Solve 24x<100: dividing both sides by 24 (positive, so the inequality direction is unchanged), x<24100=4.16.
…
- CBSE 2025Set ANNUAL1 markMCQQ.Let a is a positive integer then ∣x∣>a is(a) x<−a or x>a(b) x>−a or x<a(c) x<−a or x>−a(d) None of these
›Reveal solutionSolution
∣x∣>a (with a>0) means x lies outside the interval [−a,a], i.e. x<−a or x>a.
By definition, ∣x∣ is the distance of x from 0. ∣x∣>a means this distance exceeds a, so x must lie strictly beyond a on either side of 0:
x>aorx<−a
…
- CBSE 2025Set sz1 markMCQQ.The solution of the inequality 7x−8≥6 is :(a) [2, \infty)(b) (2, 8)(c) [4, \infty)(d) (14, \infty)
›Reveal solutionSolution
7x−8≥6⟹7x≥14⟹x≥2, so the solution set is [2,∞).
Start with 7x−8≥6.
Add 8 to both sides:
7x≥14.
Divide both sides by 7 (a positive number, so the inequality direction is unchanged): …
- CBSE 2025Set ANNUAL1 markMCQQ.If −3x+17<−13, then(a) x∈(10,∞)(b) x∈[10,∞)(c) x∈(−∞,10)(d) x∈[−10,10]
›Reveal solutionSolution
Isolate x, remembering to flip the inequality when dividing by a negative number.
−3x+17<−13
−3x<−13−17=−30 …
- CBSE 2025Set ANNUAL1 markQ.For a real number x, write the solution of the inequality 3x−7>5x−1 in interval form.
›Reveal solutionSolution
Solving 3x−7>5x−1 gives x<−3, i.e. the interval (−∞,−3).
3x−7>5x−1
3x−5x>−1+7
−2x>6
Dividing both sides by −2 (a negative number reverses the inequality):
x<−3.
…
- CBSE 2024Set ANNUAL1 markMCQQ.Solution of 3(2−x)≥4x−9 is(a) (−∞,715](b) [715,∞)(c) [−3,∞)(d) None of these
›Reveal solutionSolution
Expand the bracket, collect the x terms on one side and constants on the other, keeping the inequality direction (no sign flip since we never multiply/divide by a negative).
Start with 3(2−x)≥4x−9.
Expand: 6−3x≥4x−9
Add 3x to both sides: 6≥7x−9
Add 9 to both sides: 15≥7x
…
- CBSE 2024Set sz1 markMCQQ.The value of −12x>30 when x is a natural number is:(a) 3(b) <3(c) No solution(d) 0
›Reveal solutionSolution
−12x>30 requires x<−2.5, which no natural number can satisfy, so there is no solution.
We are given −12x>30 with x restricted to be a natural number.
Divide both sides by −12. Since we are dividing by a negative number, the inequality sign flips:
x<−1230=−2.5
…
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