5x + 1 > -24, 5x - 1 < 24 – Class 11 Mathematics

Miscellaneous Exercise · Q7

5x + 1 > -24

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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: $<$ , $>$ , $\le$ , or $\ge$ . 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: $2x = 4 \implies x = 2$ . 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.

Watch out

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
$x \ge a$	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
$x \le a$	All numbers less than or equal to $a$	Closed circle at $a$ , shade left

The solution set is usually written in interval notation:

$x > 2$ → $(2, \infty)$
$x \le -3$ → $(-\infty, -3]$

Parentheses $($ or $)$ mean the endpoint is not included. Brackets $[$ or $]$ mean it is included.

Solving a Linear Inequality: Step by Step

Solve $3x - 5 \le 7x + 3$ .

Bring variable terms to one side:

$3x - 5 - 7x \le 3$

$-4x - 5 \le 3$
Isolate the variable term:

$-4x \le 8$
Divide by the coefficient (here it's $-4$ , so reverse the sign):

$x \ge -2$

The solution is $x \ge -2$ , or in interval notation: $[-2, \infty)$ .

Tip

Always check your answer by testing a number from the solution set. For $x \ge -2$ , test $x = 0$ : $3(0)-5 = -5$ and $7(0)+3 = 3$ . Is $-5 \le 3$ ? Yes. Now test a number outside, say $x = -3$ : $3(-3)-5 = -14$ and $7(-3)+3 = -18$ . Is $-14 \le -18$ ? No — so the inequality fails, confirming our solution is correct.

Why This Matters …

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Solve each inequality separately, then find the intersection of the two solution sets. The combined solution is $-5 < x < 5$ .

When you see two inequalities presented together like this, you're being asked to find all values of $x$ that satisfy both conditions simultaneously. Think of it as finding the overlap between two sets of numbers on the number line. Each inequality carves out a region, and we want the region where both are true at once.

The strategy is straightforward: solve each inequality independently using the same algebraic moves you'd use for equations (with one crucial caveat about multiplying or dividing by negatives), then identify where the solutions overlap.

Solving the system

1. Solve the first inequality: $5x + 1 > -24$

Subtract $1$ from both sides:

5x > -24 - 1

5x > -25

Divide both sides by $5$ (since $5$ is positive, the inequality direction stays the same):

x > -5

2. Solve the second inequality: $5x - 1 < 24$

Add $1$ to both sides:

5x < 24 + 1

5x < 25

Divide both sides by $5$ :

x < 5

3. Find the intersection

We need $x$ to satisfy both $x > -5$ AND $x < 5$ . On the number line, this is the region strictly between $-5$ and $5$ .

In interval notation: $x \in (-5, 5)$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 35 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$4x + 3 < 6x + 7$
(a) $(-\infty, -2)$
(b) $(-2, \infty)$
(c) $(2, \infty)$
(d) $(-\infty, 2)$
›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:

$-4 < 2x$ …

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CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{x}{3} > \dfrac{x}{2} + 1$
(a) $(-6, \infty)$
(b) $(-\infty, -6)$
(c) $(-\infty, 6)$
(d) None of these
›Reveal solutionSolution

Clear the fractions by multiplying by the LCM (6), then isolate $x$ .

$\dfrac{x}{3} > \dfrac{x}{2} + 1$

Multiply every term by 6 (positive, direction unchanged):

$2x > 3x + 6$

Subtract $3x$ :

$-x > 6$ …

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CBSE 2026Set ANNUAL1 markMCQ
$-8 \le 5x - 3 < 7$
(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.

$-8 \le 5x - 3 < 7$

Add 3 to all parts:

$-5 \le 5x < 10$

Divide by 5 (positive):

$-1 \le x < 2$ …

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CBSE 2026Set ANNUAL1 markMCQ
$x > -4$
(a) $(-4, \infty)$
(b) $(2, 2)$
(c) $(1, 4)$
(d) $(\infty, -4)$
›Reveal solutionSolution

The solution set of $x > -4$ is the open interval $(-4, \infty)$ , 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 …

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CBSE 2026Set ANNUAL1 markMCQ
$x$
(a) $-3$
(b) $1$
(c) $4$
(d) $3$
›Reveal solutionSolution
Only $x=4$ satisfies $x+4>7$ , option (c).

Solve the inequality: $x + 4 > 7 \Rightarrow x > 3$ .

Now test each option:

$x=-3$ : $-3 < 3$ — fails

$x=1$ : $1 < 3$ — fails

$x=4$ : $4 > 3$ — satisfies ( $4+4=8>7$ ✓) …
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CBSE 2025Set ANNUAL1 markMCQ
$24x < 100$
(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), $x < \dfrac{100}{24} = 4.1\overline{6}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$a$
(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 $0$ :

$x > a \quad \text{or} \quad x < -a$

…

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CBSE 2025Set sz1 markMCQ
$7x - 8 \geq 6$
(a) [2, \infty)
(b) (2, 8)
(c) [4, \infty)
(d) (14, \infty)
›Reveal solutionSolution

$7x-8\ge 6 \implies 7x\ge14 \implies x\ge2$ , so the solution set is $[2,\infty)$ .

Start with $7x - 8 \ge 6$ .

Add 8 to both sides:

$7x \ge 14.$

Divide both sides by 7 (a positive number, so the inequality direction is unchanged): …

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CBSE 2025Set ANNUAL1 markMCQ
$-3x + 17 < -13$
(a) $x \in (10, \infty)$
(b) $x \in [10, \infty)$
(c) $x \in (-\infty, 10)$
(d) $x \in [-10, 10]$
›Reveal solutionSolution

Isolate $x$ , remembering to flip the inequality when dividing by a negative number.

$-3x+17<-13$

$-3x<-13-17=-30$ …

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CBSE 2025Set ANNUAL1 mark
$x$
›Reveal solutionSolution

Solving $3x - 7 > 5x - 1$ gives $x < -3$ , i.e. the interval $(-\infty, -3)$ .

$3x - 7 > 5x - 1$

$3x - 5x > -1 + 7$

$-2x > 6$

Dividing both sides by $-2$ (a negative number reverses the inequality):

$x < -3$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$3(2-x) \ge 4x-9$
(a) $\left(-\infty, \dfrac{15}{7}\right]$
(b) $\left[\dfrac{15}{7}, \infty\right)$
(c) $[-3, \infty)$
(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 $3(2-x) \ge 4x - 9$ .

Expand: $6 - 3x \ge 4x - 9$

Add $3x$ to both sides: $6 \ge 7x - 9$

Add $9$ to both sides: $15 \ge 7x$

…

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CBSE 2024Set sz1 markMCQ
$-12x > 30$
(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:

$x < \dfrac{30}{-12} = -2.5$

…

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←Previous question · misc-exerciseQ6 — 7 ≤ (3x + 11)/(2) ≤ 11 →Next question · misc-exerciseQ8 — 2(x - 1) < x + 5, 3(x + 2) > 2 - x

Q.5x+1>−245x + 1 > -245x+1>−24, 5x−1<245x - 1 < 245x−1<24

Solving the system

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Q. $5x + 1 > -24$ , $5x - 1 < 24$

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