Solve the linear inequality: 2(2x+3)-10 < 6(x-2) – Class 11 Mathematics

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2(2x+3)-10 < 6(x-2)

Nagaland NbseNagaland Board of School Education (Class XI) 2022Subjective· 2mImportance★★★★★

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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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Expand both sides, collect $x$ terms, and isolate $x$ .

$2(2x+3)-10 < 6(x-2)$

Expand the left side: $4x+6-10 = 4x-4$

Expand the right side: $6x-12$

So the inequality becomes: $4x-4 < 6x-12$

Add $12$ to both sides: $4x+8 < 6x$

…

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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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Q.Solve the linear inequality: 2(2x+3)−10<6(x−2)2(2x+3)-10 < 6(x-2)2(2x+3)−10<6(x−2)

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Q.Solve the linear inequality: $2(2x+3)-10 < 6(x-2)$

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