2 ≤ 3x - 4 ≤ 5 – Class 11 Mathematics

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2 \le 3x - 4 \le 5

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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

Linear inequalities are the foundation for:

Compound inequalities (like $-2 < x \le 5$ )
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.

Important

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.

A compound inequality chains two conditions on the same variable; solve each piece separately, then intersect the results. Here $x$ lies in $\boxed{[2, 3]}$ .

Why compound inequalities work this way

When you see $2 \le 3x - 4 \le 5$ , you're really reading two statements at once: " $3x - 4$ is at least $2$ " and " $3x - 4$ is at most $5$ ." The solution set is every $x$ that satisfies both conditions simultaneously. The cleanest route is to isolate $x$ by performing identical operations on all three parts of the chain, preserving the inequality signs as long as we avoid multiplying or dividing by a negative number.

Step-by-step solution

1. Add $4$ to every part.

We want to peel away the " $-4$ " from $3x$ . Adding $4$ throughout keeps the inequalities balanced:

2 + 4 \le 3x - 4 + 4 \le 5 + 4

6 \le 3x \le 9

2. Divide every part by $3$ .

Now isolate $x$ by dividing through by $3$ . Since $3 > 0$ , the inequality directions stay the same:

\frac{6}{3} \le \frac{3x}{3} \le \frac{9}{3}

2 \le x \le 3

3. Interpret the result.

This tells us $x$ must lie between $2$ and $3$ , inclusive of both endpoints. In interval notation, that's $[2, 3]$ .

Tip

You can always check boundary and interior points: $x = 2$ gives $3(2) - 4 = 2$ ✓, $x = 3$ gives $3(3) - 4 = 5$ ✓, and $x = 2.5$ gives $3(2.5) - 4 = 3.5$ , which lies in $[2, 5]$ ✓.

Watch out

Had we needed to multiply or divide by a negative number, every inequality sign would flip. For instance, if the middle term were $-3x + 4$ , dividing by $-3$ would reverse both $\le$ symbols.

✓Final answer

\boxed{2 \le x \le 3}

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 36 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$6(2x+3)+x>53-2x$
›Reveal solutionSolution

Expand and collect terms: $15x>35$ , so $x>\tfrac73$ .

Expand the left side:
$6(2x+3)+x=12x+18+x=13x+18.$
The inequality becomes
$13x+18>53-2x\Rightarrow 13x+2x>53-18\Rightarrow 15x>35\Rightarrow x>\frac{35}{15}=\frac73.$
The solution set is $\left(\dfrac73,\infty\right)$ .

✓Final answer
$The correct option is (A) .$
KEAM 2026Set eng-2026-04174 marksMCQ
$ABCD$
›Reveal solutionSolution

The perimeter is $18a+2$ ; solving $56<18a+2<92$ gives $3<a<5$ .

In the right-angled trapezium, $AB\parallel CD$ with the perpendicular leg $AD=3a$ . The horizontal offset between the parallel sides is $AB-CD=(7a+1)-(3a+1)=4a$ , so the slant side is
$BC=\sqrt{(4a)^2+(3a)^2}=\sqrt{25a^2}=5a.$
The perimeter is
$P=(7a+1)+5a+(3a+1)+3a=18a+2.$
Applying $56<P<92$ :
$56<18a+2<92\Rightarrow 54<18a<90\Rightarrow 3<a<5.$
✓Final answer
$The correct option is (E) .$
KEAM 2026Set eng-2026-04184 marksMCQ
$x$
›Reveal solutionSolution

Solving $8+3x>4(x-3)+2$ yields $x<18$ .

$8+3x>4(x-3)+2=4x-12+2=4x-10.$
Bring terms together:
$8+10>4x-3x\;\Rightarrow\;18>x.$
So the solution set is $(-\infty,18)$ .

✓Final answer
$The correct option is (C) .$
KEAM 2026Set eng-2026-04194 marksMCQ
$x$
›Reveal solutionSolution

Solve the two-sided inequality: $-14<x<-4$ or $6<x<16$ .

Upper bound $|x-1|<15$ : $-15<x-1<15\Rightarrow -14<x<16$ .

Lower bound $|x-1|>5$ : $x-1>5$ or $x-1<-5\Rightarrow x>6$ or $x<-4$ .

Intersect the two: on $x<-4$ side with $-14<x$ gives $-14<x<-4$ ; on $x>6$ side with $x<16$ gives $6<x<16$ .

✓Final answer
$The correct option is (D) .$
KEAM 2026Set eng-2026-04194 marksMCQ
$x$
›Reveal solutionSolution

Subtract 1 and analyse the sign: solution is $(-\infty,2)$ .

Rearrange.
$\frac{x-3}{x-2}-1\ge0\Rightarrow \frac{(x-3)-(x-2)}{x-2}\ge0\Rightarrow \frac{-1}{x-2}\ge0.$
Sign. The numerator $-1<0$ , so the fraction is $\ge0$ only when the denominator is negative: $x-2<0\Rightarrow x<2$ . (Equality is never attained, and $x=2$ is excluded.)

Solution set $=(-\infty,2)$ .

✓Final answer
$The correct option is (B) .$
KEAM 2026Set eng-2026-04204 marksMCQ
$\dfrac{2x - 10}{x + 2} \geq x - 5$
›Reveal solutionSolution
Factoring $2x-10=2(x-5)$ and moving everything to one side yields $-\frac{x(x-5)}{x+2}\ge0$ , i.e. $\frac{x(x-5)}{x+2}\le0$ . A sign chart gives $x<-2$ or $0\le x\le5$ .

Start from $\dfrac{2x-10}{x+2}\ge x-5$ . Since $2x-10=2(x-5)$ :
$\frac{2(x-5)}{x+2}-(x-5)\ge0\;\Rightarrow\;(x-5)\left[\frac{2}{x+2}-1\right]\ge0.$
Simplify the bracket:
$\frac{2-(x+2)}{x+2}=\frac{-x}{x+2},$
so
$(x-5)\cdot\frac{-x}{x+2}\ge0\;\Longleftrightarrow\;\frac{x(x-5)}{x+2}\le0.$
Critical points: $x=-2$ (excluded), $x=0$ , $x=5$ . Sign of $\frac{x(x-5)}{x+2}$ :

$x<-2$ : $\frac{(-)(-)}{(-)}=(-)$ ✓

$-2<x<0$ : $\frac{(-)(-)}{(+)}=(+)$ ✗

$0\le x\le5$ : $\frac{(+)(-)}{(+)}=(-)$ ✓ (endpoints give $0$ )

$x>5$ : $(+)$ ✗

Solution: $x\in(-\infty,-2)\cup[0,5]$ .

✓Final answer
$The correct option is (E) .$
KEAM 2026Set eng-2026-04204 marksMCQ
$10 < |x + 10| \leq 25$
›Reveal solutionSolution

Substitute $y=x+10$ . The double inequality $10<|y|\le25$ means $y\in[-25,-10)\cup(10,25]$ . Converting back ( $x=y-10$ ) gives $x\in[-35,-20)\cup(0,15]$ .

Let $y=x+10$ , so the condition is $10<|y|\le25$ .

This breaks into:
$|y|>10 \Rightarrow y<-10 \text{ or } y>10,$ $|y|\le25 \Rightarrow -25\le y\le25.$
Combining: $y\in[-25,-10)\cup(10,25]$ .

Now $x=y-10$ , subtract $10$ from each bound:
$[-25-10,\,-10-10)\cup(10-10,\,25-10]=[-35,-20)\cup(0,15].$
✓Final answer
$The correct option is (C) .$
KEAM 2026Set eng-2026-04204 marksMCQ
$x$
›Reveal solutionSolution
Sign-analyse $\frac{(x-15)(x-70)}{x-30}\ge 0$ over integers $1$ – $100$ ( $x=30$ excluded).

Critical points: $15,30,70$ (with $x=30$ making the denominator zero).

$x<15$ : expression $<0$ .

$15\le x<30$ : $\ge 0$ (equals $0$ at $x=15$ ). Integers $15,\dots,29$ → $15$ values.

$30<x<70$ : $<0$ .

$x\ge 70$ : $\ge 0$ (equals $0$ at $x=70$ ). Integers $70,\dots,100$ → $31$ values.

Favourable $=15+31=46$ out of $100$ , so probability $=\dfrac{46}{100}=0.46$ .

✓Final answer
$The correct option is (E) .$
KEAM 2026Set eng-2026-04214 marksMCQ
$(x - 1)(x^2 - 5x + 7) < (x - 1)$
›Reveal solutionSolution

Bringing everything to one side factors as $(x-1)(x-2)(x-3)<0$ , giving $(-\infty,1)\cup(2,3)$ .

$(x-1)(x^2-5x+7)-(x-1)<0\;\Rightarrow\;(x-1)(x^2-5x+6)<0\;\Rightarrow\;(x-1)(x-2)(x-3)<0.$
With roots $1,2,3$ , the sign of the cubic is negative on $(-\infty,1)$ and on $(2,3)$ .

Solution set: $(-\infty,1)\cup(2,3)$ .

✓Final answer
$The correct option is (C) .$
KEAM 2026Set eng-2026-04214 marksMCQ
$\left| x + \dfrac{1}{x} \right| > 2$
›Reveal solutionSolution

By AM–GM $|x+1/x|\ge2$ with equality only at $x=\pm1$ ; excluding those and $x=0$ gives $\mathbb{R}-\{-1,0,1\}$ .

For real $x\neq0$ , $\left|x+\dfrac1x\right|\ge2$ , with equality exactly when $x=\pm1$ . The strict inequality $>2$ therefore holds for all $x$ except $x=\pm1$ ; also $x=0$ is excluded from the domain.

Solution set: $\mathbb{R}-\{-1,0,1\}$ .

✓Final answer
$The correct option is (E) .$
KEAM 2026Set eng-2026-04224 marksMCQ
$\frac{21x - 6}{4} - 9 \leq 0$
›Reveal solutionSolution

Solve each inequality separately and intersect.

First: $\frac{21x-6}{4}\le9\Rightarrow 21x-6\le36\Rightarrow 21x\le42\Rightarrow x\le2$ .

Second: $\frac{x-1}{3}\ge-1\Rightarrow x-1\ge-3\Rightarrow x\ge-2$ .

Intersection: $-2\le x\le2$ , i.e. $[-2,2]$ .

✓Final answer
$The correct option is (B) .$
KEAM 2026Set eng-2026-04224 marksMCQ
$|2x - 3| < 5, x \in \mathbb{R}$
›Reveal solutionSolution

Remove the modulus with a two-sided inequality; strict $<$ gives an open interval.

$|2x-3|<5$ means $-5<2x-3<5$ .

Add $3$ : $-2<2x<8$ . Divide by $2$ : $-1<x<4$ .

Since the inequality is strict, the interval is open: $(-1,4)$ .

✓Final answer
$The correct option is (C) .$

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Q.2≤3x−4≤52 \le 3x - 4 \le 52≤3x−4≤5

Why compound inequalities work this way

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Q. $2 \le 3x - 4 \le 5$

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