6 ≤ -3(2x - 4) < 12 – Class 11 Mathematics

Miscellaneous Exercise · Q2

6 \le -3(2x - 4) < 12

Yanam CbseNCERTSubjective· 2mImportance★★★★★

44% · 41/94 Questions

✓ Free question

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.

This is a compound inequality that can be solved by isolating $x$ in the middle. The key is to handle the negative coefficient carefully — dividing by a negative flips both inequality signs. The solution is $0 < x \le 1$ .

The Core Idea

When you see an inequality like $6 \le -3(2x - 4) < 12$ , you're looking at a three-part statement: it says that the expression $-3(2x - 4)$ is simultaneously greater than or equal to 6 and less than 12. Your job is to find all $x$ that satisfy both conditions at once.

The most natural approach is to treat the middle expression as a single quantity and undo the operations that are being done to $x$ , one step at a time. But here's the trap: the coefficient $-3$ is negative. That means when you eventually divide to free $x$ , you must reverse both inequality signs. Many students forget this and get the direction wrong.

Let’s work through it cleanly.

Step-by-Step Solution

1. Write the compound inequality clearly

We have:

6 \le -3(2x - 4) < 12

The variable $x$ is buried inside the parentheses, multiplied by $-3$ . Our goal is to isolate $x$ in the middle.

2. Divide every part by $-3$ — and flip both inequality signs

Since $-3$ is negative, dividing by it reverses the direction of both inequalities. This is the single most important step.

\frac{6}{-3} \;\; \color{red}{\ge} \;\; \frac{-3(2x - 4)}{-3} \;\; \color{red}{\ge} \;\; \frac{12}{-3}

Be careful: the "less than" sign on the right also flips to "greater than". After simplifying:

-2 \;\; \ge \;\; 2x - 4 \;\; > \;\; -4

It's often easier to read if we rewrite it from smallest to largest. Flip the entire inequality around (which reverses the order but keeps the meaning):

-4 < 2x - 4 \le -2

Watch out

A common mistake is to only flip one inequality sign. Remember: when you multiply or divide a compound inequality by a negative number, every inequality sign flips. If you forget, you'll get a solution that's backwards or incomplete.

3. Add 4 to all three parts

Now we have a simpler middle: $2x - 4$ . To isolate the $2x$ term, add 4 to every part:

-4 + 4 < 2x - 4 + 4 \le -2 + 4

This gives:

0 < 2x \le 2

4. Divide every part by 2

Since 2 is positive, the inequality signs stay the same:

\frac{0}{2} < \frac{2x}{2} \le \frac{2}{2}

Which simplifies to:

0 < x \le 1

Tip

You can check your answer by picking a value inside the range, say $x = 0.5$ . Plug it into the original: $-3(2(0.5)-4) = -3(1-4) = -3(-3) = 9$ . Is $6 \le 9 < 12$ ? Yes. Now try a value just outside, like $x = 0$ : $-3(0-4) = 12$ , which is not less than 12 (it's equal, but the right inequality is strict). So $x=0$ is excluded — consistent with $0 < x$ .

✓Final answer

0 < x \le 1

←Previous question · misc-exerciseQ1 — 2 ≤ 3x - 4 ≤ 5 →Next question · misc-exerciseQ3 — -3 ≤ 4 - (7x)/(2) ≤ 18

Q.6≤−3(2x−4)<126 \le -3(2x - 4) < 126≤−3(2x−4)<12

The Core Idea

Step-by-Step Solution

🎓Unlock everything free for 14 days

Q. $6 \le -3(2x - 4) < 12$

Unlock everything free for 14 days