Solve $7x - 3 \le 5x + 11$ for real $x$.

Solve 7x - 3 ≤ 5x + 11 for real x. – Class 11 Mathematics

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7x - 3 \le 5x + 11

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Concept understanding — Linear Inequalities

Linear Inequalities: The Intuition First

You already know what an equation is: a statement that two things are exactly equal. $2x + 3 = 7$ says "twice something plus three is exactly seven." That's a tight, precise condition — only one number ( $x = 2$ ) satisfies it.

Now imagine you loosen that condition. Instead of "exactly equal to 7," what if you said "less than 7"? Or "greater than or equal to 7"? That's an inequality. You're no longer looking for a single point; you're looking for a whole range of numbers.

Note

Real life is full of inequalities: "You need at least 60% to pass" ( $\text{marks} \ge 60$ ), "The bus can carry at most 50 people" ( $\text{passengers} \le 50$ ), "Profit must be more than zero" ( $P > 0$ ). Equations are rare; inequalities are everywhere.

The Four Symbols

There are only four inequality symbols. Memorise them once:

Symbol	Meaning	Example	Reads as
$<$	less than	$x < 5$	x is less than 5
$>$	greater than	$x > 5$	x is greater than 5
$\le$	less than or equal to	$x \le 5$	x is at most 5
$\ge$	greater than or equal to	$x \ge 5$	x is at least 5

The "or equal to" versions ( $\le$ , $\ge$ ) include the boundary number itself. The strict versions ( $<$ , $>$ ) do not.

Solving Linear Inequalities: Almost Like Equations

A linear inequality looks just like a linear equation, but with an inequality sign instead of an equals sign. For example:

2x + 3 < 7

You solve it the same way you solve $2x + 3 = 7$ — with one critical difference.

The Golden Rule (and the only trap)

When you multiply or divide both sides of an inequality by a negative number, you must flip the inequality sign.

Why? Think of the number line. $3 < 5$ is true. Multiply both sides by $-1$ : $-3 < -5$ ? No — $-3$ is actually greater than $-5$ (because $-3$ is to the right on the number line). So the inequality flips: $-3 > -5$ .

Watch out

This is the single most common mistake students make. If you multiply or divide by a negative, flip the sign. If you multiply/divide by a positive, leave it alone.

Example: Solve $2x + 3 < 7$

Step 1: Subtract 3 from both sides (no sign change — subtracting is always safe).

2x < 4

Step 2: Divide both sides by 2 (positive — no flip).

x < 2

Answer: Any number less than 2 works. $x = 1.9$ , $x = 0$ , $x = -100$ — all satisfy the original inequality.

Example: Solve $-3x + 5 \ge 11$

Step 1: Subtract 5 from both sides.

-3x \ge 6

Step 2: Divide both sides by $-3$ (negative — flip the sign).

x \le -2

Answer: $x$ must be less than or equal to $-2$ .

Representing Solutions: The Number Line

The solution to an inequality is an interval (or union of intervals), not a single number. You can show it on a number line:

Open circle at a number means that number is not included ( $<$ or $>$ ).
Closed circle means it is included ( $\le$ or $\ge$ ).
Shade the region that satisfies the inequality.

For $x < 2$ : open circle at 2, shade everything to the left.

For $x \le -2$ : closed circle at -2, shade everything to the left.

The Precise Definition

A linear inequality in one variable is any inequality that can be written in one of these four forms:

ax + b < 0, \quad ax + b > 0, \quad ax + b \le 0, \quad ax + b \ge 0

where $a$ and $b$ are real numbers, and $a \neq 0$ .

The solution set is the set of all real numbers $x$ that make the inequality true. It is always an interval (or a ray) on the real number line.

Important

Unlike a linear equation (which has exactly one solution), a linear inequality has infinitely many solutions — a whole continuous range of numbers.

Two Quick Checks

Does $x = 3$ satisfy $2x - 5 > 0$ ?

$2(3) - 5 = 6 - 5 = 1 > 0$ → Yes.
Does $x = 3$ satisfy $2x - 5 \ge 0$ ?

$1 \ge 0$ → Yes (the boundary $x = 2.5$ also works here, but not in the strict version).

What Comes Next

Once you're comfortable with one variable, you'll extend to linear inequalities in two variables ( $ax + by + c > 0$ ), where the solution set becomes a half-plane on the coordinate grid. But that's a step ahead — master the one-variable case first.

Final takeaway: Inequalities are just equations with a looser condition. Solve them the same way, but remember the flip rule when multiplying/dividing by a negative. The answer is always a range, not a point.

Linear inequalities in one variable form their own dedicated chapter in the NCERT Class 11 Mathematics syllabus, and "linear inequalities rules and sign flip when multiplying by negative" is a frequently searched revision topic for CBSE board preparation. This sign-flip rule is also a classic source of small but costly errors tested in "linear inequalities important questions" for competitive exams.

Start with $7x - 3 \le 5x + 11$ . Subtract $5x$ from both sides: $2x - 3 \le 11$ . Add $3$ to both sides: $2x \le 14$ . Divide both sides by the positive number $2$ (no flip): $x \le 7$ . [!ANSWER] $x \le 7$ , i.e. $x \in (-\infty, 7]$ .

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

CBSE 2026Set ANNUAL1 mark
$ax^2 + bx + c \le 0$
›Reveal solutionSolution

$ax^2+bx+c\le0$ has degree 2 in $x$ , making it a quadratic inequality; a linear inequality would have degree 1 (e.g. $ax+b\le0$ ).

An inequality is named after the degree of the polynomial involved: degree 1 gives a linear inequality, degree 2 gives a quadratic inequality.

Here the expression $ax^2+bx+c$ has highest power $x^2$ , i.e. degree 2.

So $ax^2+bx+c\le0$ is a quadratic inequality, not linear.

✓Final answer
$False.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following is not an inequality?$
(a) $Ax + B < 0$
(b) $Ax + B \geq 0$
(c) $Ax + B \leq 5$
(d) $Ax + B = 5$
›Reveal solutionSolution

$Ax + B = 5$ is an equation, not an inequality, so it is the odd one out.

An inequality relates two expressions with $<$ , $>$ , $\le$ , or $\ge$ . Options (a) $Ax+B<0$ , (b) $Ax+B\ge0$ , and (c) $Ax+B\le5$ all use inequality symbols.

Option (d) $Ax+B=5$ uses an equals sign, so it states an equation, not an inequality.

✓Final answer
$Ax+B=5$
CBSE 2025Set ANNUAL1 mark
$Q. Write True or False: The value of an inequality remains unchanged when both of its sides are multiplied by the same positive number.$
›Reveal solutionSolution

Multiplying (or dividing) an inequality by a positive number keeps its direction unchanged; this rule flips only when multiplying by a negative number.

One of the basic rules of inequalities states: if $a < b$ and $k > 0$ , then $ka < kb$ (the direction of the inequality is preserved). Only multiplying by a negative number reverses the inequality sign.

Since the statement describes multiplying both sides by the same positive number, the inequality's value/direction remains unchanged.

✓Final answer
$True.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is atleast 61 cm. Find the minimum length of the shortest side:$
(a) 9 cm
(b) 6 cm
(c) 4 cm
(d) 10 cm
›Reveal solutionSolution

Perimeter $=7x-2\ge 61 \Rightarrow x\ge 9$ .

Let the shortest side $=x$ cm.

Longest side $=3x$ cm.

Third side $=3x-2$ cm (2 cm shorter than the longest).

Perimeter $= x + 3x + (3x-2) = 7x - 2$ .

Given perimeter is at least 61 cm:

$7x - 2 \ge 61$

$7x \ge 63$

$x \ge 9$ .

So the minimum possible length of the shortest side is 9 cm.

✓Final answer
$Minimum length of the shortest side = 9 cm — option (a).$
CBSE 2024Set ANNUAL1 markMCQ
$Q. An inequality in one variable is:$
(a) $ax + b > 0$
(b) $ax + by > 0$
(c) $ax + by + c > 0$
(d) $ax^2 + by > 0$
›Reveal solutionSolution

An inequality "in one variable" must contain exactly one variable; only $ax+b>0$ qualifies among the given options.

Step 1. Option (A) $ax+b>0$ has only $x$ as the variable — one variable.

Step 2. Options (B) $ax+by>0$ and (C) $ax+by+c>0$ contain two variables, $x$ and $y$ — these are linear inequalities in two variables.

Step 3. Option (D) $ax^2+by>0$ contains two variables and is also non-linear in $x$ .

✓Final answer
$ax + b > 0$
CBSE 2024Set ANNUAL1 mark
$Q. Write true or false: Equal numbers may be added to (or subtracted from) both sides of an inequality without affecting the sign of the inequality.$
›Reveal solutionSolution

Adding/subtracting the same quantity from both sides shifts both sides equally, so the inequality direction is preserved.

Step 1. If $x > y$ , then adding $c$ to both sides gives $x+c > y+c$ — the relation still holds in the same direction.

Step 2. Similarly, subtracting $c$ from both sides gives $x-c > y-c$ .

Step 3. This is one of the standard rules for solving linear inequalities.

✓Final answer
$True.$
CBSE 2022Set ANNUAL1 mark
$ax + by \le c$
›Reveal solutionSolution

The statement is True.

A linear inequality is an inequality that involves a linear (first-degree) expression. $ax+by\le c$ is linear in $x$ and $y$ (no squares, products of variables, etc.), and it uses an inequality sign ( $\le$ ).

So it fits the exact definition of a linear inequality.

✓Final answer
$True.$
CBSE 2022Set ANNUAL1 mark
$x \ge 0$
›Reveal solutionSolution

The region $x\ge0, y\ge0$ is the First quadrant (including the axes).

In the Cartesian plane, the region where both $x\ge0$ and $y\ge0$ is exactly the first quadrant (together with the positive x-axis and positive y-axis as its boundary).

✓Final answer
$First quadrant (Quadrant I).$

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Q.Solve 7x−3≤5x+117x - 3 \le 5x + 117x−3≤5x+11 for real xxx.

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Q.Solve $7x - 3 \le 5x + 11$ for real $x$ .

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