CBSE Class 11 Mathematics Chapter 6: Linear Inequalities NCERT Solutions

The given inequality is .

To solve for x, we divide both sides by 24. Since 24 is a positive number, the direction of the inequality sign remains unchanged.

Simplifying the fraction $\backslash frac\{100\}\{24\}$ by dividing both the numerator and the denominator by their greatest common divisor, which is 4, we get:

Now, we consider the two cases:

(i) When x is a natural number:

Natural numbers are positive integers starting from 1 (i.e., 1, 2, 3, ...). We need to find the natural numbers that are less than $\backslash frac\{25\}\{6\}$. Since $\backslash frac\{25\}\{6\}$ is approximately 4.17, the natural numbers satisfying this condition are 1, 2, 3, and 4.

Thus, the solution set when x is a natural number is $\backslash \{1,\; 2,\; 3,\; 4\backslash \}$.

(ii) When x is an integer:

Integers include positive numbers, negative numbers, and zero (i.e., ..., -3, -2, -1, 0, 1, 2, 3, ...). We need to find all integers that are less than $\backslash frac\{25\}\{6\}$ (approximately 4.17).

These integers are ..., -3, -2, -1, 0, 1, 2, 3, and 4.

Thus, the solution set when x is an integer is $\backslash \{...,\; -3,\; -2,\; -1,\; 0,\; 1,\; 2,\; 3,\; 4\backslash \}$.

The given inequality is $-12x\; >\; 30$.

To solve for x, we divide both sides by -12. When dividing an inequality by a negative number, we must reverse the direction of the inequality sign.

Simplifying the fraction $\backslash frac\{30\}\{-12\}$ by dividing both the numerator and the denominator by their greatest common divisor, which is 6, we get:

Now, we consider the two cases:

(i) When x is a natural number:

Natural numbers are positive integers (1, 2, 3, ...). We need to find natural numbers that are less than $-\; \backslash frac\{5\}\{2\}$ (which is -2.5). Since there are no positive integers less than a negative number, there is no solution in the set of natural numbers.

Thus, when x is a natural number, there is no solution to the given inequality.

(ii) When x is an integer:

Integers include positive, negative, and zero values. We need to find all integers that are less than $-\; \backslash frac\{5\}\{2\}$ (which is -2.5).

The integers satisfying this condition are ..., -5, -4, -3.

Thus, the solution set when x is an integer is $\backslash \{...,\; -5,\; -4,\; -3\backslash \}$.

The given inequality is .

First, we isolate the term with x by adding 3 to both sides of the inequality. Since 3 is positive, the inequality sign remains the same.

Next, we solve for x by dividing both sides by 5. Since 5 is a positive number, the inequality sign remains unchanged.

Now, we consider the two cases:

(i) When x is an integer:

We need to find all integers that are strictly less than 2. These integers are ..., -4, -3, -2, -1, 0, and 1.

Thus, the solution set when x is an integer is $\backslash \{...,\; -4,\; -3,\; -2,\; -1,\; 0,\; 1\backslash \}$.

(ii) When x is a real number:

When x is a real number, the solution includes all real numbers that are strictly less than 2. This can be represented as an open interval on the number line.

Thus, the solution set when x is a real number is $x\; \backslash in\; (-\backslash infty,\; 2)$.

The given inequality is $3x\; +\; 8\; >\; 2$.

To solve for x, we first subtract 8 from both sides of the inequality. Since 8 is positive, the inequality sign does not change.

$$3x\; +\; 8\; -\; 8\; >\; 2\; -\; 8$$

$$3x\; >\; -6$$

Now, we divide both sides by 3. Since 3 is a positive number, the inequality sign remains the same.

$$\backslash frac\{3x\}\{3\}\; >\; \backslash frac\{-6\}\{3\}$$

$$x\; >\; -2$$

Now, we consider the two cases:

(i) When x is an integer:

We need to find all integers that are strictly greater than -2. These integers are -1, 0, 1, 2, 3, and so on.

Thus, the solution set when x is an integer is $\backslash \{-1,\; 0,\; 1,\; 2,\; 3,\; ...\backslash \}$.

(ii) When x is a real number:

When x is a real number, the solution includes all real numbers that are strictly greater than -2. This can be represented as an open interval on the number line.

Thus, the solution set when x is a real number is $x\; \backslash in\; (-2,\; \backslash infty)$.