CBSE Class 11 Mathematics Chapter 6: Linear Inequalities - NCERT Solutions

The given inequality is .

To solve for x, we divide both sides by 24:

This simplifies to:

Note that $\backslash frac\{25\}\{6\}$ is approximately 4.17.

(i) When x is a natural number:

We need to find the natural numbers that are less than $\backslash frac\{25\}\{6\}$. The natural numbers are 1, 2, 3, 4, ...

The natural numbers less than 4.17 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:

We need to find the integers that are less than $\backslash frac\{25\}\{6\}$ (approximately 4.17).

The integers less than 4.17 include ..., -3, -2, -1, 0, 1, 2, 3, 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. Remember to reverse the inequality sign when dividing by a negative number:

This simplifies to:

Note that $-\backslash frac\{5\}\{2\}$ is -2.5.

(i) When x is a natural number:

Natural numbers are positive integers (1, 2, 3, ...). There are no natural numbers that are less than -2.5.

Thus, there is no solution when x is a natural number.

(ii) When x is an integer:

We need to find the integers that are less than -2.5.

The integers less than -2.5 are ..., -5, -4, -3.

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

The given inequality is .

First, add 3 to both sides of the inequality:

This simplifies to:

Now, divide both sides by 5:

This gives us:

(i) When x is an integer:

We need to find the integers that are less than 2.

The integers less than 2 are ..., -4, -3, -2, -1, 0, 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:

The solution is all real numbers x such that x is less than 2.

This can be represented in interval notation as $(-\backslash infty,\; 2)$.

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

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

First, subtract 8 from both sides of the inequality:

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

This simplifies to:

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

Now, divide both sides by 3:

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

This gives us:

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

(i) When x is an integer:

We need to find the integers that are greater than -2.

The integers greater than -2 are -1, 0, 1, 2, 3, ...

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

(ii) When x is a real number:

The solution is all real numbers x such that x is greater than -2.

This can be represented in interval notation as $(-2,\; \backslash infty)$.

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