NCERT Solutions for Class 11 Maths Chapter 6 - Linear Inequalities

Exercise 6.1: This exercise consists of 11 questions and introduces students to the concept of linear inequalities. The questions are based on various concepts such as representing linear inequalities on a number line, solving linear inequalities graphically, and finding the solution set of a system of linear inequalities. Students will learn how to solve linear inequalities and represent their solutions on a number line.

Exercise 6.2: This exercise consists of 16 questions and is focused on solving linear inequalities in one variable. The questions are based on various concepts such as solving linear inequalities using algebraic methods, representing the solutions of linear inequalities on a number line, and finding the intervals in which a given inequality is true. Students will learn how to solve linear inequalities in one variable using algebraic methods.

Exercise 6.3: This exercise consists of 10 questions and is focused on solving linear inequalities in two variables. The questions are based on various concepts such as graphing linear inequalities in two variables, finding the solution set of a system of linear inequalities, and representing the solution set of a system of linear inequalities graphically. Students will learn how to graph linear inequalities in two variables and find their solution sets.

Miscellaneous Exercise: This exercise consists of 12 questions and covers a variety of topics related to linear inequalities. The questions are based on various concepts such as absolute value inequalities, solving linear inequalities using intervals, and solving word problems related to linear inequalities. This exercise will help students to revise and reinforce the concepts learned in the previous exercises.

Overall, the exercises in NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities are designed to help students understand and apply the concepts of linear inequalities. The solutions to these exercises are provided in the textbook, which will help students to check their answers and understand the concepts better.

Access NCERT Solution for Class 11 Maths Chapter 6 – Linear Inequalities

Exercise 6.1

1. Solve $24x < 100$, when

Ans: The given inequality is$24x < 100$. 

$\Rightarrow \dfrac{{24x}}{{24}} < \dfrac{{100}}{{24}}$

(Dividing both sides by same positive number)

$\Rightarrow x < \dfrac{{25}}{6}$

(i) ${\text{x}}$ is a natural number- 

1,2,3 and 4 are the only natural numbers that are smaller than 25\ 6 . When x is a natural number, the above inequality's solutions are 1,2,3 and 4. As a result, the solution set in this example is( 1, 2 ,3 ,4 ).

(ii) ${\text{x}}$ is an integer-

When $\mathrm{x}$ is an integer.

Suppose the given $\mathrm{x}$ is an integer then by the integer definition we have $(-\infty, \infty)$ which means there is no end at positive and negative values and then we have $\{\ldots,-3,-2,-1,0,1,2,3, \ldots .\}\}$

Thus, the value of $\mathrm{x}$ as integers is $\{\ldots,-3,-2,-1,0,1,2,3,4\}$ after four which is a greater value.

2. Solve$ - 12x > 30$, when

Ans: The given inequality is$ - 12x > 30$. 

$\Rightarrow \dfrac{{ - 12x}}{{ - 12}} < \dfrac{{30}}{{ - 12}}$

(Dividing both sides by same negative number)

$\Rightarrow x <  - \dfrac{5}{2}$

 (i) ${\text{x}}$ is a natural number-

There isn't a natural number that isn't less than $\left( { - \dfrac{5}{2}} \right)$. As a result there is no solution to the given inequality when it is a natural number. 

 (ii) ${\text{x}}$ is an integer-

The  integers less than $\left( { - \dfrac{5}{2}} \right)$ are$ \ldots  - 5, - 4, - 3$. As a result, when an integer is an integer, the above inequality's solutions are $ \ldots .. - 5, - 4, - 3$ As a result, the solution set in this situation is $\{ ......., - 5, - 4, - 3\} $.

3. Solve $5x - 3 < 7$, when

Ans: The given inequality is $5x - 3 < 7$. 

$\Rightarrow 5x - 3 + 3 < 7 + 3$

$\Rightarrow 5x < 10$

$\Rightarrow \dfrac{{5x}}{5} < \dfrac{{10}}{5}$

$\Rightarrow x < 2$

(i) ${\text{x}}$ is an integer-

The integers less than 2 are ..., $ - 4, - 3, - 2, - 1,0,1$. As a result, the solutions to the above inequality are when is an integer$ \ldots , - 4, - 3, - 2, - 1,0,1$. As a result, the solution set in this example is $\{  \cdots , - 4, - 3, - 2, - 1,0,1\} $.

 (ii) ${\text{x}}$ is a real number-

When x is a real number, the inequality's solutions are x<2, that is, all real values x  less than 2. As a result, the given inequality's solution set is $x \in ( - \infty ,2)$.

4. Solve $3x + 8 > 2$, when

Ans: The given inequality is $3x + 8 > 2$ $3x + 8 > 2$

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

$\Rightarrow 3x >  - 6$

$\Rightarrow \dfrac{{3x}}{3} > \dfrac{{ - 6}}{3}$

$\Rightarrow x >  - 2$

(i) ${\text{x}}$ is an integer-

The integers higher than $ - 2$ are $ - 1,0,1,2, \ldots $ Hence, when is an integer, the solutions of the above inequality are  $ - 1,0,1,2, \ldots $ Hence, the solution set in this case is $\{  - 1,0,1,2, \ldots \} $ -

(ii) ${\text{x}}$ is a real number-

When x is a real number, the above inequality's solutions are all real numbers that are bigger than -2. As a result, the solution set in this situation is $( - 2,\infty )$.

5. Solve the given inequality for real $x:4x + 3 < 5x + 7$

Ans: $4x + 3 < 5x + 7$

$\Rightarrow 4x + 3 - 7 < 5x + 7 - 7$

$\Rightarrow 4x - 4 < 5x$

$\Rightarrow 4x - 4 - 4x < 5x - 4x$

$\Rightarrow  - 4 < x$

As a result, all real numbers x  bigger  than -4 are solutions to the given inequality. As a result, the given inequality's solution set is  $( - 4,\infty )$.

6. Solve the given inequality for real $x:3x - 7 > 5x - 1$

Ans: $3x - 7 > 5x - 1$

$\Rightarrow 3x - 7 + 7 > 5x - 1 + 7$

$\Rightarrow 3x > 5x + 6$

$\Rightarrow 3x - 5x > 5x + 6 - 5x$

$\Rightarrow  - 2x > 6$

$\Rightarrow \dfrac{{ - 2x}}{{ - 2}} < \dfrac{6}{{ - 2}}$

$\Rightarrow x <  - 3$

As a result, all real numbers x  less  than -3 are solutions to the given inequality. As a result, the given inequality's solution set  is  $( - \infty , - 3)$.

7. Solve the given inequality for real $x:3(x - 1) \leqslant 2(x - 3)$

Ans: $3(x - 1) \leqslant 2(x - 3)$

$\Rightarrow 3x - 3 \leqslant 2x - 6$

$\Rightarrow 3x - 3 + 3 \leqslant 2x - 6 + 3$

$\Rightarrow 3x \leqslant 2x - 3$

$\Rightarrow 3x - 2x \leqslant 2x - 3 - 2x$

$\Rightarrow x \leqslant  - 3$

As a result, any real numbers x  less  than  or equal -3 to are solutions to the specified inequality. As a result, the given inequality's solution set is  $( - \infty , - 3]$.

8. Solve the given inequality for real $x:3(2 - x) \geqslant 2(1 - x)$

Ans: $3(2 - x) \geqslant 2(1 - x)$

$\Rightarrow 6 - 3x \geqslant 2 - 2x$

$\Rightarrow 6 - 3x + 2x \geqslant 2 - 2x + 2x$

$\Rightarrow 6 - x \geqslant 2$

$\Rightarrow 6 - x - 6 \geqslant 2 - 6$

$\Rightarrow  - x \geqslant  - 4$

$\Rightarrow x \leqslant 4$

As a result, the solutions to the following inequality are all real values x  higher than or equal to 4. As a result, the given inequality's solution set is  $( - \infty ,4]$.

9. Solve the given inequality for real $x:x + \dfrac{x}{2} + \dfrac{x}{3} < 11$

Ans: $x + \dfrac{x}{2} + \dfrac{x}{3} < 11$

$\Rightarrow x\left( {1 + \dfrac{1}{2} + \dfrac{1}{3}} \right) < 11$

$\Rightarrow x\left( {\dfrac{{6 + 3 + 2}}{6}} \right) < 11$

$\Rightarrow \dfrac{{11x}}{6} < 11$

$\Rightarrow \dfrac{{11x}}{{6 \times 11}} < \dfrac{{11}}{{11}}$

$\Rightarrow \dfrac{x}{6} < 1$

$\Rightarrow x < 6$

As a result, all real numbers x  less than 6 are solutions to the specified inequality. As a result, the given inequality's solution set is  $( - \infty ,6)$.

10. Solve the given inequality for real $x:\dfrac{x}{3} > \dfrac{x}{2} + 1$

Ans: $\dfrac{x}{3} > \dfrac{x}{2} + 1$

$\Rightarrow \dfrac{x}{3} - \dfrac{x}{2} > 1$

$\Rightarrow \dfrac{{2x - 3x}}{6} > 1$

$\Rightarrow  - \dfrac{x}{6} > 1$

$\Rightarrow  - x > 6$

$\Rightarrow x <  - 6$

As a result, all real numbers less than equal the specified inequality's solution. As a result, the solution set for the given inequality is  $( - \infty , - 6)$.

11. Solve the given inequality for real $x:\dfrac{{3(x - 2)}}{5} \leqslant \dfrac{{5(2 - x)}}{3}$

Ans: $\dfrac{{3(x - 2)}}{5} \leqslant \dfrac{{5(2 - x)}}{3}$

$\Rightarrow 9(x - 2) \leqslant 25(2 - x)$

$\Rightarrow 9x - 18 \leqslant 50 - 25x$

$\Rightarrow 9x - 18 + 25x \leqslant 50$

$\Rightarrow 34x - 18 \leqslant 50$

$\Rightarrow 34x \leqslant 50 + 18$

$\Rightarrow 34x \leqslant 68$

$\Rightarrow \dfrac{{34x}}{{34}} \leqslant \dfrac{{68}}{{34}}$

$\Rightarrow x \leqslant 2$

As a result, any real numbers x  less than or equal to 2 are solutions of the above inequality, and hence the solution set of the given inequality is $( - \infty ,2]$.

12. Solve the given inequality for real $x:\dfrac{1}{2}\left( {\dfrac{{3x}}{5} + 4} \right) \geqslant \dfrac{1}{3}(x - 6)$

Ans:$\dfrac{1}{2}\left( {\dfrac{{3x}}{5} + 4} \right) \geqslant \dfrac{1}{3}(x - 6)$

$\Rightarrow 3\left( {\dfrac{{3x}}{5} + 4} \right) \geqslant 2(x - 6)$

$\Rightarrow \dfrac{{9x}}{5} + 12 \geqslant 2x - 12$

$\Rightarrow 12 + 12 \geqslant 2x - \dfrac{{9x}}{5}$

$\Rightarrow 24 \geqslant \dfrac{{10x - 9x}}{5}$

$\Rightarrow 24 \geqslant \dfrac{x}{5}$

$\Rightarrow 120 \geqslant x$

As a result, the solutions of the following inequality are all real  values  x  less than or equal to 120. As a result, the given inequality's solution set is  $( - \infty ,120]$.

13. Solve the given inequality for real $x:2(2x + 3) - 10 < 6(x - 2)$

Ans: $2(2x + 3) - 10 < 6(x - 2)$

$\Rightarrow 4x + 6 - 10 < 6x - 12$

$\Rightarrow 4x - 4 < 6x - 12$

$\Rightarrow  - 4 + 12 < 6x - 4x$

$\Rightarrow 8 < 2x$

$\Rightarrow 4 < x$

As a result, any real numbers  x  bigger than 4 are solutions to the specified inequality. As a result, the given inequality's solution set is  $(4, - \infty )$.

14. Solve the given inequality for real $x:37 - (3x + 5) \geqslant 9x - 8(x - 3)$

Ans: $37 - (3x + 5) \geqslant 9x - 8(x - 3)$

$\Rightarrow 37 - 3x - 5 \geqslant 9x - 8x + 24$

$\Rightarrow 32 - 3x \geqslant x + 24$

$\Rightarrow 32 - 24 \geqslant x + 3x$

$\Rightarrow 8 \geqslant 4x$

$\Rightarrow 2 \geqslant x$

As a result, the solutions of the following inequality are all real values  x less than or equal to 2. As a result, the given inequality's solution set is  $( - \infty ,2]$.

15. Solve the given inequality for real $x:\dfrac{x}{4} < \dfrac{{(5x - 2)}}{3} - \dfrac{{(7x - 3)}}{5}$

Ans: $\dfrac{x}{4} < \dfrac{{(5x - 2)}}{3} - \dfrac{{(7x - 3)}}{5}$

$\Rightarrow \dfrac{x}{4} < \dfrac{{5(5x - 2) - 3(7x - 3)}}{{15}}$

$\Rightarrow \dfrac{x}{4} < \dfrac{{25x - 10 - 21x + 9}}{{15}}$

$\Rightarrow \dfrac{x}{4} < \dfrac{{4x - 1}}{{15}}$

$\Rightarrow 15x < 4(4x - 1)$

$\Rightarrow 15x < 16x - 4$

$\Rightarrow 4 < 16x - 15x$

$\Rightarrow 4 < x$

As a result, any real numbers x  bigger than 4 are solutions to the specified inequality. As a result, the given inequality's solution set is  $(4,\infty )$.

16. Solve the given inequality for real $x:\dfrac{{(2x - 1)}}{3} \geqslant \dfrac{{(3x - 2)}}{4} - \dfrac{{(2 - x)}}{5}$

Ans: $\dfrac{{(2x - 1)}}{3} \geqslant \dfrac{{(3x - 2)}}{4} - \dfrac{{(2 - x)}}{5}$

$\Rightarrow \dfrac{{(2x - 1)}}{3} \geqslant \dfrac{{5(3x - 2) - 4(2 - x)}}{{20}}$

$\Rightarrow \dfrac{{(2x - 1)}}{3} \geqslant \dfrac{{15x - 10 - 8 + 4x}}{{20}}$

$\Rightarrow \dfrac{{(2x - 1)}}{3} \geqslant \dfrac{{19x - 18}}{{20}}$

$\Rightarrow 20(2x - 1) \geqslant 3(19x - 18)$

$\Rightarrow 40x - 20 \geqslant 57x - 54$

$\Rightarrow  - 20 + 54 \geqslant 57x - 40x$

$\Rightarrow 34 \geqslant 17x$

$\Rightarrow 2 \geqslant x$

As a result, the solutions of the following inequality are all real values x  less than or equal to 2. As a result, the given inequality's solution set is $( - \infty ,2]$.

17. Solve the given inequality and show the graph of the solution on number line:

$3x - 2 < 2x + 1$

Ans: $3x - 2 < 2x + 1$

$\Rightarrow 3x - 2x < 1 + 2$

The following is a graphical illustration of the solutions to the given inequality:

(image will be uploaded soon)

18. Solve the given inequality and show the graph of the solution on number line:

$5x - 3 \geqslant 3x - 5$

Ans: $5x - 3 \geqslant 3x - 5$

$\Rightarrow 5x - 3x \geqslant  - 5 + 3$

$\Rightarrow 2x \geqslant  - 2$

$\Rightarrow \dfrac{{2x}}{2} \geqslant \dfrac{{ - 2}}{2}$

$\Rightarrow x \geqslant  - 1$

The answers to the given inequality are shown graphically as follows :

(image will be uploaded soon)

19. Solve the given inequality and show the graph of the solution on number line:

$3(1 - x) < 2(x + 4)$

Ans: $3(1 - x) < 2(x + 4)$

$\Rightarrow 3 - 3x < 2x + 8$

$\Rightarrow 3 - 8 < 2x + 3x$

$\Rightarrow  - 5 < 5x$

$\Rightarrow \dfrac{{ - 5}}{5} < \dfrac{{5x}}{5}$

$\Rightarrow  - 1 < x$

The following is a graphical illustration of the solutions to the given inequality:

(image will be uploaded soon)

20. Solve the given inequality and show the graph of the solution on number line:

$\dfrac{x}{2} \geqslant \dfrac{{(5x - 2)}}{3} - \dfrac{{(7x - 3)}}{5}$

Ans:  $\dfrac{x}{2} \geqslant \dfrac{{(5x - 2)}}{3} - \dfrac{{(7x - 3)}}{5}$

$\Rightarrow \dfrac{x}{2} \geqslant \dfrac{{5(5x - 2) - 3(7x - 3)}}{{15}}$

$\Rightarrow \dfrac{x}{2} \geqslant \dfrac{{25x - 10 - 21x + 9}}{{15}}$

$\Rightarrow \dfrac{x}{2} \geqslant \dfrac{{4x - 1}}{{15}}$

$\Rightarrow 15x \geqslant 2(4x - 1)$

$\Rightarrow 15x \geqslant 8x - 2$

$\Rightarrow 15x - 8x \geqslant 8x - 2 - 8x$

$\Rightarrow 7x \geqslant  - 2$

$\Rightarrow x \geqslant  - \dfrac{2}{7}$

The following is a graphical illustration of the solutions to the given inequality.

(image will be uploaded soon)

21. Ravi obtained 70 and 75 marks in first two-unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks.

Ans: Let x represent Ravi's score on the third unit test. Because the student must have a grade point average of at least 60, $\dfrac{{70 + 75 + x}}{3} \geqslant 60$

$\Rightarrow 145 + x \geqslant 180$

$\Rightarrow x \geqslant 180 - 145$

$\Rightarrow x \geqslant 35$

As a result, the student must score at least 35 points to achieve a 60-point average.

22. To receive Grade ' ${{\text{A}}^*}$ in a course, one must obtain an average of $90{\text{marks}}$ or more in five examinations (each of 100 marks). If Sunita"s marks in first four examinations are $87,92,94$ and 95, find minimum marks that Sunita must obtain in fifth examination to get grade ' ${\text{A}}$ " in the course.

Ans: Let x represent Sunita's grade in the fifth examination. She must receive an average of 90 or above in five examinations in order to receive a grade of  in the course,

$\dfrac{{87 + 92 + 94 + 95 + x}}{5} \geqslant 90$

$\Rightarrow \dfrac{{368 + x}}{5} \geqslant 90$

$\Rightarrow 368 + x \geqslant 450$

$\Rightarrow x \geqslant 450 - 368$

$\Rightarrow x \geqslant 82$

Sunita must therefore achieve a score of at least 82 in the fifth examination.

23. Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11 .

Ans: Let x be the lesser of the two odd positive integers that follow. Then there's x + 2 as the other integer. Because both integers are less than  10 , $x + 2 < 10$

$\Rightarrow x < 10 - 2$

$\Rightarrow x < 8 \ldots .(t)$

 Furthermore, the sum of the two integers exceeds 11.$\therefore x + (x + 2) > 11$

$\Rightarrow 2x + 2 > 11$

$\Rightarrow 2x > 11 - 2$

$\Rightarrow 2x > 9$

$\Rightarrow x > \dfrac{9}{2}$

$\Rightarrow x > 4.5$

We get I and (ii) from I and (ii).

Because x is an odd number, the values 5 and 7 can be used.

As a result, the required pairs are $(5,7)$ and $(7,9)$.

24. Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23 .

Ans: Let be the smaller of the two even positive integers that follow. The other integer is x + 2

 because both integers are greater than 5, $x > 5 \ldots  \ldots (1)$

Also, the sum of the two integers is less than 23 $x + (x + 2) < 23$

$\Rightarrow 2x + 2 < 23$

$\Rightarrow 2x < 23 - 2$

$\Rightarrow 2x < 21$

$\Rightarrow x < \dfrac{{21}}{2}$

$\Rightarrow x < 10.5 \ldots  \ldots (2)$

From $(1)$ and $(2)$, we obtain $5 < x < 10.5$ 

Because x is an even number, it can have any of the following values: 6,8, or 10.

As a result, the required pairs are $(6,8),(8,10)$ and $(10,12)$.

25. The longest side of a triangle is 3 times the shortest side and the third side is $2\;{\text{cm}}$ shorter than the longest side. If the perimeter of the triangle is at least $61\;{\text{cm}}$, find the minimum length of the shortest side.

Ans: Determine the length of the triangle's shortest side  ${\text{xcm}}$. Then, length of the longest side $ = 3{\text{xcm}}$ Length of the third side $(3x - 2){\text{cm}}$ Since the perimeter of the triangle is at least $61\;{\text{cm}}$, $xcm + 3xcm + (3x - 2)cm \geqslant 61\;{\text{cm}}$

$\Rightarrow 7x - 2 \geqslant 61$

$\Rightarrow 7x \geqslant 61 + 2$

$\Rightarrow 7x \geqslant 63$

$\Rightarrow \dfrac{{7x}}{7} \geqslant \dfrac{{63}}{7}$

$\Rightarrow x \geqslant 9$

As a result, the shortest side's minimal length is  $9\;{\text{cm}}$.

26. A man wants to cut three lengths from a single piece of board of length $91\;{\text{cm}}$. The second length is to be $3\;{\text{cm}}$ longer than the shortest and the third length is to be twice as bang as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least $5\;{\text{cm}}$ longer than the second? [Hint: It $x$ is the length of the shortest board, then $x,(x + 3)$ and $2 x$ are the lengths of the second and third piece, respectively. Thus, $x = (x + 3) + 2x \leqslant 91$ and $2x \geqslant (x + 3) + 5]$

Ans: The shortest piece's length should be ${\text{xcm}}$. Then, length of the second piece and the third piece are $({\text{x}} + 3){\text{cm}}$ and $2{\text{xcm}}$ respectively. Since the three lengths are to be cut from a single piece of board of length $91\;{\text{cm}}$, $x\;{\text{cm}} + (x + 3){\text{cm}} + 2x\;{\text{cm}} \leqslant 91\;{\text{cm}}$

$\Rightarrow 4x + 3 \leqslant 91$

$\Rightarrow 4x \leqslant 91 - 3$

$\Rightarrow 4x \leqslant 88$

$\Rightarrow \dfrac{{4x}}{4} \leqslant \dfrac{{88}}{4}$

$\Rightarrow x \leqslant 22 \ldots .(1)$

 In addition, the third component is at least 5 cm as long as the second. $\therefore 2x \geqslant (x + 3) + 5$

$\Rightarrow 2x \geqslant x + 8$

$\Rightarrow x \geqslant 8$

From (1) and (2), we obtain $8 \leqslant x \leqslant 22$

As a result, the smallest board's potential length is larger than or equal to but less than or equal to $22\;{\text{cm}}$.

Exercise 6.2

1. Solve the given inequality graphically in two-dimensional plane $:\,x + y < 5$

Ans: The visual representation of the$x + y = 5$ In the diagram below, is shown as a dotted line. This line separates the two groups. $xy - $ Split the plane into two half-planes. and II Choose a point (not on the line) that lies in one of the half planes to see if it meets the stated inequality or not.

The point is chosen as $(0,0)$. It is observed that. $0 + 0 < 5$ or $0 < 5$, Because half plane II is not the solution zone of the given inequality, it is true. Furthermore, any point on the line clearly does not satisfy the provided stringent inequality. As a result, the shaded half plane I, excluding the points on the line, is the solution region of the given inequality.

This can be expressed in the following way:

(image will be uploaded soon)

2. Solve the given inequality graphically in two-dimensional plane: $2x + y \geqslant 6$

Ans: The graphical representation of $2x + y = 6$ is given in the figure below. The xy-plane is divided into two half planes, I and II, by this line. To determine whether a point (not on the line) in one of the half planes fulfils the given inequality, choose a point (not on the line) that lies in one of the half planes. The point is chosen as $(0,0)$.

It is observed that, $2(0) + 0 \geqslant 6$ or $0 \geqslant 6$, Because half plane I is not the solution zone of the given inequality, it is false. Furthermore, any point on the line must satisfy the stated inequality. As a result, the shaded half plane II, which includes the points on the line, is the solution region of the given inequality. As an example, consider the following.

(image will be uploaded soon)

3. Solve the given inequality graphically in two-dimensional plane: $3x + 4y \leqslant 12$

Ans: $3x + 4y \leqslant 12$

The graphical representation of $3x + 4y = 12$is depicted in the diagram below. The xy-plane is divided into two half planes, I and II, by this line. To determine whether a point (not on the line) in one of the half planes fulfils the given inequality, choose a point (not on the line) that lies in one of the half planes.point is chosen as $(0,0)$. As a result, the shaded half plane I, which includes the points on the line, is the solution region of the given inequality. The following is an example of how this could be represented. 

(image will be uploaded soon)

4. Solve the given inequality graphically in two-dimensional plane: $y + 8 \geqslant 2x$

Ans : The graphical representation of $y + 8 = 2x$ is depicted in the diagram below.

The xy-plane is divided into two half planes by this line. To determine whether a point (not on the line) in one of the half planes fulfils the given inequality, choose a point (not on the line) that lies in one of the half planes. The point is chosen as

$(0,0)$. 

It is observed that, $0 + 8 \geqslant 2(0)$ or $8 \geqslant 0$, As a result, the solution zone of the given inequality is not the lower half plane. Furthermore, any point on the line must satisfy the stated inequality. As a result, the half plane containing the point ( 0,0 ) and the line is the solution region of the stated inequality. The dark zone represents the solution region as follows.

(image will be uploaded soon)

5. Solve the given inequality graphically in two-dimensional plane: $x - y \leqslant 2$

Ans: The graphical representation of $x - y = 2$ is depicted in the diagram below. The xy-plane is divided into two half planes by this line. To determine whether a point (not on the line) in one of the half planes fulfils the given inequality, choose a point (not on the line) that lies in one of the half planes. 

The point is chosen as $(0,0)$. It is observed that, $0 - 0 \leqslant 2$ or $0 \leqslant 2$, As a result, the solution zone of the given inequality is not the lower half plane. Furthermore, any point on the line clearly satisfies the specified inequality. As a result, the half plane containing the point ( 0,0 ) and the line is the solution region of the stated inequality. The dark zone represents the solution region as follows.

(image will be uploaded soon)

6. Solve the given inequality graphically in two-dimensional plane: $2x - 3y > 6$

Ans: The graphical representation of $2x - 3y = 6$ is given as dotted line in the figure below. This line divides the xy-plane in two half planes. Seloct a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not. We select the point as $(0,0)$.

It is observed that. As a result, the solution zone of the given inequality is not the upper half plane. Furthermore, any point on the line clearly does not satisfy the specified inequality. As a result, the half plane that does not contain the point ( 0,0 ) including the line is the solution region of the stated inequality. The dark zone represents the solution region as follows.

(image will be uploaded soon)

$2(0) - 3(0) > 6$ or $0 > 6$, 

7. Solve the given inequality graphically in two-dimensional plane: $ - 3x + 2y \geqslant  - 6$

Ans: The graphical representation of $ - 3x + 2y =  - 6$is depicted in the diagram below. The xy-plane is divided into two half planes by this line. To determine whether a point (not on the line) in one of the half planes fulfils the given inequality, choose a point (not on the line) that lies in one of the half planes. 

The point is chosen as $(0,0)$. It is observed that, $ - 3(0) + 2(0) \geqslant  - 6$ or $0 \geqslant  - 6$, As a result, the solution zone of the given inequality is not the lower half plane. Furthermore, any point on the line must satisfy the stated inequality. As a result, the half plane containing the point ( 0, 0 ) including the line is the solation zone of the stated inequality. The dark zone represents the solution region as follows.

(image will be uploaded soon)

8. Solve the given inequality graphically in two-dimensional plane: $3y - 5x < 30$

Ans: The graphical representation of $3y - 5x = 30$ In the diagram below, is shown as a dotted line. The xy-plane is divided into two half planes by this line. To determine whether a point (not on the line) in one of the half planes fulfils the given inequality, choose a point (not on the line) that lies in one of the half planes. 

The point is chosen as $(0,0)$. It is observed that, $3(0) - 5(0) < 30$ or $0 > 30$, As a result, the upper half plane does not contain the solution region for the stated inequality. Furthermore, any point on the line clearly does not satisfy the specified inequality.

As a result, the half plane containing the point ( 0,0 ) excluding the line is the solution zone of the stated inequality. The dark zone represents the solution region as follows.

(image will be uploaded soon)

9. Solve the given inequality graphically in two-dimensional plane: $y <  - 2$

Ans: The graphical representation of $y =  - 2$ The figure below shows this as a dotted line. The xy-plane is split into two halves by this line. To determine whether or not a point (not on the line) in one of the half planes fulfils the specified inequality, choose a point (not on the line) that is in one of the half planes. 

The point is chosen as follows: $(0,0)$. It is observed that, $0 <  - 2$, which is false.

Furthermore, any point on the line clearly does not satisfy the specified inequality. As a result, the solution of the above inequality is determined by every point below the line, y = -2 (excluding all points on the line). The darker zone represents the solution region as follows:

(image will be uploaded soon)

10. Solve the given inequality graphically in two-dimensional plane: $x >  - 3$

Ans: The graphical representation of $x =  - 3$ In the diagram below, is shown as a dotted line. The xy-plane is divided into two half planes by this line. To determine whether a point (not on the line) in one of the half planes fulfils the given inequality, choose a point (not on the line) that lies in one of the half planes. 

The point is chosen as $(0,0)$. It is observed that, $0 <  - 3$,This is also true, as any point on the line clearly does not satisfy the provided inequality. As a result, the solution of the above inequality is determined by every point on the right side of the line, x = - 3 (excluding all points on the line). The darker zone represents the solution region as follows:

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Exercise 6.3

1. Solve the following system of inequalities graphically: $x \geqslant 3,y \geqslant 2$.

Ans: $x \geqslant 3 \ldots (1)$

$y \geqslant 2 \ldots  \ldots (2)$

The graph of the lines, $x = 3$ and $y = 2$, are drawn in the figure below. 

Inequality (1) represents the region on the right hand side of the line, $x = 3$ (including the line $x = 3$ ).

Inequality ( 2 ) represents the region above the line, $y = 2$ (including the line $y = 2$ ). Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as follows.

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2. Solve the following system of inequalities graphically: $3x + 2y \leqslant 12,x \geqslant 1,y \geqslant 2$

Ans: $3x + 2y \leqslant 12 \ldots  \ldots (1)$

$x \geqslant 1 \ldots (2)$

$y \geqslant 2. \ldots  \ldots (3)$

The graphs of the lines, $3x + 2y = 12,x = 1$, and $y = 2$, are drawn in the figure below. 

Inequality (1) represents the region below the line, $3x + 2y = 12$ (including the line $3x + 2y = 12$ ). 

Inequality $(2)$ represents the region on the right side of the line, $x = 1$ (including the line $x = 1$ ). 

Inequality (3) represents the region above the line, $y = 2$ (including the line $y = 2)$

As a result, the solution of the given system of linear inequalities is represented as follows by a common shaded region that includes the points on the various lines.

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3. Solve the following system of inequalities graphically: $2x + y \geqslant 6,3x + 4y \leqslant 12$

Ans: $2x + y \geqslant 6 \ldots  \ldots (1)$

$3x + 4y \leqslant 12\ldots  \ldots (2)$

The graph of the lines, $2x + y \geqslant 6$ and $3x + 4y = 12$, are drawn in the figure below. 

Inequality (1) represents the region above the line, $2x + y = 6$ (including the line $2x + y = 6$ ), and 

inequality (2) represents the region below the line, $3x + 4y = 12$ (including the line $3x + 4y = 12$ ). As a result, the solution of the given system of linear inequalities is represented as follows by a common shaded region that includes the points on the various lines.

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4. Solve the following system of inequalities graphically: $x + y \geqslant 4,2x - y > 0$

Ans: $x + y \geqslant 4 \ldots  \ldots (1)$

$2x - y > 0\ldots  \ldots (2)$

The graph of the lines, $x + y = 4$ and $2x - y = 0$, are depicted in the diagram below. The region above the line is represented by inequality (1) , $x + y = 4$ (including the line $x + y = 4$ ). It is observed that $(1,0)$ satisfies the inequality, $2x - y > 0.[2(1) - 0 = 2 > 0]$ Therefore, inequality (2) represents the half plane corresponding to the line, $2x - y = 0$, containing the point $(1,0)$ [excluding the line $2x - y > 0]$. As a result, the solution of the above system of linear inequalities is represented by thecommonshaded region as follows, which includes the points on line x + y =4 but excludes the points on line 2x - y = 0.

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5. Solve the following system of inequalities graphically: $2x - y > 1,x - 2y <  - 1$

Ans: $2 \mathrm{x}-\mathrm{y}>1 \ldots(1)$

$\mathrm{x}-2 \mathrm{y}<-1 \ldots .(2)$

The graphs of the lines, $2 \mathrm{x}-\mathrm{y}=1$ and $\mathrm{x}-2 \mathrm{y}=-1$, are drawn in the figure below. Inequality (1) represents the region below the line, $2 \mathrm{x}-\mathrm{y}=1$ ( excluding the line $2 \mathrm{x}-\mathrm{y}=1$ ). and inequality (2) represents the region above the line, $\mathrm{x}-2 \mathrm{y}=-1,($ excluding the line $\mathrm{x}-2 \mathrm{y}=-1$ ).

Hence, the solution of the given system of linear inequalities is represented by the common shaded region excluding the points on line as follows.

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6. Solve the following system of inequalities graphically: $x + y \leqslant 6,x + y \geqslant 4$

Ans: $x + y \leqslant 6 \ldots  \ldots (1)$

$x + y \geqslant 4 \ldots  \ldots (2)$

$\text { Putting } y=0 \text { in (1) } , \text { Putting } x=0 \text { in (1) }$

$x+0=6 , 0+y=6$

$x=6 , y=6$

The graph of the lines, $x + y = 6$ and $x + y = 4$, are depicted in the diagram below. The region below the line is represented by inequality (1). , $x + y = 6$ (including the line $x + y = 6$ ), and inequality (2) represents the region above the line, $x + y = 4$ (including the line $x + y = 4$ ). As a result, the solution of the given system of linear inequalities is represented as follows by a common shaded region that includes the points on the various lines.

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7. Solve the following system of inequalities graphically: $2x + y \geqslant 8,x + 2y \geqslant 10$

Ans: $2x + y = 8 \ldots  \ldots (1)$

$x + 2y = 10$   ….(2)

Putting $x=0$ in $(1)$

2(0)+y=6

0+y=6

y=8

Putting $y=0$ in $(1)$

2x+(0)=6

2x=8

$x=\dfrac{8}{2}$

x=4

The graph of the lines, $2x + y = 8$ and $x + 2y = 10$, are drawn in the figure below. Inequality (1) represents the region above the line, $x + 2y = 8$, and inequality (2) represents the region above the line, $x + 2y = 10$. As a result, the solution to the given system of linear inequalities is written as follows:

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8. Solve the following system of the inequalities graphically: $x + y \leqslant 9,y > x,x \geqslant 0$

Ans: First we solve $x+y \leq 9$

Putting $x=0$ in (1)

0+y=9

y=9

Putting $\mathrm{y}=\mathbf{0}$ in $(1)$

x+0=9

x=9

Checking for $(0,0)$

Putting $x=0, y=0$

$x+y \leq 9$

$0+0 \leq 9$

$0 \leq 9$

which is true.Hence origin lies in plane $x+y \leq 9$

So, we shade left side of line.

Now we solve $y>x$

Lets first draw graph of

$y=x$

Putting $x=0$ in (2)

$y=0$

Putting $x=2$ in $(2)$

$y=2$

Checking for $(9,0)$

Putting $x=9, y=0$

y>x

0>9

which is false

Hence $(9,0)$ does not lie in plane $x>y$

So, we shade the left side of the line. Also, $x \geq 0$

So, we shade the right side of the y axis. Hence the shaded region represents the given inequality

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9. Solve the following system of inequalities graphically: $5x + 4y \leqslant 20,x \geqslant 1,y \geqslant 2$

Ans: First we solve $5 \mathrm{x}+4 \mathrm{y} \geq 20$

Let first draw graph of

$5 \mathrm{x}+4 \mathrm{y}=20$

Putting $\mathrm{x}=0$ in (1)

$5(0)+4 y=20$

$0+4 \mathrm{y}=20$

$4 \mathrm{y}=20$

$\mathrm{y}=\frac{20}{4}$

$\mathrm{y}=5$

Putting $\mathrm{y}=0 \mathrm{in}(1)$

$5 \mathrm{x}+4(0)=20$

$5 \mathrm{x}+0=20$

$5 \mathrm{x}=20$

$\mathrm{x}=\frac{20}{5}$

$\mathrm{x}=4$

Drawing graph

Points to be plotted are $(0,5),(4,0)$

Points to be plotted are $(0,5),(4,0)$

Checking for $(0,0)$

$5 x+4 y \leq 20$

Putting $\mathrm{x}=0, \mathrm{y}=0$

$5(0)+4(0) \leq 20$

$0 \leq 20$

Which is true

Hence origin lies in plane $5 \mathrm{x}+4 \mathrm{y} \leq 20$

So, we shade left side of line

Also, $\mathrm{y} \geq 2$

So, for all values of

$\mathrm{x}$, where $\mathrm{y}=2$

Points to be plotted are $(0,2),(-1,2),(4,2)$

Also, $x \geq 1$

So, for all values of $y$, where $x=1$

Point to be plotted are $(1,0),(1,-1),(1,3)$

Hence the shaded region represents the given in equality

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10. Solve the following system of inequalities graphically: $3x + 4y \leqslant 60,x + 3y \leqslant 30,x \geqslant 0,y \geqslant 0$

Ans: Now we solve $3 x+4 y \leq 60$

Lets first draw graph of$3 x+4 y=60$

Putting $x=0$ in $(1)$

$3(0)+4 y=60$

$0+4 y=60$

4 y=60

$y=\frac{60}{4}$

y=15

Putting $\mathrm{y}=0$ in $(1)$

$3 x+4(0)=60$

$3 x+0=60$

$3 x=60$

$x=\frac{60}{3}$

x=20

Points to be plotted are

$(0,15),(20,0)$

Checking for $(0,0)$

Putting $x=0, y=0$

$3 x+4 y \leq 60$

$3(0)+4(0) \leq 60$

$0 \leq 60$

which is true

Hence origin lies in plane $3 x+4 y \leq 60$

So, we shade left side of line

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Now we solve $x+3 y \leq 30$

Lets first draw graph of

$x+3 y=30$

Putting $x=0$ in (2)

0+3 y=30

3 y=30

$y=\frac{30}{3}$

y=10

Putting $y=0$ in (2)

$x+3(0)=30$

$x+0=30$

x=30

Points to be plotted are

$(0,10),(30,0)$

Checking for $(0,0)$

Putting $x=0, y=0$

$x+3 y \leq 30$

$0+3(0) \leq 30$

$0 \leq 30$

which is true

Hence origin lies in plane $x+3 y \leq 30$

So, we shade left side of line.

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Also given that $x \geq 0, y \geq 0$

Hence the shaded region will lie in the first quadrant.

Hence the shaded region represents the given inequality.

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11. Solve the following system of inequalities graphically: $2x + y \geqslant 4,x + y \leqslant 3,2x - 3y \leqslant 6$

Ans: The set of inequalities are given as,

$2 x+y \geq 4$

$x+y \leq 3$

$2 x-3 y \leq 6$

Simplifying $2 \mathrm{x}+\mathrm{y} \geq 4$ by substituting different values of $\mathrm{x}$ and $\mathrm{y}$.

When $\mathrm{x}=0$, then, $\mathrm{y}=4$

When $\mathrm{y}=0$, then, $\mathrm{x}=2$

So, the coordinates are $(0,4)$ and $(2,0)$.

Now, if $\mathrm{x}=0$ and $\mathrm{y}=0$, then,

$0 \geq 4$

Which can never be true, so the graph will be shaded away the origin. Simplifying $\mathrm{x}+\mathrm{y} \leq 3$ by substituting different values of $\mathrm{x}$ and $\mathrm{y}$.

When $\mathrm{x}=0$, then, $\mathrm{y}=3$

When $\mathrm{y}=0$, then, $\mathrm{x}=3$

So, the coordinates are $(0,3)$ and $(3,0)$.

Now, if $\mathrm{x}=0$ and $\mathrm{y}=0$, then,

$0 \leq 3$

Which is true, so the graph will be shaded towards the origin. Simplifying $2 \mathrm{x}-3 \mathrm{y} \leq 6$ by substituting different values of $\mathrm{x}$ and $\mathrm{y}$.

When $\mathrm{x}=0$, then, $\mathrm{y}=-2$

When $\mathrm{y}=0$, then, $\mathrm{x}=3$

So, the coordinates are $(0,-2)$ and $(3,0)$.

Now, if $\mathrm{x}=0$ and $\mathrm{y}=0$, then,

$0 \leq 6$

Which is true, so the graph will be shaded towards the origin.

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12. Solve the following system of inequalities graphically:

$x - 2y \leqslant 3,3x + 4y \geqslant 12,x \geqslant 0,y \geqslant 1$

Ans: First we solve $\mathrm{x}-2 \mathrm{y} \leq 3$

Lets first draw graph of

$\mathrm{x}-2 \mathrm{y}=3$

Putting $\mathrm{x}=0$ in $(1)$

$0-2 \mathrm{y}=3$

$-2 \mathrm{y}=3$

$\mathrm{y}=\frac{3}{-2}$

$\mathrm{y}=-1.5$

Putting $\mathrm{y}=0 \mathrm{in}(1)$

$\mathrm{x}-2(0)=3$

$\mathrm{x}-0=3$

$\mathrm{x}=3$

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Points to be plotted are

$(0,6),(3,0)$

Checking for $(0,0)$

$\text { Putting } \mathrm{x}=0, \mathrm{y}=0$

$\mathrm{x}-2 \mathrm{y} \leq 3$

$0-2(0) \leq 3$

$0 \leq 3$

Which is true Hence origin lies in plane $\mathrm{x}-2 \mathrm{y} \leq 3$

So, we shade left of line

Now we solve $3 \mathrm{x}+4 \mathrm{y} \geq 12$

Lets first draw graph of

$3 x+4 y=12$

Putting $\mathrm{x}=0$ in (1)

$3(0)+4 y=12$

$0+4 \mathrm{y}=12$

$4 \mathrm{y}=12$

$\mathrm{y}=\frac{12}{4}$

$\mathrm{y}=3$

$\text { Putting } \mathrm{y}=0 \mathrm{in}(1) \\$

$3 \mathrm{x}+4(0)=12 \\$

$3 \mathrm{x}+0=12 \\$

$3 \mathrm{x}=12 \\$

$\mathrm{x}=\frac{12}{3} \\$

$\mathrm{x}=4$

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Points to be plotted are

$(0,3),(4,0)$

checking for $(0,0)$

Putting $\mathrm{x}=0, \mathrm{y}=0$

$3+4 \mathrm{y} \geq 12$

$3(0)+2(0) \geq 12$

$0 \geq 12$

which is false

Hence origin does not lie in plane $3 \mathrm{x}+2 \mathrm{y}>6$

So, we shade rights side of line.

Also $\mathrm{y} \geq 1$

So, for all values $\mathrm{x}$

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Points to be plotted are

$(0,1),(-1,1),(4,1)$

Also $\mathrm{x} \geq 0$

So, the shaded region will lie on the right side of y-axis Hence the shaded region represents the given inequality.

13. Solve the following system of inequalities graphically:

$4x + 3y \leqslant 60,y \geqslant 2x,x \geqslant 3,x,y \geqslant 0$

Ans: $4x + 3y \leqslant 60 \ldots  \ldots $(1)

$y \geqslant 2x \ldots  - (2)$

$x \geqslant 3$……….(3)

Putting $x=0$ in (1)

$4(0)+3 y=60$

$0+3 y=60$

$3 y=60$

$y=\frac{60}{3}$

$y=20$

Putting $y=0$ in $(1)$

$4 x+3(0)=60$

$4 x+0=60$

$4 x=60$

$x=\frac{60}{4}$

$x=15$

Points to be plotted are $(0,20),(15,0)$

Checking for $(0,0)$

$\text { Putting } x=0, y=0$

$4 x+3 y \leq 60$

$4(0)+3(0) \leq 60$

$0 \leq 60$

which is true

Hence origin lies in plane $3 x+4 y \leq 60$

So, we shade the left side of the line.

Now we solve $y \geq 2 x$

Putting $x=0$ in (1)

y=2(0)

y=0

Putting $x=2$ in (1)

$y=2(2)$

y=4

Points to be plotted are $(0,20),(15,0)$

Checking for $(0,15)$

Putting x=0, y=15

$y \geq 2 x$

$15 \geq 2(0)$

$15 \geq 0$

which is true

Hence $(0,15)$ lies in plane $y \geq 2 x$

So, we shade left side of line

Also, $x \geq 3$

So, for all values of $y, x=3$

Given $x \geq 3$ so we shade right side of line.

Also, given

$x, y \geq 0$

So, the shaded region will line in $1^{\text {st }}$ quadrant

Hence the shaded region represents the given inequality.

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14. Solve the following system of inequalities graphically:

$3x + 2y \leqslant 150,x + 4y = 80,x \leqslant 15,y \geqslant 0,x \geqslant 0$

Ans: $3x + 2y \leqslant 150$……..(1)

$x + 4y = 80 \ldots  \ldots (2)$

$x \leqslant 15 \ldots (3)$

$3 x+2 y \leq 150$

$x+4 y \leq 80$

$x \leq 15$

$y \geq 0$

$x \geq 0$

Draw $3 x+2 y=150$

check $(0,0)$

Draw $x+4 y=80$

check $(0,0)$

Draw $x=15$

take left sides of $x$

$x \geq 0, y \geq 0$, hence only 1 st Quadrant

Intersection area is shaded.

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15. Solve the following system of inequalities graphically:

$x + 2y \leqslant 10,x + y \geqslant 1,x - y \leqslant 0,x \geqslant 0,y \geqslant 0$

Ans: $x + 2y \leqslant 10$……….(1)

$x + y \geqslant 1 \ldots $…..(2)

$x - y \leqslant 0$……..$(3)$

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The graph of the lines, $x + 2y = 10,x + y = 1$ and $x - y = 0$, are drawn in the figure below. Inequality (1) represents the region below the line, $x + 2y = 10$ (including the line $x + 2y = 10)$. Inequality (2) represents the region above the line, $x + y = 1$ (including the line $x + y = 1$ ). Inequality (3) represents the region above the line, $x - y = 0$ (including the line $x - y = 0$ ). Since $x \geqslant 0$ and $y \geqslant 0$, The solution of the following system of linear inequalities is represented by every point in the common shaded region in the first quadrant, including points on the relevant lines and axes.

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Miscellaneous Exercise

1. Solve the inequality $2 \leqslant 3x - 4 \leqslant 5$

Ans: $2 \leqslant 3x - 4 \leqslant 5$

$\Rightarrow 2 + 4 \leqslant 3x - 4 + 4 \leqslant 5 + 4$

$\Rightarrow 6 \leqslant 3x \leqslant 9$

$\Rightarrow 2 \leqslant x \leqslant 3$

As a result, the solutions of the following inequality are all real values higher than or equal to 2 but less than or equal to 3. For the given inequality, the solution set is [2,3].

2. Solve the inequality $6 \leqslant  - 3(2x - 4) < 12$

Ans: $6 \leqslant  - 3(2x - 4) < 12$

$\Rightarrow 2 \leqslant  - (2x - 4) < 4$

$\Rightarrow  - 2 \geqslant 2x - 4 >  - 4$

$\Rightarrow 4 - 2 \geqslant 2x > 4 - 4$

$\Rightarrow 2 \geqslant 2x > 0$

$\Rightarrow 1 \geqslant x > 0$

As a result, the set of solutions for the given inequality is [1 ,0 ).

3. Solve the inequality $ - 3 \leqslant 4 - \dfrac{{7x}}{2} \leqslant 18$

Ans: $ - 3 \leqslant 4 - \dfrac{{7x}}{2} \leqslant 18$

$\Rightarrow  - 3 - 4 \leqslant  - \dfrac{{7x}}{2} \leqslant 18 - 4$

$\Rightarrow  - 7 \leqslant  - \dfrac{{7x}}{2} \leqslant 14$

$\Rightarrow 7 \geqslant \dfrac{{7x}}{2} \geqslant  - 14$

$\Rightarrow 1 \geqslant \dfrac{x}{2} \geqslant  - 2$

$\Rightarrow 2 \geqslant x \geqslant  - 4$

As a result, the set of solutions for the given inequality is $[ - 4,2]$.

4. Solve the inequality $ - 15 < \dfrac{{3(x - 2)}}{5} \leqslant 0$

Ans: $ - 15 < \dfrac{{3(x - 2)}}{5} \leqslant 0$

$\Rightarrow  - 75 < 3(x - 2) \leqslant 0$

$\Rightarrow  - 25 < x - 2 \leqslant 0$

$\Rightarrow  - 25 + 2 < x \leqslant 2$

$\Rightarrow  - 23 < x \leqslant 2$

As a result, the set of solutions for the given inequality is (-23, 2]

5. Solve the inequality $ - 12 < 4 - \dfrac{{3x}}{{ - 5}} \leqslant 2$

Ans: \[ - 12 < 4 - \dfrac{{3x}}{{ - 5}} \leqslant 2\]

  $\Rightarrow  - 12 - 4 < \dfrac{{ - 3x}}{{ - 5}} \leqslant 2 - 4$

$\Rightarrow  - 16 < \dfrac{{3x}}{5} \leqslant  - 2$

$\Rightarrow  - 80 < 3x \leqslant  - 10$

$\Rightarrow \dfrac{{ - 80}}{3} < x \leqslant \dfrac{{ - 10}}{3}$

As a result, the set of solutions for the given inequality is $\left( {\dfrac{{ - 80}}{3},\dfrac{{ - 10}}{3}} \right]$.

6. Solve the inequality $7 \leqslant \dfrac{{(3x + 11)}}{2} \leqslant 11$

Ans: $7 \leqslant \dfrac{{(3x + 11)}}{2} \leqslant 11$

$\Rightarrow 14 \leqslant 3x + 11 \leqslant 22$

$\Rightarrow 14 - 11 \leqslant 3x \leqslant 22 - 11$

$\Rightarrow 3 \leqslant 3x \leqslant 11$

$\Rightarrow 1 \leqslant x \leqslant \dfrac{{11}}{3}$

As a result, the set of solutions for the given inequality is $\left[ {1,\dfrac{{11}}{3}} \right]$.

7. Solve the inequalities and represent the solution graphically on number line:

$5x + 1 >  - 24,5x - 1 < 24$

Ans: $5x + 1 >  - 24 \Rightarrow 5x >  - 25$

$\Rightarrow x >  - 5 \ldots .(1)$

$5x - 1 < 24 \Rightarrow 5x < 25$

$\Rightarrow x < 5$

From (1) and ( 2 ), The solution set for the given system of inequalities can be deduced to be$( - 5,5)$. On a number line, the solution to the above system of inequalities can be expressed as 

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8. Solve the inequalities and represent the solution graphically on number line:

$2(x - 1) < x + 5,3(x + 2) > 2 - x$

Ans: $2(x - 1) < x + 5 \Rightarrow 2x - 2 < x + 5 \Rightarrow 2x - x < 5 + 2$

$\Rightarrow x < 7$

(1) $3(x + 2) > 2 - x \Rightarrow 3x + 6 > 2 - x \Rightarrow 3x + x > 2 - 6$

$\Rightarrow 4x >  - 4$

$\Rightarrow x >  - 1 \ldots  \ldots (2)$

From (1) and (2), The solution set for the given system of inequalities can be deduced to be $( - 1,7)$. On a number line, the solution to the above system of inequalities can be expressed as

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9. Solve the following inequalities and represent the solution graphically on number line:

$3x - 7 > 2(x - 6),6 - x > 11 - 2x$

Ans: $3x - 7 > 2(x - 6) \Rightarrow 3x - 7 > 2x - 12 \Rightarrow 3x - 2x >  - 12 + 7$

$\Rightarrow x >  - 5 \ldots  \ldots  \ldots .(1)$

$ 6 - x > 11 - 2x \Rightarrow  - x + 2x > 11 - 6$

$\Rightarrow x > 5$

From (1) and (2), The solution set for the given system of inequalities can be deduced to be

$(5,\infty )$. On a number line, the solution to the above system of inequalities can be expressed as

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10. Solve the inequalities and represent the solution graphically on number line:

$5(2x - 7) - 3(2x + 3) \leqslant 0,2x + 19 \leqslant 6x + 47$

Ans: $5(2x - 7) - 3(2x + 3) \leqslant 0 \Rightarrow 10x - 35 - 6x - 9 \leqslant 0 \Rightarrow 4x - 44 \leqslant 0 \Rightarrow 4x \leqslant 44$

$\Rightarrow x \leqslant 11$

$2x + 19 \leqslant 6x + 47 \Rightarrow 19 - 47 \leqslant 6x - 2x \Rightarrow  - 28 \leqslant 4x$

$\Rightarrow  - 7 \leqslant x$

From (1) and (2), The solution set for the given system of inequalities can be deduced to be $[ - 7,11]$. On a number line, the solution to the above system of inequalities can be expressed as

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11. A solution is to be kept between ${68^\circ }{\text{F}}$ and ${77^\circ }{\text{F}}$. What is the range in temperature in degree Celsius (C) if the Celsius/Fahrenheit (F) conversion formula is given by $F = \dfrac{9}{5}C + 32?$

Ans: Because the solution must be preserved somewhere in the middle, ${68^\circ }{\text{F}}$ and ${77^\circ }{\text{F}},68 < F < 77$ Putting $F = \dfrac{9}{5}C + 32$, we obtain $68 < \dfrac{9}{5}C + 32 < 77$

$\Rightarrow 68 - 32 < \dfrac{9}{5}C < 77 - 32$

$\Rightarrow 36 < \dfrac{9}{5}C < 45$

$\Rightarrow 36 \times \dfrac{5}{9} < C < 45 \times \dfrac{5}{9}$

$\Rightarrow 20 < C < 25$

As a result, the required temperature range in degrees Celsius is between ${20^\circ }C$ and ${25^\circ }C$.

12. A solution of $8\% $ boric acid is to be diluted by adding a $2\% $ boric acid solution to it. The resulting mixture is to be more than $4\% $ but less than $6\% $ boric acid. If we have 640 litres of the $8\% $ solution, how many litres of the $2\% $ solution will have to be added?

Ans: Let $2\% $ of $x$ litres of boric acid solution is required to be added. Then, total mixture $ = (x + 640)$ litres.

 This resulting mixture is to be more than $4\% $ but less than $6\% $ boric acid. 

$\therefore \quad 2 \%$ of $x+8 \%$ of $640>4 \%$ of $(x+640)$

And, $2 \%$ of $x+8 \%$ of $640<6 \%$ of $(x+640)$`

$\Rightarrow 2x + 5120 > 4x + 2560$

$\Rightarrow 5120 - 2560 > 4x - 2x$

$\Rightarrow 5120 - 2560 > 2x$

$\Rightarrow 2560 > 2x$

$\Rightarrow 1280 > x$

$2\% x + 8\% $ of $640 < 6\% $ of $(x + 640)$ $\dfrac{2}{{100}}x + \dfrac{8}{{100}}(640) < \dfrac{6}{{100}}(x + 640)$

$\Rightarrow 2x + 5120 < 6x + 3840$

$\Rightarrow 5120 - 3840 < 6x - 2x$

$\Rightarrow 5120 - 3840 < 6x - 2x$

$\Rightarrow 1280 < 4x$

$\Rightarrow 320 < x$

$\therefore 320 < x < 1280$

As a result, the total amount of boric acid solution to be added must be greater than 320 litres but less than 1280 litres.

13. How many litres of water will have to be added to 1125 litres of the $45\% $ solution of acid so that the resulting mixture will contain more than $25\% $ but less than $30\% $ acid content?

Ans: Allow for the addition of x litres of water. The entire mixture is then calculated $ = (x + 1125)$litres It is clear that the amount of acid in the final mixture is excessive. $45\% $ of 1125 litres. The resulting mixture will have a higher concentration of $25\% $ but less than $30\% $ acid content.

$\therefore 30\% $ of $(1125 + x) > 45\% $ of 1125

And, $25\% $ of $(1125 + x) < 45\% $ of 1125 

$\Rightarrow \dfrac{{30}}{{100}}(1125 + x) > \dfrac{{45}}{{100}} \times 1125$

$\Rightarrow 30(1125 + x) > 45 \times 1125$

$\Rightarrow 30 \times 1125 + 30x > 45 \times 1125$

$\Rightarrow 30 > 45 \times 1125 - 30 \times 1125$

$\Rightarrow 30x > (45 - 30) \times 1125$

$\Rightarrow x > \dfrac{{15 \times 1125}}{{30}} = 562.5$

$25\% $ of $(1125 + x) < 45\% $ of 1125 $\Rightarrow \dfrac{{25}}{{100}}(1125 + x) < \dfrac{{45}}{{100}} \times 1125$

$\Rightarrow 25(1125 + x) > 45 \times 1125$

$\Rightarrow 25 \times 1125 + 25x > 45 \times 1125$

$\Rightarrow 25x > 45 \times 1125 - 25 \times 1125$

$\Rightarrow 25x > (45 - 25) \times 1125$

$\Rightarrow x > \dfrac{{20 \times 1125}}{{25}} = 900$

$\therefore 562.5 < x < 900$

As a result, the required number of litres of water must be greater than 562.5 but less than 900.

14. IQ of a person is given by the formula $IQ = \dfrac{{MA}}{{CA}} \times 100$, Where MA is mental age and CA is chronological age. If $80 \leqslant 1Q \leqslant 140$ for a group of 12 years old children, find the range of their mental age.

Ans: It is reported that for a group of twelve-year-olds $80 \leqslant IQ \leqslant 140 \ldots  \ldots (i)$

For a group of 12 years old children, ${\text{CA}} = 12$ years ${\text{IQ}} = \dfrac{{{\text{MA}}}}{{12}} \times 100$

Putting this value of IQ in (i), we obtain $80 \leqslant \dfrac{{{\text{MA}}}}{{12}} \times 100 \leqslant 140$

$\Rightarrow 80 \times \dfrac{{12}}{{100}} \leqslant {\text{MA}} \leqslant 140 \times \dfrac{{12}}{{100}}$

$\Rightarrow 9.6 \leqslant {\text{MA}} \leqslant 16.8$

As a result, the mental age range of the 12-year-olds has widened.$\Rightarrow 9.6 \leqslant {\text{MA}} \leqslant 16.8$.

NCERT Solutions for Class 11 Maths Chapters

• Chapter 1 - Sets

• Chapter 2 - Relations and Functions

• Chapter 3 - Trigonometric Functions

• Chapter 4 - Principle of Mathematical Induction

• Chapter 5 - Complex Numbers and Quadratic Equations

• Chapter 7 - Permutations and Combinations

• Chapter 8 - Binomial Theorem

• Chapter 9 - Sequences and Series

• Chapter 10 - Straight Lines

• Chapter 11 - Conic Sections

• Chapter 12 - Introduction to Three Dimensional Geometry

• Chapter 13 - Limits and Derivatives

• Chapter 14 - Mathematical Reasoning

• Chapter 15 - Statistics

• Chapter 16 - Probability

Chapter 6 Maths Class 11

Keeping in mind the benefit of students, we are providing the solution material. Many times, learners are unable to get hold of a substantial number of practice problems along with correct solutions. Our primary objective is to eradicate this issue and provide students with illustrative problems of Ch 6 Maths Class 11. All the NCERT Solutions for Class 11 Maths Chapter 6 can be availed from a single platform. Besides, you can easily download it without paying any amount.

Linear Inequalities Class 11 Solutions

Linear Inequalities comprise several topics and the NCERT Solution material consists of questions related to the same. Here is an overview:

• Linear Inequalities Class 11: Inequalities

NCERT Solutions for Class 11 Maths Chapter 6 PDF contains both subjective and objective problems. Learners need to fill the MCQs and work out the exercises with all the necessary steps. You must note that few questions are provided with graphs as well. Carrying out a comprehensive revision of such solution material can assist students in attaining commendable grades.

Here Are Some Examples of Problems in This Section

1. x need to be solved in the equation 24x < 100 provided when the variable is an integer and a natural number.

2. Solve and find 3 (2 – x) >= 2 (1 – x), 37 – (3x + 5) >= 9x – 8 (x – 3), 2 (2x + 3) – 10 < 6 (x – 2) and similar type equations.

• Linear Inequalities Class 11: Solutions of Linear Inequalities with the help of Algebra (One Variable) and Pictorial Representation

Here, students will get to learn about solving inequalities (one variable) and plotting them on a graph. This portion covers all kinds of linear inequalities such as compound inequalities, multi-step inequalities and two-variable inequalities. However, the primary focus will be on both graphical and algebraic methods. Questions of this section require the solution of x as well as plotting the same on a graph. Here are some examples:

Solve for the inequalities: x + y < 5, 2x + y >= 6, 3x + 4y <= 12, y + 8 >= 2, x > -3, y – 5x < 30, etc. After finding the desired solution, you need to plot and shade the correct region on the graph.

• Linear Inequalities Class 11: Two-Variable Solutions of Linear Inequalities and its Graphical Representation

In this section, you will get to understand about solving inequality word problems with two variables. Plus, students will also get familiar with the inequality subjective type questions and will be able to interpret them better. Few examples of questions from this section are as follows:

You need to represent the system of inequalities on a graph for:

1. x >= 3, y >=2

2. 3x + 2y <= 12, x >= 1, y >= 2

3. 2x + y >= 6, 3x + 4y <= 12

• Linear Inequalities Class 11: Miscellaneous Exercises

This section infuses problem-solving skills and critical thinking among learners. The solution material not only provides an extensive idea about the linear inequality problems but also tries to make solving Maths easier for them.

How Advantageous is the Class 11 Maths Chapter 6 Solutions PDF?

1. Linear Inequalities Class 11 NCERT Solutions are curated by subject matter experts and skilled tutors having years' long professional expertise.

2. This NCERT Ch 6 Class 11 Maths material can assist all the students in having a considerable grasp over the concepts and solving the sums accurately.

3. All the problem solutions are written in a step-by-step manner, using the right expressions and formulas. Plus, they strictly abide by the recent CBSE guidelines that will provide them with commendable grades in exams.

Therefore, without wasting much time, download Vedantu's NCERT Solution for Linear Inequalities Class 11 PDF. Nevertheless, if you are still facing issues while preparing or solving the problems, please get in touch with our experts. We ensure to solve all your subject related queries, and also present you reference materials if required.

We Cover All Exercises in The Chapter Given Below

NCERT Solution Class 11 Maths of Chapter 6 Exercise