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#### Question 1:

Solve: 12x < 50, when

(i) x ∈ R
(ii) x ∈ Z
(iii) x ∈ N

#### Question 2:

Solve: −4x > 30, when

(i) x ∈ R
(ii) x ∈ Z
(iii) x ∈ N

#### Question 3:

Solve: 4x − 2 < 8, when

(i) x ∈ R
(ii) x ∈ Z
(iii) x ∈ N

3x − 7 > x + 1

x + 5 > 4x − 10

3x + 9 ≥ −x + 19

#### Question 7:

$2\left(3-x\right)\ge \frac{x}{5}+4$

#### Question 8:

$\frac{3x-2}{5}\le \frac{4x-3}{2}$

#### Question 9:

−(x − 3) + 4 < 5 − 2x

#### Question 10:

$\frac{x}{5}<\frac{3x-2}{4}-\frac{5x-3}{5}$

#### Question 11:

$\frac{2\left(x-1\right)}{5}\le \frac{3\left(2+x\right)}{7}$

#### Question 12:

$\frac{5x}{2}+\frac{3x}{4}\ge \frac{39}{4}$

#### Question 13:

$\frac{x-1}{3}+4<\frac{x-5}{5}-2$

#### Question 14:

$\frac{2x+3}{4}-3<\frac{x-4}{3}-2$

#### Question 15:

$\frac{5-2x}{3}<\frac{x}{6}-5$

#### Question 16:

$\frac{4+2x}{3}\ge \frac{x}{2}-3$

#### Question 17:

$\frac{2x+3}{5}-2<\frac{3\left(x-2\right)}{5}$

#### Question 19:

$\frac{6x-5}{4x+1}<0$

#### Question 20:

$\frac{2x-3}{3x-7}>0$

#### Question 21:

$\frac{3}{x-2}<1$

$\frac{3}{x-2}<1\phantom{\rule{0ex}{0ex}}⇒\frac{3}{x-2}-1<0\phantom{\rule{0ex}{0ex}}⇒\frac{3-x+2}{x-2}<0\phantom{\rule{0ex}{0ex}}⇒\frac{-x+5}{x-2}<0\phantom{\rule{0ex}{0ex}}⇒\frac{x-5}{x-2}>0$

$x\in \left(-\infty ,2\right)\cup \left(5,\infty \right)$

#### Question 22:

$\frac{1}{x-1}\le 2$

$\frac{1}{x-1}\le 2\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x-1}-2\le 0\phantom{\rule{0ex}{0ex}}⇒\frac{1-2x+2}{x-1}\le 0\phantom{\rule{0ex}{0ex}}⇒\frac{-2x+3}{x-1}\le 0\phantom{\rule{0ex}{0ex}}⇒\frac{2x-3}{x-1}\ge 0\phantom{\rule{0ex}{0ex}}$

$x\in \left(-\infty ,1\right)\cup \left[\frac{3}{2},\infty \right)$

#### Question 23:

$\frac{4x+3}{2x-5}<6$

$\frac{4x+3}{2x-5}<6\phantom{\rule{0ex}{0ex}}⇒\frac{4x+3}{2x-5}-6<0\phantom{\rule{0ex}{0ex}}⇒\frac{4x+3-6\left(2x-5\right)}{2x-5}<0\phantom{\rule{0ex}{0ex}}⇒\frac{4x+3-12x+30}{2x-5}<0\phantom{\rule{0ex}{0ex}}⇒\frac{-8x+33}{2x-5}<0\phantom{\rule{0ex}{0ex}}⇒\frac{8x-33}{2x-5}>0\phantom{\rule{0ex}{0ex}}$

$x\in \left(-\infty ,\frac{5}{2}\right)\cup \left(\frac{33}{8},\infty \right)$

#### Question 24:

$\frac{5x-6}{x+6}<1$

$\frac{5x-6}{x+6}<1\phantom{\rule{0ex}{0ex}}⇒\frac{5x-6}{x+6}-1<0\phantom{\rule{0ex}{0ex}}⇒\frac{5x-6-x-6}{x+6}<0\phantom{\rule{0ex}{0ex}}⇒\frac{4x-12}{x+6}<0\phantom{\rule{0ex}{0ex}}⇒\frac{x-3}{x+6}<0$

#### Question 25:

$\frac{5x+8}{4-x}<2$

#### Question 26:

$\frac{x-1}{x+3}>2$

∴ $x\in \left(-7,-3\right)$

#### Question 27:

$\frac{7x-5}{8x+3}>4$

We have,

$\mathrm{x}\in \left(\frac{-17}{25},\frac{-3}{8}\right)$

#### Question 28:

$\frac{x}{x-5}>\frac{1}{2}$

$\frac{x}{x-5}>\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{x-5}-\frac{1}{2}>0\phantom{\rule{0ex}{0ex}}⇒\frac{2x-x+5}{2\left(x-5\right)}>0\phantom{\rule{0ex}{0ex}}⇒\frac{x+5}{2\left(x-5\right)}>0\phantom{\rule{0ex}{0ex}}⇒\frac{x+5}{x-5}>0$

#### Question 1:

Solve each of the following system of equations in R.

1. x + 3 > 0, 2x < 14

#### Question 2:

Solve each of the following system of equations in R.

2. 2x − 7 > 5 − x, 11 − 5x ≤ 1

#### Question 3:

Solve each of the following system of equations in R.

3. x − 2 > 0, 3x < 18

#### Question 4:

2x + 6 ≥ 0, 4x − 7 < 0

#### Question 5:

Solve each of the following system of equations in R.

5. 3x − 6 > 0, 2x − 5 > 0

#### Question 6:

Solve each of the following system of equations in R.

6. 2x − 3 < 7, 2x > −4

#### Question 7:

Solve each of the following system of equations in R.

7. 2x + 5 ≤ 0, x − 3 ≤ 0

#### Question 8:

Solve each of the following system of equations in R.

8. 5x − 1 < 24, 5x + 1 > −24

#### Question 9:

Solve each of the following system of equations in R.

9. 3x − 1 ≥ 5, x + 2 > −1

#### Question 10:

Solve each of the following system of equations in R.

10. 11 − 5x > −4, 4x + 13 ≤ −11

We have,

#### Question 11:

Solve each of the following system of equations in R.

11. 4x − 1 ≤ 0, 3 − 4x < 0

#### Question 12:

Solve each of the following system of equations in R.

12. x + 5 > 2(x + 1), 2 − x < 3 (x + 2)

#### Question 13:

Solve each of the following system of equations in R.

13. 2 (x − 6) < 3x − 7, 11 − 2x < 6 − x

#### Question 14:

Solve each of the following system of equations in R.

14.

#### Question 15:

Solve each of the following system of equations in R.

15.

#### Question 16:

Solve each of the following system of equations in R.

16.

#### Question 17:

Solve each of the following system of equations in R.

17.

#### Question 18:

Solve each of the following system of equations in R.

18. $0<\frac{-x}{2}<3$

#### Question 19:

Solve each of the following system of equations in R.

19. 10 ≤ −5 (x − 2) < 20

Hence, the interval (-2,0] is the solution of the given set of inequations.

#### Question 20:

Solve each of the following system of equations in R.

20. −5 < 2x − 3 < 5

Hence, the interval ($-$1,4) is the solution of the given set of inequaltions.

#### Question 21:

Solve each of the following system of equations in R.

Thus, the solution set of the inequation is .

Thus, the solution set of the inequation is .

The common values of x in both the inequation is .

Hence, the solution set of both the inequation is .

#### Question 1:

Solve $\left|x+\frac{1}{3}\right|>\frac{8}{3}$

#### Question 2:

Solve $\left|4-x\right|+1<3$

#### Question 3:

Solve $\left|\frac{3x-4}{2}\right|\le \frac{5}{12}$

#### Question 4:

Solve $\frac{\left|x-2\right|}{x-2}>0$

#### Question 5:

Solve $\frac{1}{\left|x\right|-3}<\frac{1}{2}$

#### Question 6:

Solve $\frac{\left|x+2\right|-x}{x}<2$

#### Question 7:

Solve $\left|\frac{2x-1}{x-1}\right|>2$

#### Question 8:

Solve $\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\ge 6$

#### Question 9:

Solve $\frac{\left|x-2\right|-1}{\left|x-2\right|-2}\le 0$                                                                                                                                                       [NCERT EXEMPLAR]

#### Question 10:

Solve $\frac{1}{\left|x\right|-3}\le \frac{1}{2}$                                                                                                                                                        [NCERT EXEMPLAR]

#### Question 11:

Solve $\left|x+1\right|+\left|x\right|>3$                                                                                                                                                   [NCERT EXEMPLAR]

#### Question 12:

Solve $1\le \left|x-2\right|\le 3$                                                                                                                                               [NCERT EXEMPLAR]

#### Question 13:

Solve $\left|3-4x\right|\ge 9$                                                                                                                                  [NCERT EXEMPLAR]

#### Question 1:

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

Let the smaller odd positive integer be x. Then, the other odd positive integer shall be x + 2.
Therefore, as per the given conditions:

#### Question 2:

Find all pairs of consecutive odd natural number, both of which are larger than 10, such that their sum is less than 40.

Let x be the smaller of the two odd natural numbers. Then, the other odd natural number will be x + 2.
Therefore, as per the given conditions:

#### Question 3:

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

Let x be the smaller even integer. Then, the other even integer shall be x + 2.
Therefore, as per the given condition:

#### Question 4:

The marks scored by Rohit in two tests were 65 and 70. Find the minimum marks he should score in the third test to have an average of at least 65 marks.

Let x be the minimum  marks he scores in the third test.

#### Question 5:

A solution is to be kept between 86° and 95°F. What is the range of temperature in degree Celsius, if the Celsius (C)/ Fahrenheit (F) conversion formula is given by $F=\frac{9}{5}C+32$.

Suppose the temperature of the solution is x degree Celsius.
x in Fahrenheit = $\frac{9}{5}x+32$
Then, as per the given condition:

#### Question 6:

A solution is to be kept between 30°C and 35°C. What is the range of temperature in degree Fahrenheit?

Let x degree Fahrenheit be the temperature of the solution.

#### Question 7:

To receive grade 'A' in a course, one must obtain an average of 90 marks or more in five papers each of 100 marks. If Shikha scored 87, 95, 92 and 94 marks in first four paper, find the minimum marks that she must score in the last paper to get grade 'A' in the course.

Let x be the minimum marks scored in the last paper.

#### Question 8:

A company manufactures cassettes and its cost and revenue functions for a week are respectively, where x is the number of cassettes produced and sold in a week. How many cassettes must be sold for the company to realize a profit?

To realise profit, revenue must be greater than the cost.

#### Question 9:

The longest side of a triangle is three times the shortest side and third side is 2 cm shorter than the longest side if the perimeter of the triangles at least 61 cm, find the minimum length of the shortest-side.

Let the shortest side of the triangle be x cm.
Then, the longest side will be 3x and the third side will be 3x − 2.

#### Question 10:

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?

Let x litres of water be added to the 1125 litres of 45% solution of the acid.
Total quantity of mixture is (1125+x) litres.
Total acid content in 1125 litres of mixture = 45% of 1125

#### Question 11:

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 there are 640 litres of the 8% solution, how many litres of 2% solution will have to be added?

Suppose x litres of 2% solution is added in the existing solution of 8% of boric acid.
Resulting mixture = (640 + x) L
Therefore, as per given conditions:

#### Question 12:

The water acidity in a pool is considered normal when the average pH reading of three daily measurements is between 7.2 and 7.8. If the first two pH reading are 7.48 and 7.85, find the range of pH value for the third reading that will result in the acidity level being normal.

Let x be the third  pH value.

#### Question 1:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

1. x + 2yy ≤ 0

Converting the given inequation to equation, we obtain x + 2y $-$ y = 0, i.e x + y = 0
Putting y = 0 and x = 0 in this equation, we obtain x = 0 and y = 0.
So, this line intersects the x-axis and the y-axis at (0,0).
We draw the line of the equation x + y = 0
Now we take a point (1, 1) ( any point which does not lie on the line x + y = 0 )
(1, 1) does not satisfy the inequality. So, the region not containing (1, 1)
is represented by the following figure.

Hence, the shaded region represents the in equation.

#### Question 2:

Represent to solution set of each of the following in equations graphically in two dimensional plane:

2. x + 2y ≥ 6

Converting the in equation to equation, we obtain x + 2y = 6, i.e x + 2y $-$ 6 = 0.
Putting y = 0 and x = 0 in this equation, we obtain x = 6 and y = 3.
So, this line meets x-axis at (6,0) and y-axis at (0,3).
We plot these points and join them by a thick line. This divides the xy plane into two parts.
To determine the region represented by the given inequality, consider point O(0,0). Clearly,
(0,0) does not satisfy inequality $x+2y\ge 6$.
So, the region that does not contain the origin is represented by the given inequality.
Hence, the shaded region is the solution to the in equation.

#### Question 3:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

3. x + 2 ≥ 0

Converting the inequation to equation, we obtain x + 2 = 0, i.e x = $-$2.
Clearly, it is a parallel line to y-axis at a distance of $-$2 units from it. This line divides the xy plane into two parts, viz LHS  of x = $-$2 and RHS of x = $-$2. To determine the region represented by the given inequality, consider point O(0,0). Clearly, (0,0)  does not satisfy the inequality. So, the region that does not contain the origin is represented by the given inequality.
Hence, the shaded region is the solution to the inequation.

#### Question 4:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

4. x − 2y < 0

Converting the inequation to equation, we obtain x $-$ 2y = 0
Putting y = 0 and x = 0 in this equation, we obtain x = 0 and y = 0 respectively. So, this line meets the x-axis at (0,0) and the y- axis at (0,0).
If x = 1, then y = 1/2.
So, we have another point (1,1/2).
We plot these points and join them by a thin line. This divides the xy plane into two parts.
We take a point (0, 2) and it does not satisfy the inequation.

Therefore, we shade the region which is opposite to the point (0, 2)
The shaded region is the solution to the inequation.

#### Question 5:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

5. −3x + 2y ≤ 6

Converting the inequation to equation, we obtain $-$3x + 2y = 6, i.e $-$3x + 2y $-$ 6 = 0
Putting y = 0 and x = 0 in this equation, we obtain x = $-$2 and y = 3 respectively. So, this line meets the x-axis at ($-$2,0) and the y-axis at (0,3). We plot these points and join them by a thick line.This divides the xy plane in two parts. To determine the region represented by the given inequality, consider point O(0,0). Clearly, (0,0) satisfy the inequality $-3x+2y\le 6$. So, the region containing the origin is represented by the given inequality.
Hence, the shaded region is the solution to the inequation.

#### Question 6:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

6. x ≤ 8 − 4y

Converting the inequation to equation, we obtain x + 4y $-$ 8 = 0
Putting y = 0 and x = 0 in this equation, we obtain x = 8 and y = 2 respectively. So, this line meets the x-axis at (8,0) and y-axis at (0,2). We plot these points and join them by a thick line. This divides the xy plane into two parts. To determine the region represented by the given inequality, consider point O(0,0). Clearly, (0,0) satisfies the inequality. So, the region containing the origin is represented by the given inequality.
Hence, the shaded region is the solution to the inequation.

#### Question 7:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

7. 0 ≤ 2x − 5y + 10

Converting the inequation to equation, we obtain 2x $-$ 5y+10 = 0
Putting y = 0 and x = 0 in this equation, we obtain x = $-$ 5 and y = 2 respectively.
So, this line meets the x-axis at ($-$5,0) and the y-axis at (0,2).
We plot these points and join them by a thick line.
This divides the xy plane into two parts. To determine the region represented by the given inequality, consider point O(0,0).
Clearly, (0,0) satisfies the inequality.
So, the region containing the origin is represented by the given inequality.
Hence, the shaded region is the solution to the inequation.

#### Question 8:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

8. 3y ≥ 6 − 2x

Converting the inequation to equation, we obtain 3y+2x $-$ 6 = 0
Putting y = 0 and x = 0 in this equation, we obtain x = 3 and y = 2 respectively.
So. this line meets the x-axis at (3,0) and y-axis at (0,2).
We plot these points and join them by a thick line.
This divides the xy plane into two parts. To determine the region represented by the given inequality, consider point O(0,0).
Clearly, (0,0) does not satisfy the inequality.
So, the region not containing the origin is represented by the given inequality.
Hence, the shaded region is the solution to the inequation.

#### Question 9:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

9. y ≥ 2x − 8

Converting the inequation to equation, we obtain 2x $-$ y $-$ 8 =0
Putting y = 0 and x = 0 in this equation, we obtain x = 4 and y = $-$ 8 respectively.
So, this line meets the x-axis at (4,0) and y-axis at (0, $-$8).
We plot these points and join them by a thick line.This divides the xy plane into two parts.
To determine the region represented by the given inequality, consider point O(0,0).
Clearly, (0,0) does satisfy the inequality.
So, the region containing the origin is represented by the given inequality.
Hence, the shaded region is the solution to the inequation.

#### Question 10:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

10. 3x − 2yx + y − 8

Converting the inequation to equation, we obtain 3x $-$ 2y $-$ x $-$ y + 8 = 0, i.e 2x $-$ 3y + 8= 0
Putting  y =0 and x = 0 in this equation, we obtain x$-$ 4 and y = 8/3 respectively.
So, this line meets the x-axis at ($-$4, 0) and y-axis at (0,8/3).
We plot these points and join them by a thick line.
This divides the xy plane into two parts.
To determine the region represented by the given inequality, consider point O(0,0).
Clearly, (0,0) does not satisfy the inequality.
So, the region that does not contain the origin is represented by the given inequality.
Hence, the shaded region is the solution to the inequation.

#### Question 1:

Solve the following systems of linear inequations graphically:

(i) 2x + 3y ≤ 6, 3x + 2y ≤ 6, x ≥ 0, y ≥ 0
(ii) 2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0
(iii) xy ≤ 1, x + 2y ≤ 8, 2x + y ≥ 2, x ≥ 0, y ≥ 0
(iv) x + y ≥ 1, 7x + 9y ≤ 63, x ≤ 6, y ≤ 5, x ≥ 0, y ≥ 0
(v) 2x + 3y ≤ 35, y ≥ 3, x ≥ 2, x ≥ 0, y ≥ 0

(i) Converting the inequations to equations, we obtain:
2x + 3y = 6, 3x + 2y = 6, x = 0, y = 0

2x + 3y =6:  This line meets the x-axis at (3,0) and the y-axis at (0, 2). Draw a thick line joining these points.
We see that the origin (0, 0) satisfies the inequation 2x + 3y ≤ 6
So, the portion containing  the origin represents the solution set of the inequation 2x + 3y ≤ 6

3x+2y =6:  This line meets the x-axis at (2, 0) and the y-axis at (0, 3). Draw a thick line joining these points.
We see that the origin (0,0) satisfies the inequation 3x + 2y ≤ 6.
So, the portion containing  the origin represents the solution set of the inequation 3x + 2y ≤ 6

Clearly, x ≥ 0, y ≥ 0 represents the first quadrant.
Hence. the shaded region in the figure represents the solution set of the given set of inequations.

(ii)  Converting the inequations to equations, we obtain:
2x + 3y = 6, x + 4y = 4, x = 0, y = 0

2x + 3y =6:  This line meets the x-axis at (3, 0) and the y-axis at (0, 2). Draw a thick line joining these points.
We see that the origin (0,0) satisfies the inequation 2x + 3y ≤ 6.
So, the portion containing  the origin represents the solution set of the inequation 2x + 3y ≤ 6

x + 4y = 4:  This line meets the x-axis at (4, 0) and the y-axis at (0, 1). Draw a thick line joining these points.
We see that the origin (0,0) satisfies the inequation x + 4y ≤ 4.
So, the portion containing  the origin represents the solution set of the inequation x + 4y ≤ 4

Clearly, x ≥ 0, y ≥ 0 represents the first quadrant.
Hence, the shaded region in the figure represents the solution set of the given set of inequations.

(iii)  Converting the inequations to equations, we obtain:
x $-$ y=1, x +2y =8, 2x + y = 2, x = 0, y = 0

x $-$ y = 1:  This line meets the x-axis at (1, 0) and the y-axis at (0, $-$ 1). Draw a thick line joining these points.
We see that the origin (0, 0) satisfies the inequation x $-$ y ≤ 1 So, the portion containing  the origin represents the solution set of the inequation x $-$ y ≤ 1

x + 2y =8:  This line meets the x-axis at (8, 0) and the y-axis at (0, 4). Draw a thick line joining these points.
We see that the origin (0, 0) satisfies the inequation x + 2y ≤ 8 So, the portion containing  the origin represents the solution set of the inequation x + 2y ≤ 8

2x + y =2:  This line meets the x-axis at (1, 0) and the y-axis at (0, 2). Draw a thick line joining these points.
We see that the origin (0, 0) does not satisfy the inequation 2xy $\ge$2 So, the portion that does not contain the origin represents the solution set of the inequation 2xy $\ge$2

Clearly, x ≥ 0, y ≥ 0 represents the first quadrant.
Hence, the shaded region in the figure represents the solution set of the given set of inequations.

(iv) Converting the inequations to equations, we obtain:
x + y =1, 7x + 9y = 63, x = 6, y = 5

x + y=1:  This line meets the x-axis at (1, 0) and the y-axis at (0, 1). Draw a thick line joining these points.
We see that the origin (0, 0)  does not satisfy the inequation xy $\ge$1 So, the portion not containing  the origin represents the solution set of the inequation xy $\ge$1

7x + 9y =63:  This line meets the x-axis at (9, 0) and the y-axis at (0, 7). Draw a thick line joining these points.
We see that the origin (0, 0) satisfies the inequation 7x + 9y ≤ 63 So, the portion containing  the origin represents the solution set of the inequation 7x + 9y ≤ 63

x = 6:  This line is parallel to the x-axis at a distance 6 units from it.
We see that the origin (0, 0) satisfies the inequation x ≤ 6 So, the portion containing  the origin represents the solution set of the inequation x ≤ 6

y = 5:  This line is parallel to the y-axis at a distance 5 units from it.
We see that the origin (0,0) satisfies the inequation y ≤ 5 So, the portion containing  the origin represents the solution set of the inequation
y ≤ 5

Clearly, x ≥ 0, y ≥ 0 represents the first quadrant.
Hence, the shaded region in the figure represents the solution set of the given set of inequations.

(v) Converting the inequations to equations, we obtain:
2x + 3y = 35, x = 0, y = 0

2x + 3y = 35:  This line meets the x-axis at (17.5, 0) and the y-axis at (0, 35/3). Draw a thick line joining these points.
We see that the origin (0, 0) satisfies the inequation 2x + 3y ≤ 35 So, the portion containing  the origin represents the solution set of the inequation 2x + 3y ≤ 35

x = 2:  This line is parallel to the x-axis at a distance 2 units from it.
We see that the origin (0, 0) does not satisfy the inequation x $\ge$2 So, the portion that does not contain  the origin represents the solution set of the inequation x $\ge$2

y = 3:  This line is parallel to the y-axis at a distance 3 units from it.
We see that the origin (0, 0) does not satisfies the inequation y ≥ 3 So, the portion opposite to the origin represents the solution set of the inequation y ≥ 3

Clearly, x ≥ 0, y ≥ 0 represents the first quadrant.
Hence, the shaded region in the figure represents the solution set of the given set of inequations.

#### Question 2:

Show that the solution set of the following linear inequations is empty set:

(i) x − 2y ≥ 0, 2xy ≤ −2, x ≥ 0, y ≥ 0
(ii) x + 2y ≤ 3, 3x + 4y ≥ 12, y ≥ 1, x ≥ 0, y ≥ 0

(i) Converting the inequations to equations, we obtain:
x $-$ 2y = 0, 2x $-$ y = $-$2, x = 0, y = 0
x $-$ 2y = 0:  This line meets the x-axis at (0, 0) and y-axis at (0, 0). If x = 1, then y = 1/2,
so we have another point (1, 1/2). Draw a thick line through (0, 0) and (1, 1/2).
We see that the origin (1, 0) satisfies the inequation x + 2y ≤ 3 So, the portion containing the (1, 0) represents the solution set of the inequation x $-$ 2y ≤ 0

2x $-$ y = $-$2:  This line meets the x-axis at ($-$1, 0) and y-axis at (0, 2). Draw a thick line joining these points.
We see that the origin (0, 0) does not satisfy the inequation 2x $-$ y $\le$$-$2 So, the portion not containing the origin represents the solution set of the inequation 2x $-$ y $\le$ $-$2

Clearly, x ≥ 0, y ≥ 0 represents the first quadrant.
We see in the figure that there is no common region in all the lines. Hence, the solution set to the given set of inequations is empty.

(ii) Converting the inequations to equations, we obtain:
x + 2y = 3, 3x + 4y =12,  y = 1

x + 2y = 3:  This line meets the x-axis at (3, 0) and y-axis at (0, 3/2). Draw a thick line joining these points.
We see that the origin (0, 0) satisfies the inequation x + 2y ≤ 3. So, the portion containing the origin represents the solution set of the inequation x + 2y ≤ 3

3x + 4y =12:  This line meets the x-axis at (4, 0) and y-axis at (0, 3). Draw a thick line joining these points.
We see that the origin (0, 0) does not satisfy the inequation 3x + 4y $\ge$12 So, the portion not containing the origin represents the solution set of the inequation 3x + 4y $\ge$ 12

y = 1:  This line is parallel to x-axis at a distance of 1 unit from it.
We see that the origin (0, 0) does not satisfy the inequation y $\ge$ 1 So, the portion not containing the origin represents the solution set of the inequation y $\ge$ 1

Clearly, x ≥ 0, y ≥ 0 represents the first quadrant.
We see in the figure that there is no common region in all the lines. Hence, the solution set to the given set of inequations is empty.

#### Question 3:

Find the linear inequations for which the shaded area in Fig. 15.41 is the solution set. Draw the diagram of the solution set of the linear inequations:

Figure

Considering the line 2x + 3y = 6, we find that the shaded region and the origin (0, 0) are on the opposite side of this line and (0, 0) does not satisfy the inequation 2x + 3y $\ge$ 6 So, the first inequation is 2x + 3y $\ge$ 6

Considering the line 4x + 6y = 24, we find that the shaded region and the origin (0, 0) are on the same side of this line and (0, 0) satisfies the inequation 4x + 6y $\le$ 24 So, the corresponding inequation is 4x + 6y $\le$ 24

Considering the line x $-$ 2y = 2, we find that the shaded region and the origin (0, 0) are on the same side of this line and (0, 0) satisfies the inequation x $-$ 2y $\le$ 2 So, the corresponding inequation is x $-$ 2y $\le$ 2

Considering the line $-$3x + 2y = 3, we find that the shaded region and the origin (0, 0) are on the same side of this line and (0, 0) satisfies the inequation $-$3x + 2y $\le$ 3 So, the corresponding inequation is $-$3x + 2y $\le$ 3

Also, the shaded region is in the first quadrant. Therefore, we must have x

Thus, the linear inequations comprising the given solution set are given below:
2x + 3y $\ge$ 6, 4x + 6y $\le$ 24, x $-$ 2y $\le$ 2, $-$3x + 2y $\le$ 3, x

#### Question 4:

Find the linear inequations for which the solution set is the shaded region given in Fig. 15.42

Figure

Considering the line x + y = 4, we find that the shaded region and the origin (0, 0) are on the same side of this line and (0, 0) does not satisfy the inequation x + y $\le$ 4 So, the first inequation is x + y $\le$ 4

Considering the line y = 3, we find that the shaded region and the origin (0, 0) are on the same side of this line and (0, 0) satisfies the inequation  y $\le$ 3 So, the corresponding inequation is y $\le$ 3

Considering the line x = 3, we find that the shaded region and the origin (0, 0) are on the same side of this line and (0, 0) satisfies the inequation x $\le$ 3 So, the corresponding inequation is x $\le$ 3

Considering the line x + 5y = 4, we find that the shaded region and the origin (0, 0) are on the opposite side of this line and (0, 0) does not satisfy the inequation x + 5y $\ge 4$ So, the corresponding inequation is x + 5y $\ge 4$

Considering the line 6x + 2y = 8, we find that the shaded region and the origin (0, 0) are on the opposite side of this line and (0, 0) does not satisfy the inequation 6x + 2y $\ge 8$ So, the corresponding inequation is 6x + 2y $\ge 8$

Also the shaded region is in the first quadrant. Therefore, we must have

Thus, the linear inequations comprising the given solution set are given below:
x + y $\le$ 4, y $\le$ 3, x $\le$ 3, x + 5y $\ge 4$, 6x + 2y $\ge 8$,

#### Question 5:

Show that the solution set of the following linear in equations is an unbounded set:
x + y ≥ 9
3x + y ≥ 12
x ≥ 0, y ≥ 0

Converting the inequations to equations, we obtain:
x + y = 9, 3x + y = 12 , x = 0, y = 0

x + y = 9:  This line meets the x-axis at (9, 0) and y-axis at (0, 9). Draw a thick line through these points.
Now, we see that the origin (0, 0) does not satisfy the inequation x + y $\ge$ 9 Therefore, the potion that does not contain the origin is the solution set to the inequaltion x + y $\ge$ 9

3x + y = 12:  This line meets the x-axis at (4, 0) and y-axis at (0, 12). Draw a thick line through these points.
Now, we see that the origin (0, 0) does not satisfy the inequation 3x + y $\ge$12. Therefore, the potion that does not contain the origin is the solution set to the inequaltion 3x + y $\ge$12.

Also, x ≥ 0, y ≥ 0  represents the first quadrant. Hence, the solution set lies in the first quadrant.

We see that in this solution set, the shaded region is unbounded (infinite). Hence, the solution set to the given set of inequalities is an unbounded set.

#### Question 6:

Solve the following systems of inequations graphically:

(i) 2x + y ≥ 8, x + 2y ≥ 8, x + y ≤ 6
(ii) 12x + 12y ≤ 840, 3x + 6y ≤ 300, 8x + 4y ≤ 480, x ≥ 0, y ≥ 0
(iii) x + 2y ≤ 40, 3x + y ≥ 30, 4x + 3y ≥ 60, x ≥ 0, y ≥ 0
(iv) 5x + y ≥ 10, 2x + 2y ≥ 12, x + 4y ≥ 12, x ≥ 0, y ≥ 0

(i) Converting the inequations to equations, we obtain:
2x + y = 8, x + 2y = 8, x + y = 6

2x + y = 8:  This line meets the x-axis at (4, 0) and y-axis at (0, 8). Draw a thick line through these points.
Now, we see that the origin (0, 0) does not satisfy the inequation 2x + y  $\ge$ 8.
Therefore, the region that does not contain the origin is the solution of the inequality 2x + y $\ge$8

x + 2y = 8:  This line meets the x-axis at (8, 0) and y-axis at (0, 4). Draw a thick line through these points.
Now, we see that the origin (0, 0) does not satisfy the inequation x + 2y $\ge$ 8
Therefore, the region that does not contain the origin is the solution of the inequality x + 2y $\ge$ 8

x + y = 6:  This line meets the x-axis at (6, 0) and y-axis at (0, 6). Draw a thick line through these points.
Now, we see that the origin (0, 0) satisfies the inequation x + y $\le$6 Therefore, the region containing the origin is the solution of the inequality  x + y $\le$6

Hence, the solution to the inequalities is the intersection of the above three solutions. Thus, the shaded region represents the solution set of the given set of inequalities.

(ii) Converting the inequations to equations, we obtain:
12x + 12y = 840,  3x + 6y = 300, 8x + 4y = 480, x = 0, y = 0

12x + 12y = 840:  This line meets the x-axis at (70, 0) and y-axis at (0, 70). Draw a thick line through these points.
Now, we see that the origin (0, 0) satisfies the inequation 12x + 12y $\le$ 840
Therefore the region containing the origin is the solution of the inequality 12x + 12y $\le$ 840

3x + 6y =300:  This line meets the x-axis at (100, 0) and y-axis at (0, 50). Draw a thick line through these points.
Now, we see that the origin (0, 0) satisfies the inequation 3x + 6y $\le$300
Therefore, the region containing the origin is the solution of the inequality 3x + 6y $\le$300

8x + 4y = 480:  This line meets the x-axis at (60, 0) and y-axis at (0, 120). Draw a thick line through these points.
Now, we see that the origin (0, 0) satisfies the inequation 8x + 4y $\le$ 480 Therefore, the region containing the origin is the solution of the inequality 8x + 4y $\le$ 480

Also, x represens the first quadrant. So, the solution set must lie in the first quadrant.

Hence, the solution to the inequalities is the intersection of the above three solutions. Thus, the shaded region represents the solution set of the given set of inequalities.

(iii)  Converting the inequations to equations, we obtain:
x + 2y = 40, 3x + y = 30, 4x + 3y = 60

x + 2y = 40:  This line meets the x-axis at (40, 0) and y-axis at (0, 20). Draw a thick line through these points.
We see that the origin (0,0) satisfies the inequation x + 2y $\le$ 40
Therefore, the region containing the origin is the solution of the inequality x + 2y $\le$ 40

3x + y = 30:  This line meets the x-axis at (10, 0) and y-axis at (0, 30). Draw a thick line through these points.
We see that the origin (0, 0) does not satisfy the inequation 3x + y $\ge$ 30
Therefore, the region that does not contain the origin is the solution of the inequality 3x + y $\ge$ 30

4x + 3y = 60:  This line meets the x-axis at (15, 0) and y-axis at (0, 20). Draw a thick line through these points.
We see that the origin (0, 0) does not satisfy the inequation 4x + 3y $\ge$ 60 Therefore, the region that does not contain the origin is the solution of the inequality 4x + 3y $\ge$ 60

Also, x ≥ 0, y ≥ 0 represents the first quadrant. So, the solution set must be in the first quadrant.

Hence, the solution to the inequalities is the intersection of the above three solutions. Thus, the shaded region represents the solution set of the given set of inequalities.

(iv)  Converting the inequations to equations, we obtain:
5x + y = 10, 2x + 2y = 12, x + 4y = 12

5x + y =10:  This line meets the x-axis at (2, 0) and y-axis at (0, 10). Draw a thick line through these points.
We see that the origin (0, 0) does not satisfy the inequation 5x + y $\ge$ 10
Therefore, the region that does not contain the origin is the solution of the inequality 5x + y $\ge$10

2x + 2y = 12:  This line meets the x-axis at (6, 0) and y-axis at (0, 6). Draw a thick line through these points.
We see that the origin (0, 0) does not satisfy the inequation 2x + 2y $\ge$ 12
Therefore, the region that does not contain the origin is the solution of the inequality 2x + 2y $\ge$ 12

x + 4y = 12:  This line meets the x-axis at (12, 0) and y-axis at (0, 3). Draw a thick line through these points.
We see that the origin (0, 0) does not satisfy the inequation x + 4y $\ge$ 12
Therefore, the region that does not contain the origin is the solution of the inequality
x + 4y $\ge$ 12

Also, x ≥ 0, y ≥ 0 represents the first quadrant. So, the solution set must be in the first quadrant.

Hence, the solution to the inequalities is the intersection of the above three solutions. Thus, the shaded region represents the solution set of the given set of inequalities.
Here, the solution set is unbounded region.

#### Question 7:

Show that the following system of linear equations has no solution:

As, the points satisfying x + 2y = 3 are:

 x 3 0 5 y 0 1.5 $-$1

Also, the points satisfying 3x + 4y = 12 are:

 x 0 4 8 y 3 0 $-$3

Now, the region representing the given inequalities is as follows:

Since, there is no common region.

So, the given system of inequalities has no solution.
 x 3 0 5 y 0 1.5 $-$1

#### Question 8:

Show that the solution set of the following system of linear inequalities is an unbounded region: .

As, the solutions of the equation 2x + y = 8 are:

 x 0 4 2 y 8 0 4

As, the solutions of the equation x + 2y = 10 are:

 x 0 10 2 y 5 0 4

Now, the graph represented by the inequalities (i), (ii), (iii) and (iv) is as follows:

Since, the common shaded region is the solution set of the given set of inequalities.

So, the solution set of the given linear inequalities is an unbounded region.

#### Question 1:

Mark the correct alternative in each of the following:
If x$<$7, then
(a) $-$x$<$$-$7
(b) $-$x$\le -$7
(c) $-$x$>-$7
(d) $-$x$\ge -$7

Hence, the correct option is (c).

#### Question 2:

Mark the correct alternative in each of the following:
If − 3x$+$17$<-$13, then
(a) x$\in$(10, $\infty$)
(b) x$\in$[10, $\infty$)
(c) x$\in$($-\infty$, 10]
(d) x$\in$[$-$10, 10)

Hence, the correct option is (a).

#### Question 3:

Mark the correct alternative in each of the following:
Given that xy and are real numbers and x$<$yb$>$0, then
(a) $x}{b
(b) $x}{b\le y}{b}}$
(c) $x}{b>y}{b}}$
(d) $x}{b\ge y}{b}}$

Given that xy and are real numbers and x$<$yb$>$0.

Both sides of an inequality can be multiplied or divided by the same positive number.

Hence, the correct option is (a).

#### Question 4:

Mark the correct alternative in each of the following:
If is a real number and $\left|x\right|$$<$5, then
(a) x$\ge$5
(b) $-$5$<$x$<$5
(c) x$\le$$-$5
(d) $-$5$\le$x$\le$5

If is a real number.
$\left|x\right|$$<$5
$-$5$<$x$<$5

Hence, the correct option is (b).

#### Question 5:

Mark the correct alternative in each of the following:
If and are real numbers such that a$>$0 and $\left|x\right|$$>$a, then
(a) x$\in$($-$a$\infty$)
(b) x$\in$[$-$$\infty$a]
(c) x$\in$($-$aa)
(d) x$\in$($-$$\infty$$-$a$\cup$ (a$\infty$)

If and are real numbers such that a$>$0.

$\left|x\right|$$>$a

Hence, the correct option is (d).

#### Question 6:

Mark the correct alternative in each of the following:
If $\left|x-1\right|$$>$5, then
(a) x$\in$($-$4, 6)
(b) x$\in$[$-$4, 6]
(c) x$\in$($-$$\infty$$-$4) $\cup$ (6, $\infty$)
(d) x$\in$($-$$\infty$$-$4) $\cup$ [6. $\infty$)

Hence, the correct option is (c).

#### Question 7:

Mark the correct alternative in each of the following:
If $\left|x+2\right|$$\le$9, then
(a) x$\in$($-$7, 11)
(b) x$\in$[$-$11, 7]
(c) x$\in$($-$$\infty$$-$7) $\cup$ (11, $\infty$)
(d) x$\in$($-$$\infty$$-$7) $\cup$ [11, $\infty$)

Hence, the correct option is (b).

#### Question 8:

Mark the correct alternative in each of the following:
The inequality representing the following graph is
(a) $\left|x\right|$$<$3
(b) $\left|x\right|$$\le$3
(c) $\left|x\right|$$>$3
(d) $\left|x\right|$$\ge$3

As according to the graph,

Hence, the correct option is (b).

#### Question 9:

Mark the correct alternative in each of the following:
The linear inequality representing the solution set given in Fig. 15.44 is
(a) $\left|x\right|$$<$5
(b) $\left|x\right|$$>$5
(c) $\left|x\right|$$\ge$5
(d) $\left|x\right|$$\le$5

As according to the graph,

Hence, the correct option is (c).

#### Question 10:

Mark the correct alternative in each of the following:
The solution set of the inequation $\left|x+2\right|$$\le$5 is
(a) ($-$7, 5)
(b) [$-$7, 3]
(c) [$-$5, 5]
(d) ($-$7, 3)

Hence, the correct option is (b).

#### Question 11:

Mark the correct alternative in each of the following:
If $\frac{\left|x-2\right|}{x-2}$$\ge$0, then
(a) x$\in$[2, $\infty$)
(b) x$\in$(2, $\infty$)
(c) x$\in$($-$$\infty$, 2)
(d) x$\in$($-$$\infty$, 2]

Hence, the correct option is (b).

#### Question 12:

Mark the correct alternative in each of the following:
If $\left|x+3\right|$$\ge$10, then
(a) x$\in$($-$13, 7]
(b) x$\in$($-$13, 7)
(c) x$\in$($-$$\infty$$-$13) $\cup$ (7, $\infty$)
(d) x$\in$($-$$\infty$$-$13] $\cup$ [7, $\infty$)

Hence, the correct option is (d).

#### Question 13:

Solution of a linear inequality in variable x is represented on the number line as shown in the given figure. The solution can also be described as

(a) x ∈ (–∞, 5)
(b) x ∈ (–∞, 5]
(c) x ∈ [5, ∞)
(d) x ∈ (5, ∞)

Since, number line representing solution does not include 5.
But, include every value after 5
∴ Solution can be described by
x ∈ (5, ∞)
Hence, the correct answer is option D.

#### Question 14:

The shaded part of the number line in given figure can also be represented as

(a)

(b)

(c)

(d)

Since shaded part includes $\frac{9}{2}$ and every point after $\frac{9}{2}$
Shaded region can be represented by $\left[\frac{9}{2},\infty \right)$
Hence, the correct answer is option B.

#### Question 15:

The shaded part of the number line in the given figure can also be described as

(a) (–∞, 1) ∪ (2, ∞)
(b) (–∞, 1] ∪ [2, ∞)
(c) (1, 2)
(d) [1, 2]

Since shaded region does not include 1 and 2 but includes every value before 1 and every value after 2.
Hence, shaded region is given by (–∞, 1) ∪ (2, ∞)
Hence, the correct answer is option A.

#### Question 1:

If x ≥ –3, then x + 5 _______ 2.

If x ≥ –3
By adding 5 to both sides,
we get
x + 5 ≥ –3 + 5
i.e. x + 5 ≥ 2

#### Question 2:

If  –x ≤ –4, then 2x _______ 8.

If x ≤ –4
i.e. x4                 (multiplying by 1)
i.e. 2x 8               (multiplying by 2)

#### Question 3:

If  $\frac{1}{x-2}<0,$ then x _______ 2.

#### Question 4:

If |x – 1| ≤ 2 then –1 _______ x < 3.

If  |x – 1| ≤ 2
i.e. –2 ≤ x – 1 ≤ 2                      (By defination of modulus inequality)
Now, By adding 1 throught the inequality,
We get,
–2 + 1 ≤ x – 1 + 1 ≤ 2 + 1
i.e –1 ≤ x ≤ 3

#### Question 5:

If |3x – 7| > 2, then x _______ $\frac{5}{3}$ or, x _______ 3.

|3x – 7| > 2
By defination of modulus inequality;
3x – 7 > 2 or 3x – 7 < –2
If 3x – 7 > 2
By adding 7 to both sides,
We get,
3x > 9
i.e. x > 9
If 3x – 7 < – 2
Now, by adding 7 to both sides,
We get,
3x – 7 + 7 <  – 2 + 7
i.e. 3x < 5

#### Question 6:

If – 4x ≥ 2, then x _______ –3.

Since, –4x ≥ 2

Now, multiply both sides by –1,
We get,
$x\le \frac{-1}{2}$

#### Question 7:

If $-\frac{3x}{4}\le -3,$ then x _______ 4.

#### Question 8:

If x > y and z < 0, then –xz _______ –yz.

#### Question 9:

The solution set of the inequation |x + 1| < 3 is __________.

#### Question 10:

The solution set of the inequation |x + 2| > 5 is

#### Question 11:

If $\frac{x-3}{\left|x-3\right|}\ge 0,$ then x belongs to the interval ___________.

#### Question 12:

The solution set of the inequation $\frac{\left|x\right|+1}{\left|x\right|-1}<0$ is ___________.

#### Question 1:

Write the solution of the inequation $\frac{{x}^{2}}{x-2}>0$.

.

#### Question 2:

Write the solution set of the inequation $x+\frac{1}{x}\ge 2$.

#### Question 3:

Write the set of values of x satisfying the inequation (x2 − 2x + 1) (x − 4) < 0.

We have:Therefore, the solution of the given inequality is x(2,)

herefore, the solution of the given inequality is x(2,)

#### Question 4:

Write the solution set of the equation |2 − x| = x − 2.

#### Question 5:

Write the set of values of x satisfying |x − 1| ≤ 3 and |x − 1| ≥ 1.

#### Question 6:

Write the solution set of the inequation $\left|\frac{1}{x}-2\right|>4$

#### Question 7:

Write the number of integral solutions of $\frac{x+2}{{x}^{2}+1}>\frac{1}{2}$.

#### Question 8:

Write the set of values of x satisfying the inequations 5x + 2 < 3x + 8 and $\frac{x+2}{x-1}<4$.

#### Question 9:

Write the solution of set of $\left|x+\frac{1}{x}\right|>2$.