**Solve the following system of inequalities graphically:**

**Question 1. x ≥ 3, y ≥ 2**

**Solution.**

x ≥ 3, y ≥ 2

**(i)** AB represents the line x = 3

Putting x = 0 in x ≥ 3

0 ≥ 3, which is not true.

Therefore, origin does not lie in the region of x ≥ 3

Its graph is shaded on the right side of AB.

**(ii)** CD represents the line y = 2

Putting y = 0 in y ≥ 2 0 ≥ 2, which is not true.

Therefore, origin does not lie in the region of y ≥ 2.

Its graph is shaded above the CD.

Solution of system x ≥ 3 and y ≥ 2 is shown as the shaded region.

**Question 2. 3x + 2y ≤ 12, x ≥ 1, y ≥ 2**

**Solution.**

Inequalities are 3x + 2y ≤ 12, x ≥ 1, y ≥ 2

**(i)** The line l_{1} : 3x + 2y = 12 passes through (4, 0), (0, 6)

AB represents the line, 3x + 2y = 12.

Consider the inequality 3x + 2y ≤ 12

Putting x = 0, y = 0 in 3x + 2y ≤ 12

0 + 0 ≤ 12, which is true.

Therefore, origin lies in the region 3x + 2y ≤ 12

∴ The region lying below the line AB including all the points lying on it.

**(ii)** The line l_{2} : x = 1 passes through (1, 0). This line is represented by EF. Consider the inequality x ≥ 1

Putting x = 0, 0 ≥ 1, which is not true.

Therefore, origin does not lie in the region of x ≥ 1

The region lies on the right of EF and the points on EF from the inequality x ≥ 1.

**(iii)** The line l_{3} : y = 2 passes through (0, 2). This line is represented by CD.

Consider the inequality y ≥ 2

Putting y = 0 in y ≥ 2, we get 0 ≥ 2 which is false.

∴ Origin does not lie in the region of y ≥ 2. y ≥ 2 is represented by the region above CD and all the points on this line. Hence, the region satisfying the inequalities.

3x + 2y ≥ 12, x ≥ 1, y ≥ 2 is the APQR.

**Question 3. 2x + y ≥ 6, 3x + 4y ≤ 12**

**Solution.**

The inequalities are 2x + y ≥ 6, 3x + 4y ≤ 12

**(i)** The line l_{1} : 2x + y = 6 passes through (3, 0), (0, 6)

AB represents the line 2x + y = 6

Putting x = 0, y = 0 in 2x + y ≥ 6 0 ≥ 6, which is false.

∴ Origin does not lie in the region of 2x + y ≥ 6 Therefore, the region lying above the line AB and all the points on AB represents the inequality 2x + y ≥ 6

**(ii)** The line l_{2} : 3x + 4y = 12 passes through (4, 0) and (0, 3).

This line is represented by CD.

Consider the inequality 3x + 4y ≤ 12

Putting x = 0, y = 0 in 3x + 4y ≤ 12, we get 0 ≤ 12, which is true.

∴ 3x + 4y ≤ 12 represents the region below the line CD (towards origin) and all the points lying on it.

The common region is the solution of 2x + 3y ≥ 6 are 3x + 4y ≤ 12 represented by the

shaded region in the graph.

**Question 4. x + y ≥ 4, 2x – y > 0**

**Solution.**

The inequalities are , x + y ≥ 4, 2x – y > 0

**(i)** The line l_{1}: x + y = 4 passes through (4, 0) and (0, 4). This line is represented by AB. Consider the inequality x + y ≥ 4

Putting x = 0, y = 0 in x + y ≥ 4, we get 0 ≥ 4, which is false.

Origin does not lie in this region.

Therefore, ,r + y > 4 is represented by the region above the line x + y = 4 and all points lying on it.

**(ii)** The line l_{2} : 2x – y = 0 passes through (0, 0) and (1, 2).

This line is represented by CD.

Consider the inequality 2x – y > 0

Putting x = 1, y = 0, we get 2 > 0, which is true

This shows (1, 0) lies in the region.

i.e. region lying below the line 2x – y = 0

represents 2x — y > 0

∴ The common region to both inequalities is shaded region as shown in the figure.

**Question 5. 2x – y> 1, x – 2y < -1**

**Solution.**

The inequalities are 2x – y > 1 and x – 2y < -1

**(i)** Let us draw the graph of line

l_{1} : 2x – y = 1, passes through and

(0, -1) which is represented by AB. Consider the inequality 2x – y > 1.

Putting x = y = 0, we get 0 > 1, which is false.

Therefore, origin does not lie in region of 2x – y > 1 i.e., 2x – y > 1 represents the area below the line AB excluding all the points lying on 2x – y = 1.

**(ii)** Let us draw the graph of the line

l_{2} : x – 2y = -1, passes through (-1, 0) and (0,1/2) which is represented by CD.

Consider the inequality x – 2y < -1

Putting x = y = 0, we have 0 < -1, which is false.

Therefore, origin does not lie in region of x – 2y < -1 i.e., x – 2y < -1 represents the area above the line CD excluding all the points lying on x – 2y = -1

⇒ The common region of both the inequality is the shaded region as shown in figure.

**Question 6. x + y ≤ 6, x + y ≥ 4**

**Solution.**

The inequalities are

x + y ≤ 6 and x + y ≥ 4

**(i)** The line l_{1}: x + y = 6 passes through (6, 0) and (0, 6). It is represented by AB.

Consider the inequality x + y ≤ 6

Putting x = 0, y = 0 in x + y ≤ 6 0 ≤ 6, which is true.

∴ Origin lies in the region of x + y ≤ 6

∴ x + y ≤ 6 is represented by the region below the line x + y = 6 and all the points lying on it.

**(ii)** The line l_{2} : x + y = 4 passes through (4, 0) and (0, 4). It is represented by CD.

Consider the inequality x + y ≥ 4

Putting x = 0, y = 0 in x + y ≥ 4 or, 0 + 0 ≥ 4, which is false.

∴ Origin does not lie in the region of x + y ≥ 4

∴ x + y ≥ 4 is represented by the region above the line x + y = 4 and all the points lying on it.

∴ The solution region is the shaded region between AB and CD as shown in the figure.

**Question 7. 2x + y ≥ 8, x + 2y ≥ 10**

**Solution.**

The inequalities are 2x + y ≥ 8 and x + 2y ≥ 10

**(i)** Let us draw the graph of the line

l_{1} : 2x + y = 8 passes through (4, 0) and (0, 8) which is represented by AB.

Consider the inequality 2x + y ≥ 8

Putting x = y = 0, we get 0 ≥ 8, which is false.

∴ Origin does not lie in the region of 2x + y ≥ 8.

i.e., 2x + y ≥ 8 represents the area above the line AB and all the points lying on 2x + y = 8.

**(ii)** Let us draw the graph of line

l_{2} : x + 2y = 10, passes through (10, 0) and (0, 5) which is represented by CD. Consider the inequality x + 2y ≥ 10

Putting x = 0, y = 0, we have 0 ≥ 10, which is false.

∴ The origin does not lie in region of x + 2y ≥ 10

i.e. x + 2y ≥ 10 represents the area above the line CD and all the points lying on x + 2y = 10.

⇒ The common region of both the inequality is the shaded region as shown in the figure.

**Question 8. x + y ≤ 9, y > x, x ≥ 0**

**Solution.**

The inequalities are x + y ≤ 9, y > x and x ≥ 0

**(i)** Consider the inequality x + y ≤ 9

The line l_{1} : x + y = 9 passes through (9, 0) and (0, 9). AB represents this line.

Putting x = 0, y = 0 in x + y ≤ 9

0 + 0 = 0 ≤ 9, which is true.

Origin lies in this region. i.e., x + y ≤ 9 represents the area below the line AB and all the points lying on x + y = 9.

**(ii)** The line l_{2} : y = x, passes through the origin and (2, 2).

∴ CD represents the line y = x

Consider the inequality y – x > 0

Putting x = 0,y = 1 in y – x > 0

1 – 0 > 0, which is true.

∴ (0, 1) lies in this region.

The inequality y > x is represented by the region above the line CD, excluding all the points lying on y – x = 0.

**(iii)** The region x ≥ 0 lies on the right of y-axis.

∴ The common region of the inequalities is the region bounded by ΔPQO is the solution of x + y ≤ 9, y > x, x ≥ 0.

**Question 9. 5x + 4y ≤ 20, x ≥ 1, y ≥ 2**

**Solution.**

The inequalities are 5x + 4y ≤ 20, x ≥ 1, y ≥ 2

**(i)** The line l_{1} : 5x + 4y = 20 passes through (4, 0) and (0, 5). This line is represented by AB.

Consider the inequality 5x + 4y ≤ 20

Putting x = 0, y = 0

0 + 0 = 0 ≤ 20, which is true.

The origin lies in this region, i.e., region below the line 5x + 4y = 20 and all the points lying on it belong to 5x + 4y ≤ 20.

**(ii)** The line l_{2} : y = 2, line is parallel to x-axis at a distance 2 from the origin. It is represented by EF. Putting y = 0, 0 ≥ 2 is not true.

Origin does not lie in this region.

Region above y = 2 represents the inequality y ≥ 2 including the points lying on it.

**(iii)** The line l_{3} : x = 1, line parallel to y-axis at a distance 1 from the origin. It is represented by CD. Putting x = 0in x – 1 ≥ 0

-1 ≥ 0, which is not true.

Origin does not lie in this region.

∴ The region on the right of x = 1 and all the points lying on it belong to x ≥ 1.

∴ Shaded area bounded by ΔPQR is the solution of given inequalities.

**Question 10. 3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0**

**Solution.**

The inequalities are

We first draw the graphs of lines

l_{1} : 3x + 4y = 60, l_{2} : x + 3y – 30, x = 0 and y = 0.

**(i)** The line 3x + 4y = 60 passes through (20, 0) and (0,15) which is represented by AB. Consider the inequality 3x + 4y ≤ 60, putting x = 0, y = 0 in 3x + 4y ≤ 60, we get 0 + 0 ≤ 60, which is true.

∴ 3x + 4y ≤ 60 represents the region below AB and all the points on AB.

**(ii)** Further, x + 3y = 30 passes through (0,10) and (30, 0), CD represents this line.

Consider the inequality x + 3y ≤ 30

Putting x = 0, y = 0 in x + 3y < 30, w’e get 0 < 30 is true.

∴Origin lies in the region x + 3y ≤ 30. This inequality represents the region below it and the line itself.

Thus, we note that inequalities (1) and (2) represent the two regions below the respective lines (including the lines).

**Inequality (3)** represents the region on the right of y-axis and the i/-axis itself.

**Inequality (4) **represents the region above x-axis and the x-axis itself.

**∴** Shaded area in the figure is the solution area.

**Question 11. 2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6**

**Solution.**

We have the inequalities :

We first draw the graphs of lines

l_{1} : 2x + y = 4, l_{2} : x + y = 3 and l_{3} : 2x – 3y = 6

**(i)** 2x + y = 4, passes through (2, 0) and (0, 4) which represents by AB

Consider the inequality 2x + y ≥ 4

Putting x = 0, y = 0 in 2x + y ≥ 4, we get 0 ≥ 4 is false.

∴ Origin does not lie in the region of 2x + y ≥ 4 This inequality represents the region above the line AB and all the points on the line AB.

**(ii)** Again, x + y = 3 is represented by the line CD, passes through (3, 0) and (0, 3). Consider the inequality x + y ≤ 3, putting x = 0,y = 0 in x + y ≤ 3, we get 0 ≤ 3 is true.

∴ Origin lies in the region of x + y ≤ 3

∴ x + y ≤ 3 represents the region below the line CD and all the points on the line CD.

**(iii)** Further, 2x – 3y = 6 is represented by EF passes through (0, -2) and (3, 0).

Consider the inequality 2x – 3y ≤ 6, putting x = 0, y = 0 in 2x – 3y ≤ 6, we get 0 ≤ 6, which is true.

∴ Origin lies in it.

∴ 2x – 3y ≤ 6 represents the region above the line EF and all the points on the line EF.

∴ Shaded triangular area in the figure is the solution of given inequalities.

**Question 12. x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1**

**Solution.**

The inequalities are

x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1

**(i)** The line l_{1}: x – 2y = 3 passes through (3, 0) and

This is represented by AB. Consider the inequality x – 2y ≤ 3, putting x = 0, y = 0 we get 0 ≤ 3, which is true.

⇒ Origin lies in the region of x – 2y ≤ 3.

Region on the above of this line and including its points represents x – 2y ≤ 3

**(ii)** The line l_{2} : 3x + 4y = 12 passes through (4, 0) and (0, 3). CD represents this line. Consider the inequality 3x + 4y ≥12 putting x = 0, y = 0, we get 0 ≥ 12 which is false.

∴ Origin does not lie in the region of 3x + 4y ≥ 12.

The region above the line CD and including points of the line CD represents 3x + 4y ≥ 12.

**(iii)** x ≥ 0 is the region on the right of Y-axis and all the points lying on it.

**(iv)** The line l_{3} : y = 1 is the line parallel to X-axis at a distance 1 from it. Consider y ≥ 1 or y – 1 ≥ 0, putting y = 0 in y -1 ≥ 0

We get -1 ≱ 0, origin does not lie in the region.

y ≥ 1 is the region above y = 1 and the points lying on it.

∴ The shaded region shown in figure represents the solution of the given inequalities.

**Question 13. 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0**

**Solution.**

The inequalities are 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0

**(i)** The line l_{1} : 4x + 3y = 60 passes through (15, 0), (0, 20) and it is represented by AB. Consider the inequality 4x + 3y ≤ 60

Putting x = 0, y = 0.

0 + 0 = 0 ≤ 60 which is true,

therefore, origin lies in this region.

Thus, region is below the line AB and the points lying on the line AB represents the inequality 4x + 3y ≤ 60.

**(ii)** The line l_{2} : y = 2x passes through (0, 0). It is represented by CD.

Consider the inequality y ≥ 2x. Putting x = 0, y = 5 in y – 2x ≥ 0 5 ≥ 0 is true.

∴ (0, 5) lies in this region.

Region lying above the line CD and including the points on the line CD represents y ≥ 2x

**(iii)** x ≥ 3 is the region lying on the right of line l_{3} : x = 3 and points lying on x = 3 represents the inequality x ≥ 3.

∴ The shaded area APQR in which x ≥ 0 and y ≥ 0 is true for each point, is the solution of given inequalities.

**Question 14. 3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0**

**Solution.**

The inequalities are 3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0

**(i)** The line l_{1} : 3x + 2y = 150 passes through the points (50,0) and (0, 75). AB represents the line. Consider the inequality 3x + 2y ≤ 150.

Putting x = 0, y = 0 in 3x + 2y ≤ 150

⇒ 0 ≤ 150 which is true, shows that origin lies in this region.

The region lying below the line AB and the points lying on AB represents the inequality 3x + 2y < 150.

**(ii)** The line l_{2} : x + 4y = 80 passes through the points (80, 0), (0, 20). This is represented by CD.

Consider the inequality x + 4y ≤ 80 putting x = 0, y = 0, we get 0 ≤ 80, which is true.

⇒ Region lying below the line CD and the points on the line CD represents the inequality x + 4y ≤ 80

**(iii)** x ≤ 15 is the region lying on the left to

l_{3} : x = 15 represented by EF and the points lying on EF.

**(iv)** x ≥ 0 is the region lying on the right side of Y-axis and all the points on Y-axis.

**(v)** y ≥ 0 is the region lying above the X-axis and all the points on X-axis.

Thus, the shaded region in the figure is the solution of the given inequalities.

**Question 15. x+2y ≤ 10 , x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0**

**Solution.**

The inequalities are x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0

**(i)** l_{1} : x + 2y = 10 passes through (10, 0) and (0, 5). The line AB represents this equation. Consider the inequality x + 2y ≤ 10 putting x = 0, y = 0, we get 0 ≤ 10 which is true.

∴ Origin lies in the region of x + 2y ≤ 10.

∴ Region lying below the line AB and the points lying on it represents x + 2y ≤ 10

**(ii)** l_{2} : x + y = 1 passes through (1, 0) and (0, 1). Thus line CD represents this equation. Consider the inequality x + y ≥ 1 putting x = 0, y = 0, we get 0 ≥ 1, which is not true. Origin does not lie in the region of x + y ≥ 1.

∴ The region lying above the line CD and the points lying on it represents the inequality x + y ≥1

**(iii)** l_{3} : x – y = 0, passes through (0, 0). This is being represented by EF.

Consider the inequality x – y ≤0, putting x = 0, y = 1, We get 0 – 1 ≤ 0 which is true

⇒ (0,1) lies on x – y ≤ 0

The region lying above the line EF and the points lying on it represents the inequality x – y ≤ 0.

**(iv)** x ≥ 0 is the region lying on the right of Y-axis and the points lying on x = 0.

**(v)** y ≥ 0 is the region above X-axis, and the points lying on y = 0.

∴ The shaded area in the figure represents the given inequalities.