Question

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) x − y ≤ 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) 2x + 3y ≤ 6, 3x + 2y ≤ 6, x ≥ 0, y ≥ 0

(ii) 2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0

(iii) x − y ≤ 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

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Solution

(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 2x + y $\ge $2 So, the portion that does not contain the origin represents the solution set of the inequation 2x + y $\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 x + y $\ge $1 So, the portion not containing the origin represents the solution set of the inequation x + y $\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.

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 2x + y $\ge $2 So, the portion that does not contain the origin represents the solution set of the inequation 2x + y $\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 x + y $\ge $1 So, the portion not containing the origin represents the solution set of the inequation x + y $\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.

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