**Question:**

Solve the following systems of linear inequations graphically:

(i) 2*x* + 3*y* ≤ 6, 3*x* + 2*y* ≤ 6, *x* ≥ 0, *y* ≥ 0

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

(iii) *x* − *y* ≤ 1, *x* + 2*y* ≤ 8, 2*x* + *y* ≥ 2, *x* ≥ 0, *y* ≥ 0

(iv) *x* + *y* ≥ 1, 7*x* + 9*y* ≤ 63, *x* ≤ 6, *y* ≤ 5, *x* ≥ 0, *y* ≥ 0

(v) 2*x* + 3*y* ≤ 35, *y* ≥ 3, *x* ≥ 2, *x* ≥ 0, *y* ≥ 0

**Solution:**

(i) Converting the inequations to equations, we obtain:

2*x* + 3*y* = 6, 3*x* + 2*y* = 6, *x* = 0, *y* = 0

2*x* + 3*y* =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 2*x* + 3*y* ≤ 6

So, the portion containing the origin represents the solution set of the inequation 2*x* + 3*y* ≤ 6

3*x*+2*y* =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 3*x* + 2*y* ≤ 6.

So, the portion containing the origin represents the solution set of the inequation 3*x* + 2*y* ≤ 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:

2*x* + 3*y* = 6, *x* + 4*y* = 4, *x* = 0, *y* = 0

2*x* + 3*y* =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 2*x* + 3*y* ≤ 6.

So, the portion containing the origin represents the solution set of the inequation 2*x* + 3*y* ≤ 6

*x* + 4*y* = 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* + 4*y* ≤ 4.

So, the portion containing the origin represents the solution set of the inequation *x* + 4*y* ≤ 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:

$x-y=1, x+2 y=8,2 x+y=2, x=0, y=0$

$x-y=1$ : 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 $x-y \leq 1$ So, the portion containing the origin represents the solution set of the inequation $x-y \leq 1$

$x+2 y=8$ : 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 $x+2 y \leq 8$ So, the portion containing the origin represents the solution set of the inequation $x+2 y \leq 8$

$2 x+y=2$ : 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 $2 x+y \geq 2$ So, the portion that does not contain the origin represents the solution set of the inequation $2 x+y \geq 2$

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, 7*x* + 9*y* = 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 $x+y \geq 1$ So, the portion not containing the origin represents the solution set of the inequation $x+y \geq 1$

$7 x+9 y=63$ : 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 $7 x+9 y \leq 63$ So, the portion containing the origin represents the solution set of the inequation $7 x+9 y \leq 63$

$x=6$ : 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 $x \leq 6$ So, the portion containing the origin represents the solution set of the inequation $x \leq$ 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 $y \leq 5$ So, the portion containing the origin represents the solution set of the inequation $y \leq 5$

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:

2*x* + 3*y* = 35, *x* = 0, *y* = 0

2*x* + 3*y* = 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 2*x* + 3*y* ≤ 35 So, the portion containing the origin represents the solution set of the inequation 2*x* + 3*y* ≤ 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 $x \geq 2$ So, the portion that does not contain the origin represents the solution set of the inequation $x \geq 2$

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.