Solve the following systems of inequations graphically:
Question:

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

Solution:

(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 $2 x+y \geq 8$.

Therefore, the region that does not contain the origin is the solution of the inequality $2 x+y \geq 8$

$x+2 y=8$ : 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 $x+2 y \geq 8$

Therefore, the region that does not contain the origin is the solution of the inequality $x+2 y \geq 8$

$x+y=6$ : 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 $x+y \leq 6$ Therefore, the region containing the origin is the solution of the inequality $x$ $+y \leq 6$

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 $12 x+12 y \leq 840$

Therefore the region containing the origin is the solution of the inequality $12 x+12 y \leq 840$

$3 x+6 y=300$ : 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 $3 x+6 y \leq 300$

Therefore, the region containing the origin is the solution of the inequality $3 x+6 y \leq 300$

$8 x+4 y=480$ : 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 $8 x+4 y \leq 480$ Therefore, the region containing the origin is the solution of the inequality $8 x+4 y \leq 480$

Also, $x \geq 0, y \geq 0$ 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 $x+2 y \leq 40$

Therefore, the region containing the origin is the solution of the inequality $x+2 y \leq 40$

$3 x+y=30$ : 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 $3 x+y \geq 30$

Therefore, the region that does not contain the origin is the solution of the inequality $3 x+y \geq 30$

$4 x+3 y=60$ : 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 $4 x+3 y \geq 60$ Therefore, the region that does not contain the origin is the solution of the inequality $4 x+3 y \geq 60$

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 + =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 $5 x+y \geq 10$

Therefore, the region that does not contain the origin is the solution of the inequality $5 x+y \geq 10$

$2 x+2 y=12$ : 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 $2 x+2 y \geq 12$

Therefore, the region that does not contain the origin is the solution of the inequality $2 x+2 y \geq 12$

$x+4 y=12$ : 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 $x+4 y \geq 12$

Therefore, the region that does not contain the origin is the solution of the inequality $x+4 y \geq 12$

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.

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