Question

this is from linear inequalities grade 11

Answer 1

We observe that the shaded region and the origin are on the same side of the line 3x + 2y = 48.

For (0, 0), we have 3(0) + 2(0) – 48 < 0. So, the shaded region satisfies the inequality 3x + 2y≤ 48.

Also, the shaded region and the origin are on the same side of the line x+y = 20.

For (0,0), we have 0 + 0 – 20 < 0. So, the shaded region satisfies the inequality x +y ≤ 20.

Also, the shaded region lies in the first quadrant. So, x ≥0,y≥0.

Thus, the linear in equation corresponding to the given solution set are 3x + 2y ≤ 48, x + y ≤ 20 and x ≥ 0, y ≥ 0.

Answer 2

Answer:

We observe that the shaded region and the origin are on the same side of the line 3x + 2y = 48.

For (0, 0), we have 3(0) + 2(0) – 48 < 0. So, the shaded region satisfies the inequality 3x + 2y≤ 48.

Also, the shaded region and the origin are on the same side of the line x+y = 20.

For (0,0), we have 0 + 0 – 20 < 0. So, the shaded region satisfies the inequality x +y ≤ 20.

Also, the shaded region lies in the first quadrant. So, x ≥0,y≥0.

Thus, the linear in equation corresponding to the given solution set are 3x + 2y ≤ 48, x + y ≤ 20 and x ≥ 0, y ≥ 0.

Step-by-step explanation:

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