tentukan daerah penyelesaian syatem pertidaksamaan berikut ini a.x+4y ≤12,6x-5y≤30,8x+5y≥40,x≥0,y≥0. dijawab ya kakak​

Jawaban:

x + 4y ≤ 12.

x + 4y = 12

x + 0 = 12

x = 12 (12, 0)

x + 4y ≤ 12

x + 4y = 12

0 + 4y = 12

4y = 12

y = 12/4

y = 3 (0, 3)

6x – 5y ≤ 30

6x – 5y = 30

6x – 0 = 30

6x = 30

x = 30/6

x = 5 (5, 0)

6x – 5y ≤ 30

6x – 5y = 30

0 – 5y = 30

-5y = 30

y = 30/-5

y = -6 (0, -6)

8x + 5y ≥ 40

8x + 5y = 40

8x + 0 = 40

8x = 40

x = 40/8

x = 5 (5, 0)

8x + 5y ≥ 40

8x + 5y = 40

0 + 5y = 40

5y = 40

y = 40/5

y = 8 (0, 8)