Berdasarkan grafik:
Arsiran di sebelah atas sb. x maka y ≥ 0
Arsiran di sebelah kanan sb. y maka x ≥ 0
Garis 1:
(x₁, y₁) = (4, 0)
(x₂, y₂) = (0, 8)
Arah arsiran ke kanan, berarti lebih besar sama dengan (≥)
[tex]\sf \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{x-x_{1}}{x_{2}-x_{1}}\\\\\frac{y-0}{8-0}=\frac{x-4}{0-4}\\ \\ \frac{y}{8} =\frac{x-4}{-4} \\\\y = \frac{8(x-4)}{-4} \\\\y= -2(x-4)\\\\y=-2x+8\\\\2x+y\ge8[/tex]
Garis 2:
(x₁, y₁) = (9, 0)
(x₂, y₂) = (0, 5)
Arah arsiran ke kiri, berarti lebih kecil sama dengan (≤)
[tex]\sf \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{x-x_{1}}{x_{2}-x_{1}}\\\\\frac{y-0}{5-0}=\frac{x-9}{0-9}\\\\\frac{y}{5} = \frac{x-9}{-9} \\\\y= \frac{5(x-9)}{-9} \\\\-9y = 5x-45\\\\5x+9y \le 45[/tex]
Jadi, HP = {5x + 9y ≤ 45; 2x + y ≥ 8; x ≥ 0; y ≥ 0}.
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Verified answer
Sistem Pertidaksamaan Linear Dua Variabel (SPLDV)
Berdasarkan grafik:
Arsiran di sebelah atas sb. x maka y ≥ 0
Arsiran di sebelah kanan sb. y maka x ≥ 0
Garis 1:
(x₁, y₁) = (4, 0)
(x₂, y₂) = (0, 8)
Arah arsiran ke kanan, berarti lebih besar sama dengan (≥)
[tex]\sf \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{x-x_{1}}{x_{2}-x_{1}}\\\\\frac{y-0}{8-0}=\frac{x-4}{0-4}\\ \\ \frac{y}{8} =\frac{x-4}{-4} \\\\y = \frac{8(x-4)}{-4} \\\\y= -2(x-4)\\\\y=-2x+8\\\\2x+y\ge8[/tex]
Garis 2:
(x₁, y₁) = (9, 0)
(x₂, y₂) = (0, 5)
Arah arsiran ke kiri, berarti lebih kecil sama dengan (≤)
[tex]\sf \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{x-x_{1}}{x_{2}-x_{1}}\\\\\frac{y-0}{5-0}=\frac{x-9}{0-9}\\\\\frac{y}{5} = \frac{x-9}{-9} \\\\y= \frac{5(x-9)}{-9} \\\\-9y = 5x-45\\\\5x+9y \le 45[/tex]
Jadi, HP = {5x + 9y ≤ 45; 2x + y ≥ 8; x ≥ 0; y ≥ 0}.