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A. Tujuan pembelajaran: Peserta didik secara berkelompok memahami dari contoh yang
diberikan mengenai cara menentukan sistem pertidaksamaan linear dua variabel dari
daerah penyelesaian dengan tepat.
B. Perintah:
1. Pahami CONTOH yang diberikan, diskusikan dengan kelompokmu bagaimana cara
menentukan sistem pertidaksamaan linear dua variabel dari daerah penyelesaian yang
diberikan dalam contoh.
2. Selesaikan SOAL yang disediakan tanpa melihat contoh. Setelah selesai, periksa
kembali apakah jawaban sudah sesuai. Jawaban tidak harus ditulis sama persis dengan
contoh, boleh disingkat, boleh lebih panjang, boleh menerapkan cara yang berbeda.
Pelajari Contoh 1 (5 menit)
Daerah yang tidak diarsir (Bersih) pada grafik dibawah merupakan daerah penyelesaian
dari suatu sistem pertidaksamaan. Tentukan sistem pertidaksamaan tersebut.
LEMBAR KERJA SISWA (LKS)
Penyelesaian:
Langkah 1: Tentukan titik potong dengan sumbu X dan sumbu Y = (2,0) dan (0,4)
Langkah 2: Buatlah persamaan garisnya
𝑏𝑥 + 𝑎𝑦 = 𝑎. 𝑏
4𝑥 + 2𝑦 = 4.2
4𝑥 + 2𝑦 = 8 ↔ 2𝑥 + 𝑦 = 4
Langkah 3: Uji titik O (0,0) yang terletak pada daerah penyelesaian = 2(0) + 0 ≤ 4 sehingga
diperoleh pertidaksamaan 2𝑥 + 𝑦 ≤ 4
Langkah 4: Daerah yang tidak diarsir (bersih) terletak di:
sebelah kanan sumbu Y, maka 𝑥 ≥ 0
sebelah atas sumbu X, maka 𝑦 ≥ 0
Langkah 5: Simpulan
Jadi, sistem pertidaksamaannya adalah 2𝑥 + 𝑦 ≤ 4; 𝑥 ≥ 0; 𝑦 ≥ 0; 𝑥, 𝑦 ∈ 𝑹
Pelajari Contoh 2
Daerah yang tidak diarsir (Bersih) pada grafik dibawah merupakan daerah penyelesaian
dari suatu sistem pertidaksamaan. Tentukan sistem pertidaksamaan tersebut.
Penyelesaian:
Langkah 1: Tentukan persamaan garis 1 (g1) melalui (4,0) dan (0,4), sehingga persamaannya
adalah 4𝑥 + 4𝑦 = 16 ↔ 𝑥 + 𝑦 = 4
Uji titik O (0,0) yang terletak pada daerah penyelesaian, 0 + 0 ≤ 4, sehingga
diperoleh 𝑥 + 𝑦 ≤ 4
Langkah 2: Tentukan persamaan garis 1 (g1) melalui (6,0) dan (0,2), sehingga persamaannya
adalah 2𝑥 + 6𝑦 = 12 ↔ 𝑥 + 3𝑦 = 6
Uji titik O (0,0) yang terletak pada daerah penyelesaian, 0 + 3(0) ≤ 6, sehingga
diperoleh 𝑥 + 3𝑦 ≤ 6
Langkah 3: Daerah yang diarsir terletak di:
sebelah kanan sumbu Y, maka 𝑥 ≥ 0
sebelah atas sumbu X, maka 𝑦 ≥ 0
Langkah 4: Simpulan
Jadi, sistem pertidaksamaannya adalah 𝑥 + 𝑦 ≤ 4; 𝑥 + 3𝑦 ≤ 6; 𝑥 ≥ 0; 𝑦 ≥ 0; 𝑥, 𝑦 ∈ 𝑹
Kerjakan soal
Soal 1a.
Daerah yang tidak diarsir (Bersih) pada grafik dibawah merupakan daerah penyelesaian
dari suatu sistem pertidaksamaan. Tentukan sistem pertidaksamaan tersebut.
Soal 1b
Daerah yang tidak diarsir (Bersih) pada grafik dibawah merupakan daerah penyelesaian
dari suatu sistem pertidaksamaan. Tentukan sistem pertidaksamaan tersebut.
soal 1c
Daerah yang tidak diarsir (Bersih) pada grafik dibawah merupakan daerah penyelesaian
dari suatu sistem pertidaksamaan. Tentukan sistem pertidaksamaan tersebut.
soal 2a
Daerah yang tidak diarsir (Bersih) pada grafik dibawah merupakan daerah penyelesaian
dari suatu sistem pertidaksamaan. Tentukan sistem pertidaksamaan tersebut.
soal 2b
Daerah yang tidak diarsir (Bersih) pada grafik dibawah merupakan daerah penyelesaian
dari suatu sistem pertidaksamaan. Tentukan sistem pertidaksamaan tersebut.

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Lks 2 dikonversi

  • 1. A. Tujuan pembelajaran: Peserta didik secara berkelompok memahami dari contoh yang diberikan mengenai cara menentukan sistem pertidaksamaan linear dua variabel dari daerah penyelesaian dengan tepat. B. Perintah: 1. Pahami CONTOH yang diberikan, diskusikan dengan kelompokmu bagaimana cara menentukan sistem pertidaksamaan linear dua variabel dari daerah penyelesaian yang diberikan dalam contoh. 2. Selesaikan SOAL yang disediakan tanpa melihat contoh. Setelah selesai, periksa kembali apakah jawaban sudah sesuai. Jawaban tidak harus ditulis sama persis dengan contoh, boleh disingkat, boleh lebih panjang, boleh menerapkan cara yang berbeda. Pelajari Contoh 1 (5 menit) Daerah yang tidak diarsir (Bersih) pada grafik dibawah merupakan daerah penyelesaian dari suatu sistem pertidaksamaan. Tentukan sistem pertidaksamaan tersebut. LEMBAR KERJA SISWA (LKS)
  • 2. Penyelesaian: Langkah 1: Tentukan titik potong dengan sumbu X dan sumbu Y = (2,0) dan (0,4) Langkah 2: Buatlah persamaan garisnya 𝑏𝑥 + 𝑎𝑦 = 𝑎. 𝑏 4𝑥 + 2𝑦 = 4.2 4𝑥 + 2𝑦 = 8 ↔ 2𝑥 + 𝑦 = 4 Langkah 3: Uji titik O (0,0) yang terletak pada daerah penyelesaian = 2(0) + 0 ≤ 4 sehingga diperoleh pertidaksamaan 2𝑥 + 𝑦 ≤ 4 Langkah 4: Daerah yang tidak diarsir (bersih) terletak di: sebelah kanan sumbu Y, maka 𝑥 ≥ 0 sebelah atas sumbu X, maka 𝑦 ≥ 0 Langkah 5: Simpulan Jadi, sistem pertidaksamaannya adalah 2𝑥 + 𝑦 ≤ 4; 𝑥 ≥ 0; 𝑦 ≥ 0; 𝑥, 𝑦 ∈ 𝑹 Pelajari Contoh 2 Daerah yang tidak diarsir (Bersih) pada grafik dibawah merupakan daerah penyelesaian dari suatu sistem pertidaksamaan. Tentukan sistem pertidaksamaan tersebut.
  • 3. Penyelesaian: Langkah 1: Tentukan persamaan garis 1 (g1) melalui (4,0) dan (0,4), sehingga persamaannya adalah 4𝑥 + 4𝑦 = 16 ↔ 𝑥 + 𝑦 = 4 Uji titik O (0,0) yang terletak pada daerah penyelesaian, 0 + 0 ≤ 4, sehingga diperoleh 𝑥 + 𝑦 ≤ 4 Langkah 2: Tentukan persamaan garis 1 (g1) melalui (6,0) dan (0,2), sehingga persamaannya adalah 2𝑥 + 6𝑦 = 12 ↔ 𝑥 + 3𝑦 = 6 Uji titik O (0,0) yang terletak pada daerah penyelesaian, 0 + 3(0) ≤ 6, sehingga diperoleh 𝑥 + 3𝑦 ≤ 6 Langkah 3: Daerah yang diarsir terletak di: sebelah kanan sumbu Y, maka 𝑥 ≥ 0 sebelah atas sumbu X, maka 𝑦 ≥ 0 Langkah 4: Simpulan Jadi, sistem pertidaksamaannya adalah 𝑥 + 𝑦 ≤ 4; 𝑥 + 3𝑦 ≤ 6; 𝑥 ≥ 0; 𝑦 ≥ 0; 𝑥, 𝑦 ∈ 𝑹
  • 4. Kerjakan soal Soal 1a. Daerah yang tidak diarsir (Bersih) pada grafik dibawah merupakan daerah penyelesaian dari suatu sistem pertidaksamaan. Tentukan sistem pertidaksamaan tersebut. Soal 1b Daerah yang tidak diarsir (Bersih) pada grafik dibawah merupakan daerah penyelesaian dari suatu sistem pertidaksamaan. Tentukan sistem pertidaksamaan tersebut.
  • 5. soal 1c Daerah yang tidak diarsir (Bersih) pada grafik dibawah merupakan daerah penyelesaian dari suatu sistem pertidaksamaan. Tentukan sistem pertidaksamaan tersebut. soal 2a Daerah yang tidak diarsir (Bersih) pada grafik dibawah merupakan daerah penyelesaian dari suatu sistem pertidaksamaan. Tentukan sistem pertidaksamaan tersebut.
  • 6. soal 2b Daerah yang tidak diarsir (Bersih) pada grafik dibawah merupakan daerah penyelesaian dari suatu sistem pertidaksamaan. Tentukan sistem pertidaksamaan tersebut.