{"id":2116,"date":"2026-04-03T15:04:56","date_gmt":"2026-04-03T15:04:56","guid":{"rendered":"https:\/\/rumahpintar.co.id\/blog\/?p=2116"},"modified":"2026-04-08T19:03:54","modified_gmt":"2026-04-08T19:03:54","slug":"soal-sistem-persamaan-linear-dua-variabel","status":"publish","type":"post","link":"https:\/\/rumahpintar.co.id\/blog\/soal-sistem-persamaan-linear-dua-variabel\/","title":{"rendered":"15 Contoh Soal Sistem Persamaan Linear Dua Variabel (SPLDV)"},"content":{"rendered":"

Bimbel Rumah Pintar<\/strong><\/a> – Soal Sistem Persamaan Linear Dua Variabel (SPLDV) adalah kumpulan dua atau lebih persamaan linear yang memiliki dua variabel (biasanya x<\/span>x<\/span><\/span><\/span><\/span>\u00a0dan\u00a0y<\/span>y<\/span><\/span><\/span><\/span>) dan harus diselesaikan secara bersamaan. Prinsip dasar SPLDV adalah mencari pasangan nilai\u00a0(x,y)<\/span>(<\/span>x<\/span>,<\/span>y<\/span>)<\/span><\/span><\/span><\/span>\u00a0yang memenuhi semua persamaan yang diberikan.<\/p>\n

\\Tiga metode utama yang wajib dikuasai adalah metode substitusi<\/strong>\u00a0(menyatakan satu variabel ke variabel lain),\u00a0metode eliminasi<\/strong>\u00a0(menghilangkan salah satu variabel dengan menjumlah\/mengurang persamaan), dan\u00a0metode campuran<\/strong>\u00a0(eliminasi lalu substitusi). Kunci keberhasilan terletak pada ketelitian dalam operasi aljabar dan kemampuan memilih metode yang paling efisien untuk setiap jenis soal.<\/p>\n

Mengapa latihan soal SPLDV sangat penting?<\/strong>\u00a0Soal TKA Matematika SMP hampir selalu menghadirkan SPLDV dalam berbagai bentuk, mulai dari menentukan nilai parameter dari solusi yang diketahui, menyelesaikan sistem dengan koefisien pecahan, hingga soal cerita tentang harga barang, umur, dan kecepatan.<\/p>\n

\n

15 Contoh Soal Sistem Persamaan Linear Dua Variabel dan Pembahasannya<\/h2>\n

 <\/p>\n

Soal 1 (Mencari Parameter dari Solusi yang Diketahui)<\/h3>\n

Diketahui SPLDV:<\/p>\n

{2x+ay=4bx\u22122y=\u22121<\/span>{<\/span><\/span>2x<\/span>+<\/span>a<\/span>y<\/span>=<\/span>4b<\/span>x<\/span>\u2212<\/span>2y<\/span>=<\/span>\u22121<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

mempunyai solusi\u00a0(x,y)=(5,\u22122)<\/span>(<\/span>x<\/span>,<\/span>y<\/span>)<\/span>=<\/span><\/span>(<\/span>5<\/span>,<\/span>\u2212<\/span>2<\/span>)<\/span><\/span><\/span><\/span>. Tentukan nilai\u00a0a<\/span>a<\/span><\/span><\/span><\/span>\u00a0dan\u00a0b<\/span>b<\/span><\/span><\/span><\/span>.<\/p>\n

Pembahasan:<\/strong><\/p>\n

    \n
  • \n

    Substitusi\u00a0x=5,y=\u22122<\/span>x<\/span>=<\/span><\/span>5<\/span>,<\/span>y<\/span>=<\/span><\/span>\u2212<\/span>2<\/span><\/span><\/span><\/span>\u00a0ke persamaan (1):
    \n2(5)+a(\u22122)=4<\/span>2<\/span>(<\/span>5<\/span>)<\/span>+<\/span><\/span>a<\/span>(<\/span>\u2212<\/span>2<\/span>)<\/span>=<\/span><\/span>4<\/span><\/span><\/span><\/span>
    \n10\u22122a=4<\/span>10<\/span>\u2212<\/span><\/span>2<\/span>a<\/span>=<\/span><\/span>4<\/span><\/span><\/span><\/span>
    \n\u22122a=\u22126\u2192a=3<\/span>\u2212<\/span>2<\/span>a<\/span>=<\/span><\/span>\u2212<\/span>6<\/span>\u2192<\/span><\/span>a<\/span>=<\/span><\/span>3<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

  • \n

    Substitusi ke persamaan (2):
    \nb(5)\u22122(\u22122)=\u22121<\/span>b<\/span>(<\/span>5<\/span>)<\/span>\u2212<\/span><\/span>2<\/span>(<\/span>\u2212<\/span>2<\/span>)<\/span>=<\/span><\/span>\u2212<\/span>1<\/span><\/span><\/span><\/span>
    \n5b+4=\u22121<\/span>5<\/span>b<\/span>+<\/span><\/span>4<\/span>=<\/span><\/span>\u2212<\/span>1<\/span><\/span><\/span><\/span>
    \n5b=\u22125\u2192b=\u22121<\/span>5<\/span>b<\/span>=<\/span><\/span>\u2212<\/span>5<\/span>\u2192<\/span><\/span>b<\/span>=<\/span><\/span>\u2212<\/span>1<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ul>\n

    Jawaban:\u00a0a=3<\/span>a<\/span>=<\/span><\/span>3<\/span><\/span><\/span><\/span>,\u00a0b=\u22121<\/span>b<\/span>=<\/span><\/span>\u2212<\/span>1<\/span><\/span><\/span><\/span><\/strong><\/p>\n

    \n

    Soal 2 (Metode Eliminasi)<\/h3>\n

    Selesaikan SPLDV:<\/p>\n

    {2x+3y=12x\u2212y=1<\/span>{<\/span><\/span>2x<\/span>+<\/span>3y<\/span>=<\/span>12x<\/span>\u2212<\/span>y<\/span>=<\/span>1<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

    Pembahasan (eliminasi):<\/strong><\/p>\n

      \n
    • \n

      Eliminasi\u00a0x<\/span>x<\/span><\/span><\/span><\/span>: persamaan (2) dikali 2 \u2192\u00a02x\u22122y=2<\/span>2<\/span>x<\/span>\u2212<\/span><\/span>2<\/span>y<\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

    • \n

      Kurangkan dengan persamaan (1):
      \n(2x+3y)\u2212(2x\u22122y)=12\u22122<\/span>(<\/span>2<\/span>x<\/span>+<\/span><\/span>3<\/span>y<\/span>)<\/span>\u2212<\/span><\/span>(<\/span>2<\/span>x<\/span>\u2212<\/span><\/span>2<\/span>y<\/span>)<\/span>=<\/span><\/span>12<\/span>\u2212<\/span><\/span>2<\/span><\/span><\/span><\/span>
      \n5y=10\u2192y=2<\/span>5<\/span>y<\/span>=<\/span><\/span>10<\/span>\u2192<\/span><\/span>y<\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

    • \n

      Substitusi\u00a0y=2<\/span>y<\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span>\u00a0ke persamaan (2):\u00a0x\u22122=1\u2192x=3<\/span>x<\/span>\u2212<\/span><\/span>2<\/span>=<\/span><\/span>1<\/span>\u2192<\/span><\/span>x<\/span>=<\/span><\/span>3<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ul>\n

      Jawaban:\u00a0x=3<\/span>x<\/span>=<\/span><\/span>3<\/span><\/span><\/span><\/span>,\u00a0y=2<\/span>y<\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span><\/strong><\/p>\n

      \n

      Soal 3 (Metode Substitusi)<\/h3>\n

      Selesaikan SPLDV:<\/p>\n

      {3x+y=102x\u2212y=5<\/span>{<\/span><\/span>3x<\/span>+<\/span>y<\/span>=<\/span>102x<\/span>\u2212<\/span>y<\/span>=<\/span>5<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

      Pembahasan (substitusi):<\/strong><\/p>\n

        \n
      • \n

        Dari persamaan (1):\u00a0y=10\u22123x<\/span>y<\/span>=<\/span><\/span>10<\/span>\u2212<\/span><\/span>3<\/span>x<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

      • \n

        Substitusi ke persamaan (2):
        \n2x\u2212(10\u22123x)=5<\/span>2<\/span>x<\/span>\u2212<\/span><\/span>(<\/span>10<\/span>\u2212<\/span><\/span>3<\/span>x<\/span>)<\/span>=<\/span><\/span>5<\/span><\/span><\/span><\/span>
        \n2x\u221210+3x=5<\/span>2<\/span>x<\/span>\u2212<\/span><\/span>10<\/span>+<\/span><\/span>3<\/span>x<\/span>=<\/span><\/span>5<\/span><\/span><\/span><\/span>
        \n5x\u221210=5\u21925x=15\u2192x=3<\/span>5<\/span>x<\/span>\u2212<\/span><\/span>10<\/span>=<\/span><\/span>5<\/span>\u2192<\/span><\/span>5<\/span>x<\/span>=<\/span><\/span>15<\/span>\u2192<\/span><\/span>x<\/span>=<\/span><\/span>3<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

      • \n

        Substitusi\u00a0x=3<\/span>x<\/span>=<\/span><\/span>3<\/span><\/span><\/span><\/span>\u00a0ke\u00a0y=10\u22123(3)=10\u22129=1<\/span>y<\/span>=<\/span><\/span>10<\/span>\u2212<\/span><\/span>3<\/span>(<\/span>3<\/span>)<\/span>=<\/span><\/span>10<\/span>\u2212<\/span><\/span>9<\/span>=<\/span><\/span>1<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ul>\n

        Jawaban:\u00a0x=3<\/span>x<\/span>=<\/span><\/span>3<\/span><\/span><\/span><\/span>,\u00a0y=1<\/span>y<\/span>=<\/span><\/span>1<\/span><\/span><\/span><\/span><\/strong><\/p>\n

        \n

        Soal 4 (Soal Cerita \u2013 Harga Barang)<\/h3>\n

        Harga 2 kg apel dan 3 kg jeruk adalah Rp90.000. Harga 1 kg apel dan 2 kg jeruk adalah Rp55.000. Berapa harga 1 kg apel?<\/p>\n

        Pembahasan:<\/strong><\/p>\n

          \n
        • \n

          Misal\u00a0a<\/span>a<\/span><\/span><\/span><\/span>\u00a0= harga 1 kg apel,\u00a0j<\/span>j<\/span><\/span><\/span><\/span>\u00a0= harga 1 kg jeruk<\/p>\n<\/li>\n

        • \n

          Persamaan:<\/p>\n

          {2a+3j=90.000a+2j=55.000<\/span>{<\/span><\/span>2a<\/span>+<\/span>3j<\/span>=<\/span>90.000a<\/span>+<\/span>2j<\/span>=<\/span>55.000<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/li>\n

        • \n

          Eliminasi\u00a0a<\/span>a<\/span><\/span><\/span><\/span>: persamaan (2) dikali 2 \u2192\u00a02a+4j=110.000<\/span>2<\/span>a<\/span>+<\/span><\/span>4<\/span>j<\/span>=<\/span><\/span>110.000<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

        • \n

          Kurangkan dengan persamaan (1):
          \n(2a+4j)\u2212(2a+3j)=110.000\u221290.000<\/span>(<\/span>2<\/span>a<\/span>+<\/span><\/span>4<\/span>j<\/span>)<\/span>\u2212<\/span><\/span>(<\/span>2<\/span>a<\/span>+<\/span><\/span>3<\/span>j<\/span>)<\/span>=<\/span><\/span>110.000<\/span>\u2212<\/span><\/span>90.000<\/span><\/span><\/span><\/span>
          \nj=20.000<\/span>j<\/span>=<\/span><\/span>20.000<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

        • \n

          Substitusi\u00a0j=20.000<\/span>j<\/span>=<\/span><\/span>20.000<\/span><\/span><\/span><\/span>\u00a0ke persamaan (2):
          \na+2(20.000)=55.000\u2192a+40.000=55.000\u2192a=15.000<\/span>a<\/span>+<\/span><\/span>2<\/span>(<\/span>20.000<\/span>)<\/span>=<\/span><\/span>55.000<\/span>\u2192<\/span><\/span>a<\/span>+<\/span><\/span>40.000<\/span>=<\/span><\/span>55.000<\/span>\u2192<\/span><\/span>a<\/span>=<\/span><\/span>15.000<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ul>\n

          Jawaban: Harga 1 kg apel = Rp15.000<\/strong><\/p>\n

          \n

          Soal 5 (Soal Cerita \u2013 Umur)<\/h3>\n

          Umur Ayah 3 kali umur Anak. Lima tahun yang lalu, jumlah umur mereka 40 tahun. Berapa umur Ayah sekarang?<\/p>\n

          Pembahasan:<\/strong><\/p>\n

            \n
          • \n

            Misal\u00a0A<\/span>A<\/span><\/span><\/span><\/span>\u00a0= umur Ayah sekarang,\u00a0K<\/span>K<\/span><\/span><\/span><\/span>\u00a0= umur Anak sekarang<\/p>\n<\/li>\n

          • \n

            Persamaan:<\/p>\n

            {A=3K(A\u22125)+(K\u22125)=40<\/span>{<\/span><\/span>A<\/span>=<\/span>3K<\/span>(<\/span>A<\/span>\u2212<\/span>5)<\/span>+<\/span>(<\/span>K<\/span>\u2212<\/span>5)<\/span>=<\/span>40<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/li>\n

          • \n

            Sederhanakan persamaan (2):\u00a0A+K\u221210=40\u2192A+K=50<\/span>A<\/span>+<\/span><\/span>K<\/span>\u2212<\/span><\/span>10<\/span>=<\/span><\/span>40<\/span>\u2192<\/span><\/span>A<\/span>+<\/span><\/span>K<\/span>=<\/span><\/span>50<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

          • \n

            Substitusi\u00a0A=3K<\/span>A<\/span>=<\/span><\/span>3<\/span>K<\/span><\/span><\/span><\/span>:\u00a03K+K=50\u21924K=50\u2192K=12,5<\/span>3<\/span>K<\/span>+<\/span><\/span>K<\/span>=<\/span><\/span>50<\/span>\u2192<\/span><\/span>4<\/span>K<\/span>=<\/span><\/span>50<\/span>\u2192<\/span><\/span>K<\/span>=<\/span><\/span>12<\/span>,<\/span>5<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

          • \n

            A=3\u00d712,5=37,5<\/span>A<\/span>=<\/span><\/span>3<\/span>\u00d7<\/span><\/span>12<\/span>,<\/span>5<\/span>=<\/span><\/span>37<\/span>,<\/span>5<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ul>\n

            Jawaban: Umur Ayah = 37,5 tahun<\/strong><\/p>\n

            \n

            Soal 6 (SPLDV dengan Parameter \u2013 Mencari Nilai)<\/h3>\n

            Diketahui SPLDV:<\/p>\n

            {3x+2y=7x\u2212y=p<\/span>{<\/span><\/span>3x<\/span>+<\/span>2y<\/span>=<\/span>7x<\/span>\u2212<\/span>y<\/span>=<\/span>p<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

            Jika\u00a0x=2<\/span>x<\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span>, tentukan nilai\u00a0y<\/span>y<\/span><\/span><\/span><\/span>\u00a0dan\u00a0p<\/span>p<\/span><\/span><\/span><\/span>.<\/p>\n

            Pembahasan:<\/strong><\/p>\n

              \n
            • \n

              Substitusi\u00a0x=2<\/span>x<\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span>\u00a0ke persamaan (1):
              \n3(2)+2y=7\u21926+2y=7\u21922y=1\u2192y=0,5<\/span>3<\/span>(<\/span>2<\/span>)<\/span>+<\/span><\/span>2<\/span>y<\/span>=<\/span><\/span>7<\/span>\u2192<\/span><\/span>6<\/span>+<\/span><\/span>2<\/span>y<\/span>=<\/span><\/span>7<\/span>\u2192<\/span><\/span>2<\/span>y<\/span>=<\/span><\/span>1<\/span>\u2192<\/span><\/span>y<\/span>=<\/span><\/span>0<\/span>,<\/span>5<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

            • \n

              Substitusi\u00a0x=2,y=0,5<\/span>x<\/span>=<\/span><\/span>2<\/span>,<\/span>y<\/span>=<\/span><\/span>0<\/span>,<\/span>5<\/span><\/span><\/span><\/span>\u00a0ke persamaan (2):
              \n2\u22120,5=p\u2192p=1,5<\/span>2<\/span>\u2212<\/span><\/span>0<\/span>,<\/span>5<\/span>=<\/span><\/span>p<\/span>\u2192<\/span><\/span>p<\/span>=<\/span><\/span>1<\/span>,<\/span>5<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ul>\n

              Jawaban:\u00a0y=0,5<\/span>y<\/span>=<\/span><\/span>0<\/span>,<\/span>5<\/span><\/span><\/span><\/span>,\u00a0p=1,5<\/span>p<\/span>=<\/span><\/span>1<\/span>,<\/span>5<\/span><\/span><\/span><\/span><\/strong><\/p>\n

              \n

              Soal 7 (SPLDV dengan Pecahan)<\/h3>\n

              Selesaikan:<\/p>\n

              {x2+y3=5×3+y2=5<\/span>{<\/span><\/span>2<\/span><\/span>x<\/span><\/span><\/span>\u200b<\/span><\/span>+<\/span>3<\/span><\/span>y<\/span><\/span><\/span>\u200b<\/span><\/span>=<\/span>53<\/span><\/span>x<\/span><\/span><\/span>\u200b<\/span><\/span>+<\/span>2<\/span><\/span>y<\/span><\/span><\/span>\u200b<\/span><\/span>=<\/span>5<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

              Pembahasan:<\/strong><\/p>\n

                \n
              • \n

                Kalikan persamaan (1) dengan 6:\u00a03x+2y=30<\/span>3<\/span>x<\/span>+<\/span><\/span>2<\/span>y<\/span>=<\/span><\/span>30<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

              • \n

                Kalikan persamaan (2) dengan 6:\u00a02x+3y=30<\/span>2<\/span>x<\/span>+<\/span><\/span>3<\/span>y<\/span>=<\/span><\/span>30<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

              • \n

                Eliminasi\u00a0x<\/span>x<\/span><\/span><\/span><\/span>: (1) \u00d7 2 =\u00a06x+4y=60<\/span>6<\/span>x<\/span>+<\/span><\/span>4<\/span>y<\/span>=<\/span><\/span>60<\/span><\/span><\/span><\/span>, (2) \u00d7 3 =\u00a06x+9y=90<\/span>6<\/span>x<\/span>+<\/span><\/span>9<\/span>y<\/span>=<\/span><\/span>90<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

              • \n

                Kurangkan:\u00a0(6x+9y)\u2212(6x+4y)=90\u221260\u21925y=30\u2192y=6<\/span>(<\/span>6<\/span>x<\/span>+<\/span><\/span>9<\/span>y<\/span>)<\/span>\u2212<\/span><\/span>(<\/span>6<\/span>x<\/span>+<\/span><\/span>4<\/span>y<\/span>)<\/span>=<\/span><\/span>90<\/span>\u2212<\/span><\/span>60<\/span>\u2192<\/span><\/span>5<\/span>y<\/span>=<\/span><\/span>30<\/span>\u2192<\/span><\/span>y<\/span>=<\/span><\/span>6<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

              • \n

                Substitusi\u00a0y=6<\/span>y<\/span>=<\/span><\/span>6<\/span><\/span><\/span><\/span>\u00a0ke\u00a03x+2(6)=30\u21923x+12=30\u21923x=18\u2192x=6<\/span>3<\/span>x<\/span>+<\/span><\/span>2<\/span>(<\/span>6<\/span>)<\/span>=<\/span><\/span>30<\/span>\u2192<\/span><\/span>3<\/span>x<\/span>+<\/span><\/span>12<\/span>=<\/span><\/span>30<\/span>\u2192<\/span><\/span>3<\/span>x<\/span>=<\/span><\/span>18<\/span>\u2192<\/span><\/span>x<\/span>=<\/span><\/span>6<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ul>\n

                Jawaban:\u00a0x=6<\/span>x<\/span>=<\/span><\/span>6<\/span><\/span><\/span><\/span>,\u00a0y=6<\/span>y<\/span>=<\/span><\/span>6<\/span><\/span><\/span><\/span><\/strong><\/p>\n

                \n

                Soal 8 (Menentukan Jenis Penyelesaian)<\/h3>\n

                Apakah SPLDV berikut memiliki penyelesaian? Jelaskan.<\/p>\n

                {2x+3y=64x+6y=12<\/span>{<\/span><\/span>2x<\/span>+<\/span>3y<\/span>=<\/span>64x<\/span>+<\/span>6y<\/span>=<\/span>12<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

                Pembahasan:<\/strong><\/p>\n

                  \n
                • \n

                  Persamaan (2) =\u00a02\u00d7(2x+3y)=2\u00d76<\/span>2<\/span>\u00d7<\/span><\/span>(<\/span>2<\/span>x<\/span>+<\/span><\/span>3<\/span>y<\/span>)<\/span>=<\/span><\/span>2<\/span>\u00d7<\/span><\/span>6<\/span><\/span><\/span><\/span>\u00a0\u2192\u00a04x+6y=12<\/span>4<\/span>x<\/span>+<\/span><\/span>6<\/span>y<\/span>=<\/span><\/span>12<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                • \n

                  Kedua persamaan sebanding (koefisien dan konstanta kelipatan 2)<\/p>\n<\/li>\n

                • \n

                  Artinya kedua garis berimpit (sama)<\/p>\n<\/li>\n<\/ul>\n

                  Jawaban: Memiliki tak hingga penyelesaian (garis berimpit)<\/strong><\/p>\n

                  \n

                  Soal 9 (Soal Cerita \u2013 Jumlah dan Selisih)<\/h3>\n

                  Jumlah dua bilangan adalah 25. Selisihnya adalah 5. Tentukan kedua bilangan tersebut.<\/p>\n

                  Pembahasan:<\/strong><\/p>\n

                    \n
                  • \n

                    Misal bilangan pertama =\u00a0x<\/span>x<\/span><\/span><\/span><\/span>, bilangan kedua =\u00a0y<\/span>y<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                  • \n

                    Persamaan:<\/p>\n

                    {x+y=25x\u2212y=5<\/span>{<\/span><\/span>x<\/span>+<\/span>y<\/span>=<\/span>25x<\/span>\u2212<\/span>y<\/span>=<\/span>5<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/li>\n

                  • \n

                    Jumlahkan kedua persamaan:\u00a0(x+y)+(x\u2212y)=25+5\u21922x=30\u2192x=15<\/span>(<\/span>x<\/span>+<\/span><\/span>y<\/span>)<\/span>+<\/span><\/span>(<\/span>x<\/span>\u2212<\/span><\/span>y<\/span>)<\/span>=<\/span><\/span>25<\/span>+<\/span><\/span>5<\/span>\u2192<\/span><\/span>2<\/span>x<\/span>=<\/span><\/span>30<\/span>\u2192<\/span><\/span>x<\/span>=<\/span><\/span>15<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                  • \n

                    Substitusi\u00a0x=15<\/span>x<\/span>=<\/span><\/span>15<\/span><\/span><\/span><\/span>\u00a0ke\u00a0x+y=25\u219215+y=25\u2192y=10<\/span>x<\/span>+<\/span><\/span>y<\/span>=<\/span><\/span>25<\/span>\u2192<\/span><\/span>15<\/span>+<\/span><\/span>y<\/span>=<\/span><\/span>25<\/span>\u2192<\/span><\/span>y<\/span>=<\/span><\/span>10<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ul>\n

                    Jawaban: 15 dan 10<\/strong><\/p>\n

                    \n

                    Soal 10 (SPLDV dengan Parameter \u2013 Mencari a dan b)<\/h3>\n

                    Diketahui SPLDV:<\/p>\n

                    {2x+by=4ax\u22123y=7<\/span>{<\/span><\/span>2x<\/span>+<\/span>b<\/span>y<\/span>=<\/span>4a<\/span>x<\/span>\u2212<\/span>3y<\/span>=<\/span>7<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

                    mempunyai solusi\u00a0(x,y)=(2,\u22121)<\/span>(<\/span>x<\/span>,<\/span>y<\/span>)<\/span>=<\/span><\/span>(<\/span>2<\/span>,<\/span>\u2212<\/span>1<\/span>)<\/span><\/span><\/span><\/span>. Tentukan nilai\u00a0a<\/span>a<\/span><\/span><\/span><\/span>\u00a0dan\u00a0b<\/span>b<\/span><\/span><\/span><\/span>.<\/p>\n

                    Pembahasan:<\/strong><\/p>\n

                      \n
                    • \n

                      Substitusi\u00a0x=2,y=\u22121<\/span>x<\/span>=<\/span><\/span>2<\/span>,<\/span>y<\/span>=<\/span><\/span>\u2212<\/span>1<\/span><\/span><\/span><\/span>\u00a0ke persamaan (1):
                      \n2(2)+b(\u22121)=4\u21924\u2212b=4\u2192\u2212b=0\u2192b=0<\/span>2<\/span>(<\/span>2<\/span>)<\/span>+<\/span><\/span>b<\/span>(<\/span>\u2212<\/span>1<\/span>)<\/span>=<\/span><\/span>4<\/span>\u2192<\/span><\/span>4<\/span>\u2212<\/span><\/span>b<\/span>=<\/span><\/span>4<\/span>\u2192<\/span><\/span>\u2212<\/span>b<\/span>=<\/span><\/span>0<\/span>\u2192<\/span><\/span>b<\/span>=<\/span><\/span>0<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                    • \n

                      Substitusi ke persamaan (2):
                      \na(2)\u22123(\u22121)=7\u21922a+3=7\u21922a=4\u2192a=2<\/span>a<\/span>(<\/span>2<\/span>)<\/span>\u2212<\/span><\/span>3<\/span>(<\/span>\u2212<\/span>1<\/span>)<\/span>=<\/span><\/span>7<\/span>\u2192<\/span><\/span>2<\/span>a<\/span>+<\/span><\/span>3<\/span>=<\/span><\/span>7<\/span>\u2192<\/span><\/span>2<\/span>a<\/span>=<\/span><\/span>4<\/span>\u2192<\/span><\/span>a<\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ul>\n

                      Jawaban:\u00a0a=2<\/span>a<\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span>,\u00a0b=0<\/span>b<\/span>=<\/span><\/span>0<\/span><\/span><\/span><\/span><\/strong><\/p>\n

                      \n

                      Soal 11 (Soal Cerita \u2013 Umur dengan Selisih Waktu)<\/h3>\n

                      Tiga tahun yang lalu, umur Ayah 4 kali umur Anak. Dua tahun yang akan datang, jumlah umur mereka 49 tahun. Berapa umur Ayah sekarang?<\/p>\n

                      Pembahasan:<\/strong><\/p>\n

                        \n
                      • \n

                        Misal\u00a0A<\/span>A<\/span><\/span><\/span><\/span>\u00a0= umur Ayah sekarang,\u00a0K<\/span>K<\/span><\/span><\/span><\/span>\u00a0= umur Anak sekarang<\/p>\n<\/li>\n

                      • \n

                        Persamaan:<\/p>\n

                        {A\u22123=4(K\u22123)(A+2)+(K+2)=49<\/span>{<\/span><\/span>A<\/span>\u2212<\/span>3=<\/span>4(<\/span>K<\/span>\u2212<\/span>3)<\/span>(<\/span>A<\/span>+<\/span>2)<\/span>+<\/span>(<\/span>K<\/span>+<\/span>2)<\/span>=<\/span>49<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/li>\n

                      • \n

                        Sederhanakan (1):\u00a0A\u22123=4K\u221212\u2192A\u22124K=\u22129<\/span>A<\/span>\u2212<\/span><\/span>3<\/span>=<\/span><\/span>4<\/span>K<\/span>\u2212<\/span><\/span>12<\/span>\u2192<\/span><\/span>A<\/span>\u2212<\/span><\/span>4<\/span>K<\/span>=<\/span><\/span>\u2212<\/span>9<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                      • \n

                        Sederhanakan (2):\u00a0A+K+4=49\u2192A+K=45<\/span>A<\/span>+<\/span><\/span>K<\/span>+<\/span><\/span>4<\/span>=<\/span><\/span>49<\/span>\u2192<\/span><\/span>A<\/span>+<\/span><\/span>K<\/span>=<\/span><\/span>45<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                      • \n

                        Eliminasi: kurangkan (A + K) – (A – 4K) = 45 – (-9)
                        \n5K=54\u2192K=10,8<\/span>5<\/span>K<\/span>=<\/span><\/span>54<\/span>\u2192<\/span><\/span>K<\/span>=<\/span><\/span>10<\/span>,<\/span>8<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                      • \n

                        A=45\u221210,8=34,2<\/span>A<\/span>=<\/span><\/span>45<\/span>\u2212<\/span><\/span>10<\/span>,<\/span>8<\/span>=<\/span><\/span>34<\/span>,<\/span>2<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ul>\n

                        Jawaban: Umur Ayah = 34,2 tahun<\/strong><\/p>\n

                        \n

                        Soal 12 (Metode Campuran \u2013 Eliminasi lalu Substitusi)<\/h3>\n

                        Selesaikan SPLDV:<\/p>\n

                        {3x\u22122y=5x+3y=9<\/span>{<\/span><\/span>3x<\/span>\u2212<\/span>2y<\/span>=<\/span>5x<\/span>+<\/span>3y<\/span>=<\/span>9<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

                        Pembahasan:<\/strong><\/p>\n

                          \n
                        • \n

                          Eliminasi\u00a0x<\/span>x<\/span><\/span><\/span><\/span>: persamaan (2) dikali 3 \u2192\u00a03x+9y=27<\/span>3<\/span>x<\/span>+<\/span><\/span>9<\/span>y<\/span>=<\/span><\/span>27<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                        • \n

                          Kurangkan dengan persamaan (1):
                          \n(3x+9y)\u2212(3x\u22122y)=27\u22125<\/span>(<\/span>3<\/span>x<\/span>+<\/span><\/span>9<\/span>y<\/span>)<\/span>\u2212<\/span><\/span>(<\/span>3<\/span>x<\/span>\u2212<\/span><\/span>2<\/span>y<\/span>)<\/span>=<\/span><\/span>27<\/span>\u2212<\/span><\/span>5<\/span><\/span><\/span><\/span>
                          \n11y=22\u2192y=2<\/span>11<\/span>y<\/span>=<\/span><\/span>22<\/span>\u2192<\/span><\/span>y<\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                        • \n

                          Substitusi\u00a0y=2<\/span>y<\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span>\u00a0ke persamaan (2):\u00a0x+3(2)=9\u2192x+6=9\u2192x=3<\/span>x<\/span>+<\/span><\/span>3<\/span>(<\/span>2<\/span>)<\/span>=<\/span><\/span>9<\/span>\u2192<\/span><\/span>x<\/span>+<\/span><\/span>6<\/span>=<\/span><\/span>9<\/span>\u2192<\/span><\/span>x<\/span>=<\/span><\/span>3<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ul>\n

                          Jawaban:\u00a0x=3<\/span>x<\/span>=<\/span><\/span>3<\/span><\/span><\/span><\/span>,\u00a0y=2<\/span>y<\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span><\/strong><\/p>\n

                          \n

                          Soal 13 (Soal Cerita \u2013 Kecepatan)<\/h3>\n

                          Jarak kota A ke B adalah 240 km. Jika naik mobil dengan kecepatan\u00a0x<\/span>x<\/span><\/span><\/span><\/span>\u00a0km\/jam, waktu tempuh\u00a0y<\/span>y<\/span><\/span><\/span><\/span>\u00a0jam. Jika kecepatan ditambah 20 km\/jam, waktu berkurang 1 jam. Tentukan nilai\u00a0x<\/span>x<\/span><\/span><\/span><\/span>\u00a0dan\u00a0y<\/span>y<\/span><\/span><\/span><\/span>.<\/p>\n

                          Pembahasan:<\/strong><\/p>\n

                            \n
                          • \n

                            Persamaan (1):\u00a0x\u00d7y=240<\/span>x<\/span>\u00d7<\/span><\/span>y<\/span>=<\/span><\/span>240<\/span><\/span><\/span><\/span>\u00a0\u2192\u00a0xy=240<\/span>x<\/span>y<\/span>=<\/span><\/span>240<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                          • \n

                            Persamaan (2):\u00a0(x+20)(y\u22121)=240<\/span>(<\/span>x<\/span>+<\/span><\/span>20<\/span>)<\/span>(<\/span>y<\/span>\u2212<\/span><\/span>1<\/span>)<\/span>=<\/span><\/span>240<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                          • \n

                            Uraikan (2):\u00a0xy\u2212x+20y\u221220=240<\/span>x<\/span>y<\/span>\u2212<\/span><\/span>x<\/span>+<\/span><\/span>20<\/span>y<\/span>\u2212<\/span><\/span>20<\/span>=<\/span><\/span>240<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                          • \n

                            Substitusi\u00a0xy=240<\/span>x<\/span>y<\/span>=<\/span><\/span>240<\/span><\/span><\/span><\/span>:\u00a0240\u2212x+20y\u221220=240<\/span>240<\/span>\u2212<\/span><\/span>x<\/span>+<\/span><\/span>20<\/span>y<\/span>\u2212<\/span><\/span>20<\/span>=<\/span><\/span>240<\/span><\/span><\/span><\/span>
                            \n\u2212x+20y\u221220=0\u2192\u2212x+20y=20\u2192x=20y\u221220<\/span>\u2212<\/span>x<\/span>+<\/span><\/span>20<\/span>y<\/span>\u2212<\/span><\/span>20<\/span>=<\/span><\/span>0<\/span>\u2192<\/span><\/span>\u2212<\/span>x<\/span>+<\/span><\/span>20<\/span>y<\/span>=<\/span><\/span>20<\/span>\u2192<\/span><\/span>x<\/span>=<\/span><\/span>20<\/span>y<\/span>\u2212<\/span><\/span>20<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                          • \n

                            Substitusi ke\u00a0xy=240<\/span>x<\/span>y<\/span>=<\/span><\/span>240<\/span><\/span><\/span><\/span>:\u00a0(20y\u221220)y=240<\/span>(<\/span>20<\/span>y<\/span>\u2212<\/span><\/span>20<\/span>)<\/span>y<\/span>=<\/span><\/span>240<\/span><\/span><\/span><\/span>
                            \n20y2\u221220y\u2212240=0<\/span>20<\/span>y<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u2212<\/span><\/span>20<\/span>y<\/span>\u2212<\/span><\/span>240<\/span>=<\/span><\/span>0<\/span><\/span><\/span><\/span>\u00a0\u2192 bagi 20:\u00a0y2\u2212y\u221212=0<\/span>y<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u2212<\/span><\/span>y<\/span>\u2212<\/span><\/span>12<\/span>=<\/span><\/span>0<\/span><\/span><\/span><\/span>
                            \n(y\u22124)(y+3)=0\u2192y=4<\/span>(<\/span>y<\/span>\u2212<\/span><\/span>4<\/span>)<\/span>(<\/span>y<\/span>+<\/span><\/span>3<\/span>)<\/span>=<\/span><\/span>0<\/span>\u2192<\/span><\/span>y<\/span>=<\/span><\/span>4<\/span><\/span><\/span><\/span>\u00a0(positif)<\/p>\n<\/li>\n

                          • \n

                            x=20(4)\u221220=80\u221220=60<\/span>x<\/span>=<\/span><\/span>20<\/span>(<\/span>4<\/span>)<\/span>\u2212<\/span><\/span>20<\/span>=<\/span><\/span>80<\/span>\u2212<\/span><\/span>20<\/span>=<\/span><\/span>60<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ul>\n

                            Jawaban:\u00a0x=60<\/span>x<\/span>=<\/span><\/span>60<\/span><\/span><\/span><\/span>\u00a0km\/jam,\u00a0y=4<\/span>y<\/span>=<\/span><\/span>4<\/span><\/span><\/span><\/span>\u00a0jam<\/strong><\/p>\n

                            \n

                            Soal 14 (SPLDV dengan Koefisien Negatif)<\/h3>\n

                            Selesaikan SPLDV:<\/p>\n

                            {2x\u22123y=7\u2212x+2y=\u22124<\/span>{<\/span><\/span>2x<\/span>\u2212<\/span>3y<\/span>=<\/span>7\u2212x<\/span>+<\/span>2y<\/span>=<\/span>\u22124<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

                            Pembahasan:<\/strong><\/p>\n

                              \n
                            • \n

                              Eliminasi\u00a0x<\/span>x<\/span><\/span><\/span><\/span>: persamaan (2) dikali 2 \u2192\u00a0\u22122x+4y=\u22128<\/span>\u2212<\/span>2<\/span>x<\/span>+<\/span><\/span>4<\/span>y<\/span>=<\/span><\/span>\u2212<\/span>8<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                            • \n

                              Jumlahkan dengan persamaan (1):
                              \n(2x\u22123y)+(\u22122x+4y)=7+(\u22128)<\/span>(<\/span>2<\/span>x<\/span>\u2212<\/span><\/span>3<\/span>y<\/span>)<\/span>+<\/span><\/span>(<\/span>\u2212<\/span>2<\/span>x<\/span>+<\/span><\/span>4<\/span>y<\/span>)<\/span>=<\/span><\/span>7<\/span>+<\/span><\/span>(<\/span>\u2212<\/span>8<\/span>)<\/span><\/span><\/span><\/span>
                              \ny=\u22121<\/span>y<\/span>=<\/span><\/span>\u2212<\/span>1<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                            • \n

                              Substitusi\u00a0y=\u22121<\/span>y<\/span>=<\/span><\/span>\u2212<\/span>1<\/span><\/span><\/span><\/span>\u00a0ke persamaan (1):
                              \n2x\u22123(\u22121)=7\u21922x+3=7\u21922x=4\u2192x=2<\/span>2<\/span>x<\/span>\u2212<\/span><\/span>3<\/span>(<\/span>\u2212<\/span>1<\/span>)<\/span>=<\/span><\/span>7<\/span>\u2192<\/span><\/span>2<\/span>x<\/span>+<\/span><\/span>3<\/span>=<\/span><\/span>7<\/span>\u2192<\/span><\/span>2<\/span>x<\/span>=<\/span><\/span>4<\/span>\u2192<\/span><\/span>x<\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ul>\n

                              Jawaban:\u00a0x=2<\/span>x<\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span>,\u00a0y=\u22121<\/span>y<\/span>=<\/span><\/span>\u2212<\/span>1<\/span><\/span><\/span><\/span><\/strong><\/p>\n

                              \n

                              Soal 15 (Soal Cerita \u2013 Pembelian Barang)<\/h3>\n

                              Di sebuah toko, Rina membeli 3 buku dan 2 pensil seharga Rp26.000. Siti membeli 2 buku dan 3 pensil seharga Rp24.000. Berapa harga 1 buku dan 1 pensil masing-masing?<\/p>\n

                              Pembahasan:<\/strong><\/p>\n

                                \n
                              • \n

                                Misal\u00a0b<\/span>b<\/span><\/span><\/span><\/span>\u00a0= harga 1 buku,\u00a0p<\/span>p<\/span><\/span><\/span><\/span>\u00a0= harga 1 pensil<\/p>\n<\/li>\n

                              • \n

                                Persamaan:<\/p>\n

                                {3b+2p=26.0002b+3p=24.000<\/span>{<\/span><\/span>3b<\/span>+<\/span>2p<\/span>=<\/span>26.0002b<\/span>+<\/span>3p<\/span>=<\/span>24.000<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/li>\n

                              • \n

                                Eliminasi\u00a0b<\/span>b<\/span><\/span><\/span><\/span>: (1) \u00d7 2 =\u00a06b+4p=52.000<\/span>6<\/span>b<\/span>+<\/span><\/span>4<\/span>p<\/span>=<\/span><\/span>52.000<\/span><\/span><\/span><\/span>, (2) \u00d7 3 =\u00a06b+9p=72.000<\/span>6<\/span>b<\/span>+<\/span><\/span>9<\/span>p<\/span>=<\/span><\/span>72.000<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                              • \n

                                Kurangkan:\u00a0(6b+9p)\u2212(6b+4p)=72.000\u221252.000\u21925p=20.000\u2192p=4.000<\/span>(<\/span>6<\/span>b<\/span>+<\/span><\/span>9<\/span>p<\/span>)<\/span>\u2212<\/span><\/span>(<\/span>6<\/span>b<\/span>+<\/span><\/span>4<\/span>p<\/span>)<\/span>=<\/span><\/span>72.000<\/span>\u2212<\/span><\/span>52.000<\/span>\u2192<\/span><\/span>5<\/span>p<\/span>=<\/span><\/span>20.000<\/span>\u2192<\/span><\/span>p<\/span>=<\/span><\/span>4.000<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                              • \n

                                Substitusi\u00a0p=4.000<\/span>p<\/span>=<\/span><\/span>4.000<\/span><\/span><\/span><\/span>\u00a0ke persamaan (1):
                                \n3b+2(4.000)=26.000\u21923b+8.000=26.000\u21923b=18.000\u2192b=6.000<\/span>3<\/span>b<\/span>+<\/span><\/span>2<\/span>(<\/span>4.000<\/span>)<\/span>=<\/span><\/span>26.000<\/span>\u2192<\/span><\/span>3<\/span>b<\/span>+<\/span><\/span>8.000<\/span>=<\/span><\/span>26.000<\/span>\u2192<\/span><\/span>3<\/span>b<\/span>=<\/span><\/span>18.000<\/span>\u2192<\/span><\/span>b<\/span>=<\/span><\/span>6.000<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ul>\n

                                Jawaban: Harga 1 buku = Rp6.000, harga 1 pensil = Rp4.000<\/strong><\/p>\n

                                \n

                                Ringkasan Metode Penyelesaian SPLDV<\/h2>\n
                                \n
                                \n
                                <\/div>\n
                                <\/div>\n<\/div>\n\n\n\n\n\n\n\n
                                Metode<\/th>\nCara<\/th>\nKapan Cocok<\/th>\n<\/tr>\n<\/thead>\n
                                Substitusi<\/strong><\/td>\nNyatakan satu variabel ke variabel lain, lalu substitusi<\/td>\nSalah satu persamaan mudah diubah (koefisien 1)<\/td>\n<\/tr>\n
                                Eliminasi<\/strong><\/td>\nSamakan koefisien, jumlah\/kurangkan<\/td>\nKedua persamaan dalam bentuk baku\u00a0ax+by=c<\/span>a<\/span>x<\/span>+<\/span><\/span>b<\/span>y<\/span>=<\/span><\/span>c<\/span><\/span><\/span><\/span><\/td>\n<\/tr>\n
                                Campuran<\/strong><\/td>\nEliminasi lalu substitusi<\/td>\nCara paling umum dan efisien<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

                                Aturan Penting:<\/strong><\/p>\n

                                  \n
                                • \n

                                  Jika soal memberikan solusi\u00a0(x,y)<\/span>(<\/span>x<\/span>,<\/span>y<\/span>)<\/span><\/span><\/span><\/span>\u00a0\u2192 substitusikan langsung ke semua persamaan untuk mencari parameter<\/p>\n<\/li>\n

                                • \n

                                  Jika koefisien sebanding dan konstanta sebanding \u2192 tak hingga penyelesaian (garis berimpit)<\/p>\n<\/li>\n

                                • \n

                                  Jika koefisien sebanding tetapi konstanta tidak sebanding \u2192 tidak ada penyelesaian (garis sejajar)<\/p>\n<\/li>\n<\/ul>\n

                                  \n

                                  Sahabat Rumah Pintar<\/strong>, dengan menguasai 15 soal di atas, kalian sudah siap menyelesaikan berbagai variasi soal Sistem Persamaan Linear Dua Variabel di TKA Matematika SMP. Ingat kuncinya:\u00a0pilih metode yang paling efisien, teliti dalam operasi aljabar, dan jangan lupa memeriksa kembali solusi dengan mensubstitusikannya ke persamaan awal<\/strong>. Terus berlatih, ya!\u00a0Tetap pintar, tetap santai!<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

                                  Bimbel Rumah Pintar – Soal Sistem Persamaan Linear Dua Variabel (SPLDV) adalah kumpulan dua atau lebih persamaan linear yang memiliki dua variabel (biasanya xx\u00a0dan\u00a0yy) dan harus diselesaikan secara bersamaan. Prinsip dasar SPLDV adalah mencari pasangan nilai\u00a0(x,y)(x,y)\u00a0yang memenuhi semua persamaan yang […]<\/p>\n","protected":false},"author":1,"featured_media":2118,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"set","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[4],"tags":[231,224],"class_list":["post-2116","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-bahas-soal","tag-soal-sistem-persamaan-linear","tag-soal-tka-matematika-smp"],"_links":{"self":[{"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/posts\/2116","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/comments?post=2116"}],"version-history":[{"count":1,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/posts\/2116\/revisions"}],"predecessor-version":[{"id":2119,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/posts\/2116\/revisions\/2119"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/media\/2118"}],"wp:attachment":[{"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/media?parent=2116"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/categories?post=2116"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/tags?post=2116"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}