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Tentukan penyelesaian atau akar sistem persamaan berikut! 2 x 2 ​ + 3 y 2 ​ = 16 3 1 ​ dan 4 x 2 ​ − 3 y 2 ​ = − 4 3 1 ​

Tentukan penyelesaian atau akar sistem persamaan berikut!

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E. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sebelas Maret

Jawaban terverifikasi

Jawaban

penyelesaiannya adalah dan .

penyelesaiannya adalah x equals 4 space atau space x equals negative 4 dan y equals 5 space atau space y equals negative 5

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Pembahasan

Diketahui sistem persamaan dan . Sistem persamaan di atas disederhanakan menjadi dan Menentukan Substitusi ke persamaan Substitusi ke persamaan Substitusi ke persamaan Jadi, penyelesaiannya adalah dan .

Diketahui sistem persamaan x squared over 2 plus y squared over 3 equals 16 1 third dan x squared over 4 minus y squared over 3 equals negative 4 1 third.

Sistem persamaan di atas disederhanakan menjadi

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared over 2 plus y squared over 3 end cell equals cell 16 1 third end cell row cell left parenthesis x squared over 2 plus y squared over 3 end cell equals cell 49 over 3 right parenthesis times 6 end cell row cell 3 x squared plus 2 y squared end cell equals 98 end table    dan   table attributes columnalign right center left columnspacing 0px end attributes row cell x squared over 4 minus y squared over 3 end cell equals cell negative 4 1 third end cell row cell x squared over 4 minus y squared over 3 end cell equals cell negative 13 over 3 end cell row cell 3 x squared minus 4 y squared end cell equals cell negative 52 end cell end table 

Menentukan x squared

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x squared plus 2 y squared end cell equals 98 row cell 3 x squared end cell equals cell negative 2 y squared plus 98 end cell row cell x squared end cell equals cell fraction numerator negative 2 y squared plus 98 over denominator 3 end fraction end cell end table 

Substitusi x squared equals fraction numerator negative 2 y squared plus 98 over denominator 3 end fraction ke persamaan 3 x squared minus 4 y squared equals negative 52

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x squared minus 4 y squared end cell equals cell negative 52 end cell row cell 3 left parenthesis fraction numerator negative 2 y squared plus 98 over denominator 3 end fraction right parenthesis minus 4 y squared end cell equals cell negative 52 end cell row cell negative 2 y squared plus 98 minus 4 y squared end cell equals cell negative 52 end cell row cell negative 6 y squared end cell equals cell negative 52 minus 98 end cell row cell negative 6 y squared end cell equals cell negative 150 end cell row cell y squared end cell equals cell fraction numerator negative 150 over denominator negative 6 end fraction end cell row cell y squared end cell equals 25 row y equals cell 5 space atau space y equals negative 5 end cell end table 

Substitusi y equals 5 ke persamaan x squared equals fraction numerator negative 2 y squared plus 98 over denominator 3 end fraction

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared end cell equals cell fraction numerator negative 2 y squared plus 98 over denominator 3 end fraction end cell row cell x squared end cell equals cell fraction numerator negative 2 left parenthesis 5 right parenthesis squared plus 98 over denominator 3 end fraction end cell row cell x squared end cell equals cell fraction numerator negative 2 left parenthesis 25 right parenthesis plus 98 over denominator 3 end fraction end cell row cell x squared end cell equals cell fraction numerator negative 50 plus 98 over denominator 3 end fraction end cell row cell x squared end cell equals cell 48 over 3 end cell row cell x squared end cell equals 16 row x equals cell 4 space atau space x equals negative 4 end cell end table 

Substitusi y equals negative 5 ke persamaan x squared equals fraction numerator negative 2 y squared plus 98 over denominator 3 end fraction

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared end cell equals cell fraction numerator negative 2 y squared plus 98 over denominator 3 end fraction end cell row cell x squared end cell equals cell fraction numerator negative 2 left parenthesis negative 5 right parenthesis squared plus 98 over denominator 3 end fraction end cell row cell x squared end cell equals cell fraction numerator negative 2 left parenthesis 25 right parenthesis plus 98 over denominator 3 end fraction end cell row cell x squared end cell equals cell fraction numerator negative 50 plus 98 over denominator 3 end fraction end cell row cell x squared end cell equals cell 48 over 3 end cell row cell x squared end cell equals 16 row x equals cell 4 space atau space x equals negative 4 end cell end table 

Jadi, penyelesaiannya adalah x equals 4 space atau space x equals negative 4 dan y equals 5 space atau space y equals negative 5

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