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Carilah penyelesaian solusi dari setiap sistem persamaan dua variabel kuadrat-kuadrat berikut. 5. { 2 x 2 − 3 x y + 2 y 2 − 4 = 0 4 x 2 − 6 x y + 3 y 2 − 4 = 0 ​

Carilah penyelesaian solusi dari setiap sistem persamaan dua variabel kuadrat-kuadrat berikut.

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D. Wahyu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

Jawaban terverifikasi

Jawaban

solusi dari sistem persamaan dua variabel kuadrat ini adalah (1,2) dan (2,-1).

solusi dari sistem persamaan dua variabel kuadrat ini adalah (1,2) dan (2,-1).

Pembahasan

Diketahui: Eliminasikan persamaan (1) dan (2), sehingga Subtitusikan nilai ke persamaan (1), sehingga Jadi, solusi dari sistem persamaan dua variabel kuadrat ini adalah (1,2) dan (2,-1).

Diketahui:

2 x squared minus 3 x y plus 2 y squared minus 4 equals 0 space space... left parenthesis 1 right parenthesis 4 x squared minus 6 x y plus 3 y squared minus 4 equals 0 space space... left parenthesis 2 right parenthesis

Eliminasikan persamaan (1) dan (2), sehingga

2 x squared minus 3 x y plus 2 y squared minus 4 equals 0 space space space open vertical bar x 2 close vertical bar space space up diagonal strike 4 x squared minus 6 x y end strike plus 4 y squared minus 8 equals 0 4 x squared minus 6 x y plus 3 y squared minus 4 equals 0 space space space open vertical bar x 1 close vertical bar space space up diagonal strike 4 x squared minus 6 x y end strike plus 3 y squared minus 4 equals 0 space minus space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space top enclose space space space space space space space space space space space space space space space space space y squared space space minus 4 space equals 0 space space space space space space space space space space space end enclose space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space y squared equals 4 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space y equals plus-or-minus square root of 4 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space y equals plus-or-minus 2 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space y subscript 1 equals 2 comma space y subscript 2 equals negative 2

Subtitusikan nilai  y subscript 1 space space dan space space y subscript 2  ke persamaan (1), sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell Ketika comma space y equals 2 space colon space space 2 x squared minus 3 times left parenthesis 2 right parenthesis times x plus 2 left parenthesis 2 right parenthesis squared minus 4 end cell equals 0 row cell 2 x squared minus 6 x plus 8 minus 4 end cell equals 0 row cell 2 x squared minus 6 y plus 4 end cell equals cell 0 space left parenthesis Kedua space ruas space dibagi space 2 right parenthesis end cell row cell y squared minus 3 y plus 2 end cell equals 0 row cell left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis end cell equals 0 row cell x subscript 1 end cell equals cell 1 comma space space x subscript 2 equals 2 end cell row cell Ketika comma space y equals negative 2 space colon space space 2 x squared minus 3 times left parenthesis negative 2 right parenthesis times x plus 2 left parenthesis negative 2 right parenthesis squared minus 4 end cell equals 0 row cell 2 x squared plus 6 x plus 8 minus 4 end cell equals 0 row cell 2 x squared plus 6 y plus 4 end cell equals cell 0 space left parenthesis Kedua space ruas space dibagi space 2 right parenthesis end cell row cell x squared plus 3 y plus 2 end cell equals 0 row cell left parenthesis x plus 1 right parenthesis squared end cell equals 0 row x equals cell plus-or-minus square root of negative 1 end root end cell row blank blank cell x comma space tidak space punya space solusi end cell end table

Jadi, solusi dari sistem persamaan dua variabel kuadrat ini adalah (1,2) dan (2,-1).

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Diketahui ( − 1 , − 2 ) merupakan salah satu solusi real dari sistem persamaan berikut. { 2 a x 2 − 7 x y − 2 a y 2 = − 20 − a x 2 + 4 x y + a y 2 = 11 ​ Carilah Solusi lainnya!

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