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Diketahui  open parentheses negative 1 comma negative 2 close parentheses merupakan salah satu solusi real dari sistem persamaan berikut.

open curly brackets table attributes columnalign left end attributes row cell 2 a x squared minus 7 x y minus 2 a y squared equals negative 20 end cell row cell negative a x squared plus 4 x y plus a y squared equals 11 end cell end table close

Carilah Solusi lainnya!

D. Wahyu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

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dari 2 persamaan didapatkan tidak punya solusi.

Pembahasan

Diketahui:

2 a x squared minus 7 x y minus 2 a y squared equals negative 20 space... left parenthesis 1 right parenthesis minus a x squared plus 4 x y plus a y squared equals 11 space... left parenthesis 2 right parenthesis left parenthesis x comma y right parenthesis equals left parenthesis negative 1 comma negative 2 right parenthesis space... left parenthesis 3 right parenthesis

Terlebih dahulu, subtitusikan persamaan (3) ke salah satu dari persamaan (1) atau (2), Disini, saya pilih persamaan (1), sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 a left parenthesis negative 1 right parenthesis squared minus 7 left parenthesis negative 1 right parenthesis left parenthesis negative 2 right parenthesis minus 2 a left parenthesis negative 2 right parenthesis squared end cell equals cell negative 20 end cell row cell 2 a minus 14 minus 8 a end cell equals cell negative 20 end cell row cell negative 6 a end cell equals cell negative 20 plus 14 end cell row cell negative 6 a end cell equals cell negative 6 space left parenthesis Kedua space ruas space dikali space left parenthesis negative right parenthesis right parenthesis end cell row cell 6 a end cell equals 6 row a equals cell 6 over 6 end cell row a equals 1 end table

Selanjutnya, mencari solusi lainnya. Terlebih dahulu, subtitusikan nilai a ke persamaan (1) dan persamaan (2), sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 left parenthesis 1 right parenthesis x squared minus 7 x y minus 2 left parenthesis 1 right parenthesis y squared end cell equals cell negative 20 end cell row cell 2 x squared minus 2 y squared minus 7 x y end cell equals cell negative 20 space... left parenthesis 3 right parenthesis end cell row blank blank blank row cell left parenthesis 1 right parenthesis minus x squared plus 4 x y plus left parenthesis 1 right parenthesis y squared end cell equals 11 row cell negative x squared plus y squared plus 4 x y end cell equals cell 11... left parenthesis 4 right parenthesis end cell end table

Eliminasikan persamaan (3) dan (4), sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x squared minus 2 y squared minus 7 x y end cell equals cell negative 20 space space space space space space open vertical bar x 1 close vertical bar space space space space space space up diagonal strike 2 x squared minus 2 y squared end strike minus 7 x y equals negative 20 space end cell row cell negative x squared plus y squared plus 4 x y end cell equals cell 11 space space space space space space space space space space open vertical bar x 2 close vertical bar space space space space space up diagonal strike space minus 2 x squared plus 2 y end strike squared plus 8 x y space equals 22 space plus end cell row blank blank cell 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 space space space space space space space space space x y equals space 2 space... left parenthesis 5 right parenthesis space space space space space space space space space space space space space space end enclose end cell row blank blank cell 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 end cell end table

Subtitusikan persamaan (5) ke persamaan (3) dan persamaan (4)  sehingga

2 x squared minus 2 y squared minus 7 left parenthesis 2 right parenthesis equals negative 20 2 x squared minus 2 y squared minus 14 equals negative 20 2 x squared minus 2 y squared equals negative 20 plus 14 2 x squared minus 2 y squared equals negative 6 2 left parenthesis x squared minus y squared right parenthesis equals negative 6 x squared minus y squared equals fraction numerator negative 6 over denominator 2 end fraction x squared minus y squared equals negative 3 space... left parenthesis 5 right parenthesis

negative x squared plus y squared plus 4 left parenthesis 2 right parenthesis equals 11 minus x squared plus y squared plus 8 equals 11 minus x squared plus y squared equals 11 minus 8 minus x squared plus y squared equals 3... left parenthesis 6 right parenthesis

Dari persamaan (5) dan (6) didapatkan tidak punya solusi.

Jadi, dari 2 persamaan didapatkan tidak punya solusi.

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