Pertanyaan

Carilah penyelesaian solusi dari setiap sistem persamaan dua variabel kuadrat-kuadrat berikut. 6. { 5 x 2 + 3 x y − 2 y 2 − 34 = 0 3 x 2 + 8 x y − 3 y 2 = 0

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

6. ${5 x^{2} + 3 x y - 2 y^{2} - 34 = 0 3 x^{2} + 8 x y - 3 y^{2} = 0$

D. Wahyu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

Jawaban terverifikasi

Pembahasan

Diketahui: Eliminasikan persamaaan (1) dan (2), sehingga Subtitusikan persamaan (3) ke persamaan (2), sehingga Subtitusikan persamaan (4) ke persamaan (3), sehingga Subtitusikan persamaan (5) ke persamaan (4), sehingga Selanjutnya, subtitusikan persamaan (6) dan persamaan (5) ke persamaan (2), sehingga Subtitusikan nilai ke persamaan (6), sehingga Jadi, solusi dari sistem persamaan dua variabel kuadrat-kuadrat ini adalah .

Diketahui:

5 x squared plus 3 x y minus 2 y squared minus 34 equals 0 space... left parenthesis 1 right parenthesis 3 x squared plus 8 x y minus 3 y squared equals 0 space... space left parenthesis 2 right parenthesis

Eliminasikan persamaaan (1) dan (2), sehingga

$5 x squared plus 3 x y minus 2 y squared minus 34 equals 0 space space open vertical bar x 8 close vertical bar space space space 40 x squared plus up diagonal strike 24 x y end strike minus 16 y squared minus 272 equals 0 3 x squared plus 8 x y minus 3 y squared equals 0 space space space space space space space space space space open vertical bar x 3 close vertical bar space space space space 9 x squared plus up diagonal strike 24 x y end strike minus 9 y squared space equals 0 space space space space space 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 space space top enclose space 31 x squared minus 7 y squared minus 272 equals 0 space space space 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 minus 7 y squared equals 272 minus 31 x squared space left parenthesis Kedua space ruas space dikali left parenthesis negative right parenthesis right parenthesis 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 7 straight y squared equals negative 272 plus 31 straight x squared 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 fraction numerator negative 272 plus 31 x squared over denominator 7 end fraction... space left parenthesis 3 right parenthesis$

Subtitusikan persamaan (3) ke persamaan (2), sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x squared plus 8 x y minus 3 times open parentheses fraction numerator negative 272 plus 31 x squared over denominator 7 end fraction close parentheses end cell equals cell 0 space left parenthesis Kedua space ruas space dikali space 7 right parenthesis end cell row cell 21 x squared plus 56 x y minus 3 left parenthesis negative 272 plus 31 x squared right parenthesis end cell equals 0 row cell 21 x squared plus 56 x y plus 816 minus 93 x squared end cell equals 0 row cell negative 72 x squared plus 56 x y plus 816 end cell equals cell 0 space space left parenthesis Kedua space ruas space dibagi space minus 8 right parenthesis end cell row cell 9 x squared plus 7 x y minus 102 end cell equals 0 row cell 9 x squared end cell equals cell 102 minus 7 x y end cell row cell x squared end cell equals cell fraction numerator 102 minus 7 x y over denominator 9 end fraction space... left parenthesis 4 right parenthesis end cell end table$

Subtitusikan persamaan (4) ke persamaan (3), sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell y squared end cell equals cell fraction numerator negative 272 plus 31 times open parentheses begin display style fraction numerator 102 minus 7 x y over denominator 9 end fraction end style close parentheses space space over denominator 7 end fraction end cell row cell 7 y squared end cell equals cell negative 272 plus 31 times open parentheses fraction numerator 102 minus 7 x y over denominator 9 end fraction close parentheses space space left parenthesis Kedua space ruas space dikali space 9 right parenthesis end cell row cell 63 y squared end cell equals cell negative 2448 plus 31 times open parentheses 102 minus 7 x y close parentheses end cell row cell 63 y squared end cell equals cell negative 2248 plus 3162 minus 217 x y end cell row cell 63 y squared end cell equals cell 714 plus 217 x y space left parenthesis Kedua space ruas space dibagi space 7 right parenthesis end cell row cell 9 y squared end cell equals cell 102 plus 31 x y space end cell row cell 9 y squared minus 102 end cell equals cell 31 x y end cell row cell x y end cell equals cell fraction numerator 9 y squared minus 102 over denominator 31 end fraction space... left parenthesis 5 right parenthesis end cell row blank blank blank end table$

Subtitusikan persamaan (5) ke persamaan (4), sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell x squared end cell equals cell fraction numerator 102 plus 7 times open parentheses begin display style fraction numerator 9 y squared minus 102 over denominator 31 end fraction end style close parentheses over denominator 9 end fraction end cell row cell 9 x squared end cell equals cell 102 plus 7 times open parentheses fraction numerator 9 y squared minus 102 over denominator 31 end fraction close parentheses space left parenthesis Kedua space ruas space dikali space 31 right parenthesis end cell row cell 279 x squared end cell equals cell 3162 plus 7 times open parentheses 9 y squared minus 102 close parentheses end cell row cell 279 x squared end cell equals cell 3162 plus 63 y squared minus 714 end cell row cell 279 x squared end cell equals cell 2448 plus 63 y squared space space space left parenthesis Kedua space ruas space dibagi space 9 right parenthesis end cell row cell 31 x squared end cell equals cell 272 plus 9 y squared end cell row cell x squared end cell equals cell fraction numerator 272 plus 9 y squared over denominator 31 end fraction space... space left parenthesis 6 right parenthesis end cell end table$

Selanjutnya, subtitusikan persamaan (6) dan persamaan (5) ke persamaan (2), sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell 3 times open parentheses fraction numerator 272 plus 9 y squared over denominator 31 end fraction close parentheses plus 8 times open parentheses fraction numerator 9 y squared minus 102 over denominator 31 end fraction close parentheses minus 3 y squared end cell equals cell 0 space left parenthesis Kedua space ruas space dikali space 31 right parenthesis end cell row cell 3 times left parenthesis 272 plus 9 y squared right parenthesis plus 8 times left parenthesis 9 y squared minus 102 right parenthesis minus 93 y squared end cell equals 0 row cell up diagonal strike 816 plus 27 y squared plus 72 y squared up diagonal strike negative 816 end strike minus 93 y squared end cell equals 0 row cell 6 y squared end cell equals 0 row cell y squared end cell equals cell 0 over 6 end cell row cell y squared end cell equals 0 row y equals 0 end table$

Subtitusikan nilai ke persamaan (6), sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell x squared end cell equals cell fraction numerator 102 plus 7 times left parenthesis 0 right parenthesis over denominator 9 end fraction end cell row cell x squared end cell equals cell 102 over 9 end cell row cell x squared end cell equals cell 34 over 3 end cell row x equals cell plus-or-minus square root of 34 over 3 end root end cell row cell x subscript 1 end cell equals cell square root of 34 over 3 end root space comma space x subscript 2 equals negative square root of 34 over 3 end root end cell end table$

Jadi, solusi dari sistem persamaan dua variabel kuadrat-kuadrat ini adalah open parentheses square root of 34 over 3 end root comma 0 close parentheses space space dan space space open parentheses negative square root of 34 over 3 end root comma 0 close parentheses space .

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