Pasangkan himpunan penyelesaian yang sesuai dengan setiap sistem persamaan berikut. dan

Pertanyaan

Pasangkan himpunan penyelesaian yang sesuai dengan setiap sistem persamaan berikut.

$begin mathsize 14px style y equals x squared minus 2 x minus 2 end style$ dan $begin mathsize 14px style y equals x minus 2 end style$

.... $space space$
.... $space space$

Pembahasan Soal:

$begin mathsize 14px style y equals x squared minus 2 x minus 2 end style$ dan $begin mathsize 14px style y equals x minus 2 end style$

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals y row cell x squared minus 2 x minus 2 end cell equals cell x minus 2 end cell row cell x squared minus 3 x end cell equals 0 row cell x left parenthesis x minus 3 right parenthesis end cell equals 0 row x equals cell 0 space atau space x equals 3 end cell end table end style$

$begin mathsize 14px style y equals x plus 2 x equals 0 rightwards arrow y equals 0 plus 2 equals 2 x equals 3 rightwards arrow y equals 3 plus 2 equals 5 end style$

Jadi, himpunan penyelesaian sistem persamaan adalah {(0,2),(3,5)}.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Indah

Mahasiswa/Alumni Universitas Lampung

Terakhir diupdate 29 Maret 2021

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Pertanyaan yang serupa

Carilah penyelesaian solusi dari setiap persamaan dua variabel kuadrat-kuadrat berikut. 1.

Pembahasan Soal:

Diketahui:

$x squared minus 2 x y minus 2 y squared equals 1 space... left parenthesis 1 right parenthesis 2 x squared plus x y plus y squared equals 2 space... left parenthesis 2 right parenthesis$

Eliminasikan persamaan (1) dan (2), sehingga

$x squared minus 2 x y minus 2 y squared equals 1 space space space open vertical bar x 2 close vertical bar space space space up diagonal strike 2 x squared end strike minus 4 x y minus 4 y squared equals 2 2 x squared plus x y plus y squared equals 2 space space space space space open vertical bar x 1 close vertical bar space space up diagonal strike space 2 x squared end strike plus x y plus y squared equals 2 space minus space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space top enclose space space space space space space space space space space minus 5 x y minus 5 y squared equals 0 space space space space space space space space space space space space space space space space space end enclose space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space up diagonal strike space minus 5 x end strike y equals up diagonal strike 5 y squared space y end strike space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space x equals negative y space space space.... left parenthesis 3 right parenthesis$

Selanjutnya, subtitusikan persamaan (3) ke persamaan (1), sehingga

$left parenthesis negative y right parenthesis squared minus 2 left parenthesis negative y right parenthesis times y minus 2 y squared equals 1 y squared up diagonal strike negative 2 y squared plus 2 y squared end strike equals 1 y squared equals 1 y equals plus-or-minus square root of 1 y equals plus-or-minus 1 y subscript 1 equals 1 space comma space space y subscript 2 equals negative 1$

Subtitusikan nilai $y subscript 1 space space dan space space y subscript 2$ ke persamaan (3), sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 end cell equals cell negative left parenthesis 1 right parenthesis end cell row cell x subscript 1 end cell equals cell negative 1 end cell row blank blank blank row cell x subscript 2 end cell equals cell negative left parenthesis negative 1 right parenthesis end cell row cell x subscript 2 end cell equals 1 end table$

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

Roboguru

Diketahui merupakan salah satu solusi dari sistem persamaan linear kuadrat tidak lengkap berikut. Tentukan solusi lain dari sistem persamaan linear- kuadrat tersebut.

Pembahasan Soal:

Diketahui:

$table attributes columnalign right center left columnspacing 0px end attributes row x equals cell 1 1 third end cell row blank equals cell 4 over 3 end cell row y equals cell 1 2 over 3 end cell row blank equals cell 5 over 3 end cell row cell a x plus b y end cell equals cell 7 space space... left parenthesis 1 right parenthesis end cell row cell 4 x squared minus a x y plus y squared end cell equals cell 1 space... left parenthesis 2 right parenthesis end cell end table$

Subtitusikan nilai $x$ dan $y$ ke persamaan (1), sehingga

$a times open parentheses 4 over 3 close parentheses plus b times open parentheses 5 over 3 close parentheses equals 7 space left parenthesis dikalikan space dengan space 3 right parenthesis a times 4 plus b times 5 equals 21 4 a plus 5 b equals 21 space... left parenthesis 3 right parenthesis$

Subtitusikan nilai $x$ dan $y$ ke persamaan (2), sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell 4 times open parentheses 4 over 3 close parentheses squared minus a times open parentheses 4 over 3 close parentheses times open parentheses 5 over 3 close parentheses plus open parentheses 5 over 3 close parentheses squared end cell equals 1 row cell 4 times open parentheses 16 over 9 close parentheses minus a open parentheses 20 over 9 close parentheses plus open parentheses 25 over 9 close parentheses end cell equals 1 row cell open parentheses 64 over 9 close parentheses plus open parentheses 25 over 9 close parentheses minus open parentheses 20 over 9 a close parentheses end cell equals 1 row cell open parentheses 89 over 9 close parentheses minus open parentheses 20 over 9 a close parentheses end cell equals 1 row cell negative open parentheses 20 over 9 a close parentheses end cell equals cell 1 minus open parentheses 89 over 9 close parentheses end cell row cell negative open parentheses 20 over 9 a close parentheses end cell equals cell negative open parentheses 80 over 9 close parentheses space left parenthesis dikalikan space minus 9 space kedua space ruas right parenthesis end cell row cell 20 times a end cell equals 80 row a equals cell 80 over 20 end cell row a equals 4 end table$

Selanjutnya, subtitusikan $a equals 4$ ke persamaan (3), sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell 4 times 4 plus 5 b end cell equals 21 row cell 16 plus 5 b end cell equals 21 row cell 5 b end cell equals cell 21 minus 16 end cell row cell 5 b end cell equals 5 row b equals cell 5 over 5 end cell row b equals 1 end table$

Selanjutnya, subtitusikan kembali nilai $a$ dan $b$ ke persamaan (1), sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell a x plus b y end cell equals 7 row cell 4 times x plus 1 times y end cell equals 7 row cell 4 x plus y end cell equals 7 row y equals cell 7 minus 4 x space... left parenthesis 4 right parenthesis end cell end table$

Subtitusikan nilai $a$ , $b$ dan persamaan (4) ke persamaan (2), sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell 4 x squared minus 4 times x times left parenthesis 7 minus 4 x right parenthesis plus left parenthesis 7 minus 4 x right parenthesis squared end cell equals cell 1 space end cell row cell 4 x squared minus 4 x times left parenthesis 7 minus 4 x right parenthesis plus 49 minus 56 x plus 16 x squared end cell equals 1 row cell 4 x squared minus 28 x plus 16 x squared plus 49 minus 56 x plus 16 x squared minus 1 end cell equals 0 row cell 36 x squared minus 84 x plus 40 end cell equals cell 0 space left parenthesis Dibagi space dengan space 4 right parenthesis end cell row cell 9 straight x squared minus 21 straight x plus 10 end cell equals 0 row cell left parenthesis 3 straight x minus 2 right parenthesis left parenthesis 3 straight x minus 5 right parenthesis end cell equals 0 end table$

Solusi lainnya yaitu

$x subscript 1 equals 2 over 3 x subscript 2 equals 5 over 3$

Jadi, solusi lainnya dari persamaan $open curly brackets table attributes columnalign left end attributes row cell a x plus b y equals 7 end cell row cell 4 x squared minus a x y plus y squared equals 1 end cell end table close$ adalah

$x subscript 1 equals 2 over 3 x subscript 2 equals 5 over 3$

Roboguru

Carilah penyelesaian solusi dari sietem persamaan dua variabel kuadrat-kuadrat berikut. 2.

Pembahasan Soal:

Diketahui:

$5 x squared plus 2 x y plus y squared equals 16 space... left parenthesis 1 right parenthesis 4 x squared plus 4 x y plus 16 equals 0 space... left parenthesis 2 right parenthesis$

Eliminasikan persamaan (1) dan (2), sehingga

$5 x squared plus 2 x y plus y squared equals 16 space space space open vertical bar x 2 close vertical bar space space space 10 x squared plus up diagonal strike 4 x y end strike plus 2 y squared equals 32 4 x squared plus 4 x y plus 16 equals 0 space space space space open vertical bar x 1 close vertical bar space space space space space 4 x squared space plus up diagonal strike 4 x y end strike plus 16 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 top enclose space space space space space space space space space space 6 x squared plus 2 y squared plus 16 equals 32 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 6 x squared plus 2 y squared equals 16 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 2 y squared equals 16 minus 6 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 space space space space space space space space space y squared equals fraction numerator 16 minus 6 x squared over denominator 2 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 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 8 minus 3 x squared space... left parenthesis 3 right parenthesis$

Selanjutnya, subtitusikan persamaan (3) ke persamaan (1), sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell 5 x squared plus 2 x y plus left parenthesis 8 minus 3 x squared right parenthesis end cell equals 16 row cell 2 x squared plus 2 x y end cell equals cell 16 space Kedua space ruas space dibagi space 4 right parenthesis end cell row cell x squared plus x y end cell equals 4 row cell x y end cell equals cell 4 minus x squared space... left parenthesis 4 right parenthesis end cell end table$

Subtitusikan persamaan (4) dan persamaan (3) ke persamaan (1), sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell 5 x squared plus left parenthesis 4 minus x squared right parenthesis plus left parenthesis 8 minus 3 x squared right parenthesis end cell equals 16 row cell x squared plus 12 end cell equals 16 row cell x squared end cell equals cell 16 minus 12 end cell row cell x squared end cell equals 4 row x equals cell plus-or-minus square root of 4 end cell row x equals cell plus-or-minus 2 end cell row cell x subscript 1 end cell equals cell 2 space comma space x subscript 2 equals negative 2 end cell end table$

Subtitusikan nilai $x subscript 1 space space dan space space x subscript 2$ ke persamaan (4), sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell 2 times y subscript 1 end cell equals cell 4 minus left parenthesis 2 right parenthesis squared end cell row cell 2 y subscript 1 end cell equals cell 4 minus 4 end cell row cell 2 y subscript 1 end cell equals 0 row cell y subscript 1 end cell equals cell 0 over 2 end cell row cell y subscript 1 end cell equals 0 row blank blank blank row cell 2 times y subscript 2 end cell equals cell 4 minus left parenthesis negative 2 right parenthesis squared end cell row cell 2 y subscript 2 end cell equals cell 4 minus 4 end cell row cell 2 y subscript 2 end cell equals 0 row cell y subscript 2 end cell equals cell 0 over 2 end cell row cell y subscript 2 end cell equals 0 end table$

Jadi, solusi dari sistem persamaan dua variabel ini adalah (2,0) dan (-2,0).

Roboguru

Pasangkan himpunan penyelesaian yang sesuai dengan setiap sistem persamaan berikut. dan

Pembahasan Soal:

$begin mathsize 14px style y equals negative x squared plus 4 x minus 2 end style$ dan $begin mathsize 14px style y equals negative x plus 2 end style$

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals y row cell negative x squared plus 4 x minus 2 end cell equals cell negative x plus 2 end cell row cell x squared minus 5 x plus 4 end cell equals 0 row cell left parenthesis x minus 4 right parenthesis left parenthesis x minus 1 right parenthesis end cell equals 0 row x equals cell 4 space atau space x equals 1 end cell end table end style$

$begin mathsize 14px style y equals negative x plus 2 x equals 4 rightwards arrow y equals negative 4 plus 2 equals negative 2 x equals 1 rightwards arrow y equals negative 1 plus 2 equals 1 end style$

Jadi, himpunan penyelesaian persamaan adalah {(1,10,(4,-2)}.

Roboguru

A merchant has a square carped priced at $1 per square foot and a reactangular priced at $1,50 per square foot. The combined area of the rectangular is 112 square feet and the value of the rectangular...

Pembahasan Soal:

Diketahui:

$x equals square y equals rectangular x plus 1 comma 50 y equals 112 space... left parenthesis 1 right parenthesis x equals 8 y space... left parenthesis 2 right parenthesis$

Selanjutnya, subtitusikan persamaan (2) ke persamaan (1), sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell 8 y plus 1 comma 50 y end cell equals 112 row cell 9 comma 5 y end cell equals 112 row y equals cell fraction numerator 112 over denominator 9 comma 5 end fraction end cell row y equals cell 11 comma 78 end cell end table$

Subtitusikan nilai $y$ ke persamaan (2), sehingga

$x equals 8 times left parenthesis 11 comma 78 right parenthesis x equals 94 comma 24$

Jadi, ukuran carpet untuk persegi adalah 94,24 dan persegi panjang adalah 11,78.

Roboguru