Roboguru

Calculate the coordinate of the points of intersection between the following straight lines dan curves: 3x−2y=7dan3x2+xy−2y2=28

Pertanyaan

Calculate the coordinate of the points of intersection between the following straight lines dan curves:

3 x minus 2 y equals 7 space space dan space space 3 x squared plus x y minus 2 y squared equals 28

Pembahasan Soal:

Diketahui:

3 x minus 2 y equals 7 x equals fraction numerator 7 plus 2 y over denominator 3 end fraction... left parenthesis 1 right parenthesis 3 x squared plus x y minus 2 y squared equals 28... left parenthesis 2 right parenthesis

Subtitusikan persamaan (1) ke persamaan (2) , sehingga

3 times open parentheses fraction numerator 7 plus 2 y over denominator 3 end fraction close parentheses squared plus open parentheses fraction numerator 7 plus 2 y over denominator 3 end fraction close parentheses times y minus 2 y squared equals 28 3 times open parentheses fraction numerator 49 plus 28 y plus 4 y squared over denominator 9 end fraction close parentheses plus open parentheses fraction numerator 7 y plus 2 y squared over denominator 3 end fraction close parentheses minus 2 y squared equals 28 space left parenthesis Kedua space ruas space straight x 9 right parenthesis 3 times open parentheses 49 plus 28 y plus 4 y squared close parentheses plus 3 times open parentheses 7 y plus 2 y squared close parentheses minus 18 y squared equals 252 147 plus 84 y plus 12 y squared space plus 21 y plus 6 y squared minus 18 y squared minus 252 equals 0 105 y minus 105 105 y equals 105 y equals 105 over 105 y equals 1

Selanjutnya, subtitusikan nilai y ke persamaan (1) , sehingga

table attributes columnalign right center left columnspacing 0px end attributes row x equals cell fraction numerator 7 plus 2 left parenthesis 1 right parenthesis over denominator 3 end fraction end cell row x equals cell 9 over 3 end cell row x equals 3 end table

Jadi, titik koordinat dari persamaan 3 x minus 2 y equals 7 space space dan space space 3 x squared plus x y minus 2 y squared equals 28  adalah  (3,1).

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Adma

Mahasiswa/Alumni Universitas Riau

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

Carilah solusi (x,y) bilangan real untuk setiap persamaan kuadrat berikut dan lukiska sketsa grafik penyelesaiannya! {2x2+xy+y2=162x2+7xy+6y2=0​

Pembahasan Soal:

Diketahui:

2 x squared plus x y plus y squared equals 16... left parenthesis 1 right parenthesis 2 x squared plus 7 x y plus 6 y squared equals 0... left parenthesis 2 right parenthesis

Dari persamaan (2) dapat difaktorkan, sehingga

2 x squared plus 7 x y plus 6 y squared equals 0 left parenthesis 2 x plus 3 y right parenthesis left parenthesis x plus 2 y right parenthesis equals 0  2 x plus 3 y equals 0 3 y equals negative 2 x y equals negative 2 over 3 x  x plus 2 y equals 0 2 y equals negative x y equals negative x over 2

Nilai-nilai y disubtitusikan ke persamaan (2), sehingga

Ketika space y equals negative 2 over 3 x comma space maka

2 x squared plus 7 x times open parentheses negative 2 over 3 x close parentheses plus 6 times open parentheses negative 2 over 3 x close parentheses squared equals 0 2 x squared minus 14 over 3 x squared space plus up diagonal strike 6 squared end strike times open parentheses fraction numerator 4 over denominator up diagonal strike 9 cubed end strike end fraction x squared close parentheses equals 0 2 x squared minus 14 over 3 x squared space plus 8 over 3 x squared equals 0 space left parenthesis Kedua space ruas space dikali space 3 right parenthesis 6 x squared minus 14 x squared plus 8 x squared equals 0 minus 8 x squared plus 8 x squared equals 0 x space tidak space punya space solusi

Ketika space y equals negative x over 2 comma space maka

2 x squared plus 7 x times open parentheses fraction numerator negative x over denominator 2 end fraction close parentheses plus 6 times open parentheses negative x over 2 close parentheses squared equals 0 2 x squared minus fraction numerator 7 x squared over denominator 3 end fraction space plus up diagonal strike 6 cubed end strike times open parentheses fraction numerator x squared over denominator up diagonal strike 4 squared end strike end fraction close parentheses equals 0 2 x squared minus 14 over 3 x squared space plus fraction numerator 3 x squared over denominator 2 end fraction equals 0 space left parenthesis Kedua space ruas space dikali space 3 right parenthesis fraction numerator 6 x squared over denominator 3 end fraction minus fraction numerator 14 x squared over denominator 3 end fraction plus fraction numerator 3 x squared over denominator 2 end fraction equals 0 fraction numerator negative 8 x squared over denominator 3 end fraction plus fraction numerator 3 x squared over denominator 2 end fraction equals 0 fraction numerator negative 16 x squared plus 9 x squared over denominator 6 end fraction equals negative 8 space fraction numerator negative 7 x squared over denominator 6 end fraction equals negative 8 left parenthesis Kedua space ruas space dikali space minus 6 right parenthesis 7 x squared equals 48 x squared equals 48 over 7 x equals plus-or-minus square root of 48 over 7 end root x subscript 1 equals square root of space 48 over 7 end root comma space x subscript 2 equals negative square root of 48 over 7 end root

Selanjutnya, subtitusikan nilai

y equals negative x over 2 space ke space x subscript 1 space space dan space space x subscript 2 comma space sehingga

 y subscript 1 equals negative fraction numerator square root of begin display style 48 over 7 end style end root over denominator 2 end fraction y subscript 2 equals negative fraction numerator open parentheses negative square root of begin display style 48 over 7 end style end root close parentheses over denominator 2 end fraction equals fraction numerator square root of 48 over 7 end root over denominator 7 end fraction

Jadi, solusi adalah open parentheses square root of 48 over 7 end root comma space fraction numerator negative square root of begin display style 48 over 7 end style end root over denominator 2 end fraction close parentheses space space dan space space open parentheses negative square root of 48 over 7 end root comma space fraction numerator square root of begin display style 48 over 7 end style end root over denominator 2 end fraction close parentheses.

Grafik

0

Roboguru

Himpunan penyelesaian dari sistem persamaan linear-kuadrat {y=x2y=2x+3​  adalah ....

Pembahasan Soal:

Substitusikan persamaan pertama ke persamaan kedua. Sehingga didapat

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight x squared end cell equals cell 2 straight x plus 3 end cell row cell straight x squared minus 2 straight x minus 3 end cell equals 0 row cell left parenthesis straight x plus 1 right parenthesis left parenthesis straight x minus 3 right parenthesis end cell equals 0 row cell straight x subscript 1 end cell equals cell negative 1 space atau space straight x subscript 2 equals 3 end cell end table end style 

Setelah itu, substitusikan nilai koordinat x  yang didapatkan ke dalam salah satu persamaan untuk mendapatkan nilai koordinat y .

Misalkan digunakan persamaan yang pertama.

Sehingga untuk x1 = –1, didapat :

begin mathsize 14px style straight y subscript 1 equals straight x subscript 1 superscript 2 equals left parenthesis negative 1 right parenthesis squared equals 1 end style 


Kemudian untuk x2 = 3, didapat :

 begin mathsize 14px style straight y subscript 2 equals straight x subscript 2 superscript 2 equals 3 squared equals 9 end style 

Sehingga didapatkan penyelesaian dari sistem persamaan linear-kuadrat tersebut adalah (–1,1) dan (3,9) .

Maka, himpunan penyelesaian dari sistem persamaan linear-kuadrat tersebut adalah {(–1,1), (3,9)}

0

Roboguru

Calculate the coordinate of the points of intersection between the following straight lines dan curves: 6x−5y+2=0dan2x2+5xy−3y2=0

Pembahasan Soal:

Diketahui:

6 x minus 5 y plus 2 equals 0 minus 5 y equals negative 2 minus 6 x space left parenthesis Dikali space left parenthesis negative right parenthesis right parenthesis 5 y equals 2 plus 6 x y equals fraction numerator 2 plus 6 x over denominator 5 end fraction... left parenthesis 1 right parenthesis 2 x squared plus 5 x y minus 3 y squared equals 0... left parenthesis 2 right parenthesis

Subtitusikan persamaan (1) ke persamaan (2) , sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x squared plus up diagonal strike 5 times x times open parentheses fraction numerator 2 plus 6 x over denominator up diagonal strike 5 end fraction close parentheses minus 3 times open parentheses fraction numerator 2 plus 6 x over denominator 5 end fraction close parentheses squared end cell equals 0 row cell 2 x squared plus x times left parenthesis 2 plus 6 x right parenthesis minus 3 times open parentheses fraction numerator 4 plus 24 x plus 36 x squared over denominator 25 end fraction close parentheses end cell equals 0 row cell 2 x squared plus 2 x plus 6 x squared minus 3 times open parentheses fraction numerator 4 plus 24 x plus 36 x squared over denominator 25 end fraction close parentheses end cell equals cell 0 space left parenthesis Kedua space ruas space straight x space 25 right parenthesis end cell row cell 50 x squared plus 50 x plus 150 x squared minus 3 times left parenthesis 4 plus 24 x plus 36 x squared right parenthesis end cell equals 0 row cell 50 x squared plus 150 x squared minus 108 x squared plus 50 x minus 72 x minus 12 end cell equals 0 row cell 92 x squared minus 22 x minus 12 end cell equals cell 0 space left parenthesis Kedua space ruas space colon 2 right parenthesis end cell row cell 46 straight x squared minus 11 straight x minus 6 end cell equals 0 end table

Nilai x dapat diperoleh dengan rumus abc  yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 12 end cell equals cell fraction numerator negative b plus-or-minus square root of b squared minus 4 times a times c end root over denominator 2 a end fraction end cell end table

Dimana, nilai a equals 46 comma space b equals negative 11 comma space c equals negative 6 , sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 12 end cell equals cell fraction numerator negative left parenthesis negative 11 right parenthesis plus-or-minus square root of left parenthesis negative 11 right parenthesis squared minus 4 times 46 times negative 6 end root over denominator 2 left parenthesis 46 right parenthesis end fraction end cell row cell x subscript 12 end cell equals cell fraction numerator 11 plus-or-minus square root of 121 plus 1104 end root over denominator 92 end fraction end cell row cell x subscript 12 end cell equals cell fraction numerator 11 plus-or-minus square root of 1225 over denominator 92 end fraction end cell row cell x subscript 12 end cell equals cell fraction numerator 11 plus-or-minus 35 over denominator 92 end fraction end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 end cell equals cell fraction numerator 11 plus 35 over denominator 92 end fraction end cell row cell x subscript 1 end cell equals cell 46 over 92 end cell row cell x subscript 1 end cell equals cell 1 half end cell end table     x subscript 2 equals fraction numerator 11 minus 35 over denominator 92 end fraction x subscript 2 equals fraction numerator negative 24 over denominator 92 end fraction x subscript 2 equals fraction numerator negative 6 over denominator 23 end fraction

Subtituisikan x subscript 1 equals 1 half ke persamaan (1) yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell y subscript 1 end cell equals cell fraction numerator 2 plus 6 times open parentheses begin display style 1 half end style close parentheses over denominator 5 end fraction end cell row blank equals cell fraction numerator 2 plus 3 over denominator 5 end fraction end cell row blank equals 1 end table

Subtituisikan x subscript 2 equals negative 6 over 23 ke persamaan (1) yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell y subscript 2 end cell equals cell fraction numerator 2 plus 6 times open parentheses begin display style negative 6 over 23 end style close parentheses over denominator 5 end fraction end cell row blank equals cell fraction numerator 2 plus open parentheses negative begin display style 36 over 23 end style close parentheses over denominator 5 end fraction end cell row blank equals cell fraction numerator 2 minus begin display style 36 over 23 end style over denominator 5 end fraction end cell row blank equals cell fraction numerator begin display style 10 over 23 end style over denominator 5 end fraction end cell row blank equals cell 10 over 115 end cell row blank equals cell 2 over 23 end cell end table

Jadi, titik koordinat dari persamaan  6 x minus 5 y plus 2 equals 0 space space dan space space 2 x squared plus 5 x y minus 3 y squared equals 0  adalah  open parentheses 1 half comma 1 close parentheses space space space dan space space space open parentheses fraction numerator negative 6 over denominator 23 end fraction comma 2 over 23 close parentheses.

0

Roboguru

Calculate the coordinate of the points of intersection between the following straight lines dan curves: 4x+3y=8danx2+9y2−6xy=4

Pembahasan Soal:

Diketahui:

4 x plus 3 y equals 8 y equals fraction numerator 8 minus 4 y over denominator 3 end fraction... left parenthesis 1 right parenthesis x squared plus 9 y squared minus 6 x y equals 4... left parenthesis 2 right parenthesis

Subtitusikan persamaan (1) ke persamaan (2) , sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus 9 times open parentheses fraction numerator 8 minus 4 x over denominator 3 end fraction close parentheses squared minus 6 times x times open parentheses fraction numerator 8 minus 4 x over denominator 3 end fraction close parentheses end cell equals 4 row cell x squared plus up diagonal strike 9 times open parentheses fraction numerator 64 minus 64 x plus 16 x squared over denominator up diagonal strike 9 end fraction close parentheses minus up diagonal strike 62 times x times open parentheses fraction numerator 8 minus 4 x over denominator up diagonal strike 3 end fraction close parentheses end cell equals 4 row cell x squared plus left parenthesis 64 minus 64 x plus 16 x squared right parenthesis minus 16 x plus 8 x squared minus 4 end cell equals cell 0 space end cell row cell 25 x squared minus 80 x plus 60 end cell equals cell 0 space left parenthesis Kedua space ruas space colon 5 right parenthesis end cell row cell 5 x squared minus 16 x plus 12 end cell equals 0 row cell left parenthesis 5 x minus 6 right parenthesis left parenthesis x minus 3 right parenthesis end cell equals 0 end table

Nilai x yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell 5 x subscript 1 minus 6 end cell equals 0 row cell x subscript 1 end cell equals cell 6 over 5 end cell row cell x subscript 2 minus 2 end cell equals 0 row cell x subscript 2 end cell equals 2 end table

Selanjutnya, subtitusikan nilai  x subscript 1 equals 6 over 5  ke persamaan (1) , sehingga

table attributes columnalign right center left columnspacing 0px end attributes row y equals cell fraction numerator 8 minus 4 times open parentheses begin display style 6 over 5 end style close parentheses over denominator 3 end fraction end cell row y equals cell fraction numerator 40 minus 24 over denominator 5 end fraction end cell row y equals cell 16 over 5 end cell end table

Selanjutnya, subtitusikan nilai  x subscript 2 equals 2  ke persamaan (1) , sehingga

table attributes columnalign right center left columnspacing 0px end attributes row y equals cell fraction numerator 8 minus 4 times open parentheses begin display style 2 end style close parentheses over denominator 3 end fraction end cell row y equals cell 0 over 5 end cell row y equals 0 end table

Jadi, titik koordinat dari persamaan 4 x plus 3 y equals 8 space space dan space space x squared plus 9 y squared minus 6 x y equals 4  adalah  open parentheses 6 over 5 comma 16 over 5 close parentheses space space dan space space open parentheses 2 comma 0 close parentheses.

0

Roboguru

Jumlah nilai x yang memenuhi sistem persamaan {x2+xy−y2=−4x+2y=2​ adalah ...

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus x y minus y squared end cell equals cell negative 4 space...... open parentheses 1 close parentheses end cell row cell x plus 2 y end cell equals cell 2 space space........ open parentheses 2 close parentheses end cell end table end style 

Dari persamaan (2) diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 2 y end cell equals 2 row cell 2 y end cell equals cell 2 minus x end cell row y equals cell fraction numerator 2 minus x over denominator 2 end fraction space....... space open parentheses 3 close parentheses end cell end table end style 

Substitusi persamanan (3) ke persamaan (1), diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus x y minus y squared end cell equals cell negative 4 end cell row cell x squared plus x open parentheses fraction numerator 2 minus x over denominator 2 end fraction close parentheses minus open parentheses fraction numerator 2 minus x over denominator 2 end fraction close parentheses squared end cell equals cell negative 4 end cell row cell x squared plus fraction numerator 2 x minus x squared over denominator 2 end fraction minus open parentheses fraction numerator 2 minus x over denominator 2 end fraction close parentheses squared end cell equals cell negative 4 end cell row cell x squared plus fraction numerator 2 x minus x squared over denominator 2 end fraction minus fraction numerator 4 minus 4 x plus x squared over denominator 4 end fraction end cell equals cell negative 4 end cell row cell 4 open parentheses x squared plus fraction numerator 2 x minus x squared over denominator 2 end fraction minus fraction numerator 4 minus 4 x plus x squared over denominator 4 end fraction close parentheses end cell equals cell 4 open parentheses negative 4 close parentheses end cell row cell 4 x squared plus 4 x minus 2 x squared minus 4 plus 4 x minus 4 x squared end cell equals cell negative 16 end cell row cell negative 2 x squared plus 8 x minus 4 end cell equals 0 end table end style 

Dengan menggunakan rumus jumlah akar-akar persamaan kuadrat, maka jumlah nilai undefined yang memenuhi sistem persamaan tersebut adalah 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row straight a equals cell negative 2 end cell row straight b equals 8 row cell x subscript 1 plus x subscript 2 end cell equals cell fraction numerator negative straight b over denominator straight a end fraction end cell row blank equals cell fraction numerator negative 8 over denominator negative 2 end fraction end cell row blank equals 4 end table end style 

Jadi jumlah nilai undefined yang memenuhi sistem persamaan tersebut adalah begin mathsize 14px style x subscript 1 plus x subscript 2 equals 4 end style

0

Roboguru

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