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Pertanyaan

Tentukan nilai x dan y dari persamaan linier 2  variabel dengan menggunakan:

a. matriks

b. aturan cramer

c. eliminasi

d. subsitusi

e. gabungan

 open curly brackets table row cell 2 x minus 5 y equals negative 2 end cell row cell negative 3 x plus 4 y equals negative 4 end cell end table close

M. Nasrullah

Master Teacher

Mahasiswa/Alumni Universitas Negeri Makassar

Jawaban terverifikasi

Pembahasan

Perhatikan perhitungan berikut:

-bentuk matriks:

A X equals B rightwards arrow X equals A to the power of negative 1 end exponent times B 

Pada soal diketahui:

open curly brackets table row cell 2 x minus 5 y equals negative 2 end cell row cell negative 3 x plus 4 y equals negative 4 end cell end table close rightwards arrow open parentheses table row 2 cell negative 5 end cell row cell negative 3 end cell 4 end table close parentheses open parentheses table row x row y end table close parentheses equals open parentheses table row cell negative 2 end cell row cell negative 4 end cell end table close parentheses

sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row x row y end table close parentheses end cell equals cell fraction numerator 1 over denominator 8 minus 15 end fraction open parentheses table row 4 5 row 3 2 end table close parentheses open parentheses table row cell negative 2 end cell row cell negative 4 end cell end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator negative 7 end fraction open parentheses table row 4 5 row 3 2 end table close parentheses open parentheses table row cell negative 2 end cell row cell negative 4 end cell end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator negative 14 end fraction open parentheses table row cell negative 8 minus 20 end cell row cell negative 6 minus 8 end cell end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator negative 7 end fraction open parentheses table row cell negative 28 end cell row cell negative 14 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell bevelled fraction numerator negative 28 over denominator negative 7 end fraction end cell row cell bevelled fraction numerator negative 14 over denominator negative 7 end fraction end cell end table close parentheses end cell row blank equals cell open parentheses table row 4 row 2 end table close parentheses end cell end table 

Jadi, nilai x dan y masing-masing adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table dan table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table .

b. aturan cramer

open curly brackets table row cell 2 x minus 5 y equals negative 2 end cell row cell negative 3 x plus 4 y equals negative 4 end cell end table close rightwards arrow open parentheses table row 2 cell negative 5 end cell row cell negative 3 end cell 4 end table close parentheses open parentheses table row x row y end table close parentheses equals open parentheses table row cell negative 2 end cell row cell negative 4 end cell end table close parentheses
table attributes columnalign right center left columnspacing 0px end attributes row D equals cell open vertical bar table row 2 cell negative 5 end cell row cell negative 3 end cell 4 end table close vertical bar end cell row blank equals cell 8 minus 15 end cell row blank equals cell negative 7 end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell D subscript x end cell equals cell open vertical bar table row cell negative 2 end cell cell negative 5 end cell row cell negative 4 end cell 4 end table close vertical bar end cell row blank equals cell negative 8 minus 20 end cell row blank equals cell negative 28 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell D subscript y end cell equals cell open vertical bar table row 2 cell negative 2 end cell row cell negative 3 end cell cell negative 4 end cell end table close vertical bar end cell row blank equals cell negative 8 minus 6 end cell row blank equals cell negative 14 end cell end table

sehingga,

x equals D subscript x over D equals fraction numerator negative 28 over denominator negative 7 end fraction equals 4 y equals D subscript y over D equals fraction numerator negative 14 over denominator negative 7 end fraction equals 2 

Jadi, nilai x dan y masing-masing adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table dan table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table.

c. eliminasi

open curly brackets table row cell 2 x minus 5 y equals negative 2 end cell row cell negative 3 x plus 4 y equals negative 4 end cell end table close

Langsung kita eliminasi  x pada kedua persamaan tersebut:

 table row cell 2 x minus 5 y equals negative 2 end cell cell open vertical bar cross times 3 close vertical bar end cell row cell negative 3 x plus 4 y equals negative 4 end cell cell open vertical bar cross times 2 close vertical bar end cell row blank blank row blank blank row blank blank row blank blank end table table row cell 6 x minus 15 y equals negative 6 end cell row cell negative 6 x plus 8 y equals negative 8 end cell row cell negative 7 y equals negative 14 end cell row cell y equals fraction numerator negative 14 over denominator negative 7 end fraction end cell row cell y equals 2 end cell end table plus 

eliminasi  y pada kedua persamaan tersebut:

table row cell 2 x minus 5 y equals negative 2 end cell cell open vertical bar cross times 4 close vertical bar end cell row cell negative 3 x plus 4 y equals negative 4 end cell cell open vertical bar cross times 5 close vertical bar end cell row blank blank row blank blank row blank blank row blank blank end table table row cell 8 x minus 20 y equals negative 8 end cell row cell negative 15 x plus 20 y equals negative 20 end cell row cell negative 7 y equals negative 28 end cell row cell y equals fraction numerator negative 28 over denominator negative 7 end fraction end cell row cell y equals 4 end cell end table plus

Jadi, nilai x dan y masing-masing adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table dan table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table .

d. subsitusi

table row cell 2 x minus 5 y equals negative 2 end cell row cell 2 x equals negative 2 plus 5 y end cell row cell x equals fraction numerator negative 2 plus 5 y over denominator 2 end fraction.... open parentheses 1 close parentheses end cell row blank row cell negative 3 x plus 4 y equals negative 4............ open parentheses 2 close parentheses end cell end table

Subtitusi persamaan 1 ke persamaan 2

table attributes columnalign right center left columnspacing 0px end attributes row cell negative 3 open parentheses fraction numerator negative 2 plus 5 y over denominator 2 end fraction close parentheses plus 4 y end cell equals cell negative 4 end cell row cell negative 3 open parentheses negative 2 plus 5 y close parentheses plus 8 y end cell equals cell negative 8 end cell row cell 6 minus 15 y plus 8 y end cell equals cell negative 8 end cell row cell negative 7 y end cell equals cell negative 8 minus 6 end cell row y equals cell fraction numerator negative 14 over denominator negative 7 end fraction end cell row y equals 2 end table

Subtitusi nilai y ke persamaan 1

 table attributes columnalign right center left columnspacing 0px end attributes row x equals cell fraction numerator negative 2 plus 5 y over denominator 2 end fraction end cell row x equals cell fraction numerator negative 2 plus 5 open parentheses 2 close parentheses over denominator 2 end fraction end cell row x equals cell fraction numerator negative 2 plus 10 over denominator 2 end fraction end cell row x equals cell 8 over 2 end cell row x equals 4 end table

Jadi, nilai x dan y masing-masing adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table dan table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table.

e. gabungan (eliminasi-subtitusi)

Eliminasi  x pada kedua persamaan tersebut:

table row cell 2 x minus 5 y equals negative 2 end cell cell open vertical bar cross times 3 close vertical bar end cell row cell negative 3 x plus 4 y equals negative 4 end cell cell open vertical bar cross times 2 close vertical bar end cell row blank blank row blank blank row blank blank row blank blank end table table row cell 6 x minus 15 y equals negative 6 end cell row cell negative 6 x plus 8 y equals negative 8 end cell row cell negative 7 y equals negative 14 end cell row cell y equals fraction numerator negative 14 over denominator negative 7 end fraction end cell row cell y equals 2 end cell end table plus

Subtitusi nilai y ke persamaan 1

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus 5 y end cell equals cell negative 2 end cell row cell 2 x minus 5 open parentheses 2 close parentheses end cell equals cell negative 2 end cell row cell 2 x end cell equals cell negative 2 plus 10 end cell row cell 2 x end cell equals 8 row x equals cell 8 over 2 end cell row x equals 4 end table

Dengan demikian, nilai x dan y masing-masing adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table dan table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table.

 

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