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Himpunan penyelesaian dari sistem persamaan linear tiga variabel adalah . Nilai ....

Pertanyaan

Himpunan penyelesaian dari sistem persamaan linear tiga variabel

open curly brackets table attributes columnalign left end attributes row cell size 14px x size 14px plus size 14px y size 14px minus size 14px z size 14px equals size 14px 24 end cell row cell size 14px 2 size 14px x size 14px minus size 14px y size 14px plus size 14px 2 size 14px z size 14px equals size 14px 4 end cell row cell size 14px x size 14px plus size 14px 2 size 14px y size 14px minus size 14px 3 size 14px z size 14px equals size 14px 36 end cell end table close

adalah undefined. Nilai size 14px x size 14px space size 14px colon size 14px space size 14px y size 14px space size 14px colon size 14px space size 14px z size 14px equals....

  1. size 14px 2 size 14px space size 14px colon size 14px space size 14px 7 size 14px space size 14px colon size 14px space size 14px 1

  2. size 14px 2 size 14px space size 14px colon size 14px space size 14px 5 size 14px space size 14px colon size 14px space size 14px 4

  3. size 14px 2 size 14px space size 14px colon size 14px space size 14px 5 size 14px space size 14px colon size 14px space size 14px 1

  4. size 14px 1 size 14px space size 14px colon size 14px space size 14px 5 size 14px space size 14px colon size 14px space size 14px 2

  5. size 14px 1 size 14px space size 14px colon size 14px space size 14px 2 size 14px space size 14px colon size 14px space size 14px 5 size 14px space 

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell Eliminasi space 1 end cell row blank blank cell fraction numerator x plus y minus z equals 24 2 x minus y plus 2 z equals 4 over denominator 3 x plus z equals 28 end fraction plus end cell row blank blank blank row blank blank cell Eliminasi space 2 end cell row cell 2 x minus y plus 2 z end cell equals cell 4 space space space space k a l i k a n space 2 end cell row cell x plus 2 y minus 3 z end cell equals 36 row blank blank blank row blank blank cell fraction numerator 4 x minus 2 y plus 4 z equals 8 x plus 2 y minus 3 z equals 36 over denominator 5 x plus z equals 44 end fraction plus end cell row blank blank blank row blank blank cell Eliminasi space 3 end cell row blank blank cell fraction numerator 3 x plus z equals 28 5 x plus z equals 44 over denominator negative 2 x equals negative 16 x equals 8 end fraction minus end cell row blank blank blank row blank blank cell Subtitusi space 1 end cell row cell 3 x plus z end cell equals 28 row cell 3 left parenthesis 8 right parenthesis plus z end cell equals 28 row z equals 4 row blank blank blank row blank blank cell Subtitusi space 2 end cell row cell x plus y minus z end cell equals 24 row cell 8 plus y minus 4 end cell equals 24 row y equals 20 row blank blank blank row blank blank cell Perbandingan space x comma y comma z end cell row blank blank cell x space space colon space space y space space colon space space z end cell row blank blank cell 8 space space colon space 20 space colon space space 4 end cell row blank blank cell 2 space space colon space space 5 space space colon space space 1 end cell end table end style   

 

Jadi jawaban yang tepat adalah C

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 07 Februari 2021

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