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Pertanyaan

Himpunan penyelesaian dari persamaan linear di bawah ini adalah ... .

1 over straight x plus 1 over straight y minus 1 over straight z equals 4  2 over straight x minus 3 over straight y plus 1 over straight z equals 0  1 over straight z minus 1 over straight y equals negative 2

  1. left parenthesis 2 comma straight space 1 comma straight space minus 1 right parenthesis

  2. open parentheses negative 2 comma straight space 1 comma straight space 1 close parentheses

  3. left parenthesis 1 half comma straight space 1 comma straight space minus 1 right parenthesis

  4. open parentheses negative 1 half comma straight space minus 1 comma straight space 1 close parentheses

  5. left parenthesis 1 half comma straight space 1 comma straight space 1 right parenthesis

F. Pratama

Master Teacher

Mahasiswa/Alumni Universitas Putra Indonesia YPTK Padang

Jawaban terverifikasi

Pembahasan

1 over straight x plus 1 over straight y minus 1 over straight z equals 4 horizontal ellipsis left parenthesis straight asterisk times right parenthesis  2 over straight x minus 3 over straight y plus 1 over straight z equals 0 horizontal ellipsis left parenthesis ** right parenthesis  1 over straight z minus 1 over straight y equals negative 2 horizontal ellipsis left parenthesis ** straight asterisk times right parenthesis

  • Misalkan persamaan (***) :

1 over straight z minus 1 over straight y equals negative 2  1 over straight z equals negative 2 plus 1 over straight y

  • Substitusi pada persamaan (*) :

1 over straight x plus 1 over straight y minus 1 over straight z equals 4  1 over straight x plus 1 over straight y minus open parentheses negative 2 plus 1 over straight y close parentheses equals 4  1 over straight x plus 1 over straight y plus 2 minus 1 over straight y equals 4  1 over straight x equals 2  straight x equals 1 half

  • Substitusi pada persamaan (**) :

2 over straight x minus 3 over straight y plus 1 over straight z equals 0  fraction numerator 2 over denominator 1 half end fraction minus 3 over straight y plus open parentheses negative 2 plus 1 over straight y close parentheses equals 0  4 minus 3 over straight y minus 2 plus 1 over straight y equals 0  2 equals 2 over straight y  straight y equals 2 over 2 equals 1

  • Substitusi nilai x dan y :

1 over straight z minus 1 over straight y equals negative 2  1 over straight z minus 1 over 1 equals negative 2  1 over straight z equals negative 1  straight z equals fraction numerator 1 over denominator negative 1 end fraction equals negative 1

Maka himpunan penyelesaiannya adalah  left parenthesis 1 half straight space comma straight space 1 straight space comma straight space minus 1 right parenthesis

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