Tentukan nilai yang memenuhi persamaan berikut :

Pertanyaan

Tentukan nilai $x$ yang memenuhi persamaan berikut :

$open vertical bar x minus 3 close vertical bar minus open vertical bar 2 x minus 8 close vertical bar equals 5$

Pembahasan Soal:

Definisi dari nilai mutlak adalah :

$open vertical bar x close vertical bar equals open curly brackets table row cell negative x comma space x less than 0 end cell row cell x comma space x greater or equal than 0 end cell end table close$

Dari soal diperoleh persamaan berikut :

$open vertical bar x minus 3 close vertical bar minus open vertical bar 2 x minus 8 close vertical bar equals 5$

Berdasarkan definisi nilai mutlak :

$open vertical bar x minus 3 close vertical bar equals open curly brackets table row cell negative open parentheses x minus 3 close parentheses comma space x minus 3 less than 0 end cell row cell x minus 3 comma space x minus 3 greater or equal than 0 end cell end table close$

$open vertical bar x minus 3 close vertical bar equals open curly brackets table row cell 3 minus x comma space x less than 3 end cell row cell x minus 3 comma space x greater or equal than 3 end cell end table close$

dan

$open vertical bar 2 x minus 8 close vertical bar equals open curly brackets table row cell negative open parentheses 2 x minus 8 close parentheses comma space 2 x minus 8 less than 0 end cell row cell 2 x minus 8 comma space 2 x minus 8 greater or equal than 0 end cell end table close$

$open vertical bar 2 x minus 8 close vertical bar equals open curly brackets table row cell 8 minus 2 x comma space x less than 4 end cell row cell 2 x minus 8 comma space x greater or equal than 4 end cell end table close$

Berdasarkan batas-batas nilai $x$ diatas, kita gunakan garis bilangan di bawah ini :

Berdasarkan garis bilangan diatas, kita bagi menjadi 3 daerah yaitu :

Jika $x less than 3$ , maka :

$table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x minus 3 close vertical bar minus open vertical bar 2 x minus 8 close vertical bar end cell equals 5 row cell negative left parenthesis x minus 3 right parenthesis plus 2 x minus 8 end cell equals 5 row cell negative x plus 3 plus 2 x minus 8 end cell equals 5 row x equals 10 end table$

Karena syarat yang harus terpenuhi adalah $x less than 3$ maka nilai $x equals 10$ tidak memenuhi.

Jika $3 less or equal than x less or equal than 4$ , maka :

$table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x minus 3 close vertical bar minus open vertical bar 2 x minus 8 close vertical bar end cell equals 5 row cell x minus 3 plus 2 x minus 8 end cell equals 5 row cell 3 x end cell equals 16 row x equals cell 16 over 3 end cell end table$

Karena syarat yang harus terpenuhi adalah $3 less or equal than x less or equal than 4$ maka nilai $x equals 16 over 3$ tidak memenuhi.

Jika $x greater than 3$ , maka :

$table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x minus 3 close vertical bar minus open vertical bar 2 x minus 8 close vertical bar end cell equals 5 row cell x minus 3 minus left parenthesis 2 x minus 8 right parenthesis end cell equals 5 row cell x minus 3 minus 2 x plus 8 end cell equals 5 row x equals 0 end table$

Karena syarat yang harus terpenuhi adalah $x greater than 3$ maka nilai $x equals 0$ tidak memenuhi.

Dengan demikian, tidak ada nilai $x$ yang memenuhi.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Gandhi

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 12 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Hitunglah nilai yang memenuhi persamaan berikut.

Pembahasan Soal:

Ingat definisi nilai mutlak:

$open vertical bar x close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell x space space space space space space comma space x greater or equal than 0 end cell row cell negative x space space space comma space x less than 0 end cell end table close$

Maka dapat kita peroleh:

$open vertical bar 2 x minus 1 close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell 2 x minus 1 space space space space space space space space comma space x greater or equal than 1 half end cell row cell negative left parenthesis 2 x minus 1 right parenthesis space space comma space x less than 1 half end cell end table close$

dan

$open vertical bar 3 x minus 2 close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell 3 x minus 2 space space space space space space space space comma space x greater or equal than 2 over 3 end cell row cell negative left parenthesis 3 x minus 2 right parenthesis space space comma space x less than 2 over 3 end cell end table close$

Berdasarkan definisi nilai mutlak tersebut, maka terdapat tiga kemungkinan kasus penyelesaian yaitu:

1. Untuk $x less than 1 half$ :

$table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x minus 1 close vertical bar plus open vertical bar 3 x minus 2 close vertical bar end cell equals 5 row cell negative left parenthesis 2 x minus 1 right parenthesis plus left parenthesis negative left parenthesis 3 x minus 2 right parenthesis right parenthesis end cell equals 5 row cell negative 2 x plus 1 minus 3 x plus 2 end cell equals 5 row cell negative 5 x plus 3 end cell equals 5 row cell negative 5 x end cell equals 2 row x equals cell negative 2 over 5 end cell end table$

Diperoleh $x equals negative 2 over 5$ , artinya bahwa $x less than 1 half$ (memenuhi). Sehingga $x equals negative 2 over 5$ merupakan solusi.

2. Untuk $1 half less or equal than x less than 2 over 3$ :

$table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x minus 1 close vertical bar plus open vertical bar 3 x minus 2 close vertical bar end cell equals 5 row cell 2 x minus 1 plus left parenthesis negative left parenthesis 3 x minus 2 right parenthesis right parenthesis end cell equals 5 row cell 2 x minus 1 minus 3 x plus 2 end cell equals 5 row cell negative x plus 1 end cell equals 5 row cell negative x end cell equals 4 row x equals cell negative 4 end cell end table$

Nilai $x equals negative 4$ tidak memenuhi batasan $1 half less or equal than x less than 2 over 3$ , maka $x equals negative 4$ bukan merupakan solusi.

3. Untuk $x greater or equal than 2 over 3$ :

$table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x minus 1 close vertical bar plus open vertical bar 3 x minus 2 close vertical bar end cell equals 5 row cell 2 x minus 1 plus 3 x minus 2 end cell equals 5 row cell 5 x minus 3 end cell equals 5 row cell 5 x end cell equals 2 row x equals cell 2 over 5 end cell end table$

Nilai $x equals 2 over 5$ tidak memenuhi batasan $x greater or equal than 2 over 3$ , maka $x equals 2 over 5$ bukan merupakan solusi.

Dengan demikian, nilai $x$ yang memenuhi persamaan $open vertical bar 2 x minus 1 close vertical bar plus open vertical bar 3 x minus 2 close vertical bar equals 5$ adalah $x equals negative 2 over 5$ .

Roboguru

Tentukan himpunan penyelesaiannya!

Pembahasan Soal:

Ingat kembali definisi nilai mutlak untuk setiap bilangan real $x$ adalah sebagai berikut.

$open vertical bar f open parentheses x close parentheses close vertical bar equals open curly brackets table attributes columnalign center left end attributes row cell plus f open parentheses x close parentheses comma end cell cell untuk space f open parentheses x close parentheses greater or equal than 0 end cell row cell negative f open parentheses x close parentheses comma end cell cell untuk space f open parentheses x close parentheses less than 0 end cell end table close$

Akan ditentukan himpunan penyelesaian dari $open vertical bar 3 x plus 1 close vertical bar plus open vertical bar 2 x minus 3 close vertical bar equals 5$ .

Terlebih dahulu jabarkan $open vertical bar 3 x plus 1 close vertical bar plus open vertical bar 2 x minus 3 close vertical bar$ menggunakan definisi nilai mutlak diperoleh.

$begin mathsize 12px style table attributes columnalign right center left columnspacing 2px end attributes row blank blank cell open vertical bar 3 x plus 1 close vertical bar plus open vertical bar 2 x minus 3 close vertical bar end cell row blank equals cell open curly brackets table attributes columnalign center left end attributes row cell open parentheses 3 x plus 1 close parentheses plus open parentheses 2 x minus 3 close parentheses comma space end cell cell untuk space 3 x plus 1 greater or equal than 0 space dan space 2 x minus 3 greater or equal than 0 end cell row cell open parentheses 3 x plus 1 close parentheses minus open parentheses 2 x minus 3 close parentheses comma end cell cell untuk space 3 x plus 1 greater or equal than 0 space dan space 2 x minus 3 less than 0 end cell row cell negative open parentheses 3 x plus 1 close parentheses plus open parentheses 2 x minus 3 close parentheses comma end cell cell untuk space 3 x plus 1 less than 0 space dan space 2 x minus 3 greater or equal than 0 end cell row cell negative open parentheses 3 x plus 1 close parentheses minus open parentheses 2 x minus 3 close parentheses comma end cell cell untuk space 3 x plus 1 less than 0 space dan space 2 x minus 3 less than 0 end cell end table close end cell row blank equals cell open curly brackets table attributes columnalign center left end attributes row cell 3 x plus 1 plus 2 x minus 3 comma space end cell cell untuk space x greater or equal than negative 1 third space dan space x greater or equal than 3 over 2 end cell row cell 3 x plus 1 minus 2 x plus 3 comma end cell cell untuk space x greater or equal than negative 1 third space dan space x less than 3 over 2 end cell row cell negative 3 x minus 1 plus 2 x minus 3 comma end cell cell untuk space x less than negative 1 third space dan space x greater or equal than 3 over 2 end cell row cell negative 3 x minus 1 minus 2 x plus 3 comma end cell cell untuk space x less than negative 1 third space dan space x less than 3 over 2 end cell end table close end cell row blank equals cell open curly brackets table attributes columnalign center left end attributes row cell 5 x minus 2 comma space end cell cell untuk space x greater or equal than 3 over 2 end cell row cell x plus 4 comma end cell cell untuk space minus 1 third less or equal than x less than 3 over 2 end cell row cell negative x minus 4 comma end cell cell untuk space x less than negative 1 third space dan space x greater or equal than 3 over 2 space left parenthesis tidak space memenuhi right parenthesis end cell row cell negative 5 x plus 2 comma end cell cell untuk space x less than negative 1 third end cell end table close end cell end table end style$

Diperoleh yang memenuhi syarat adalah $5 x minus 2$ untuk $x greater or equal than 3 over 2$ , $x plus 4$ untuk $negative 1 third less or equal than x less than 3 over 2$ , dan $negative 5 x plus 2$ untuk $x less than negative 1 third$ .

Nilai $x$ pada $5 x minus 2$ untuk $x greater or equal than 3 over 2$ .

$table attributes columnalign right center left columnspacing 2px end attributes row cell open vertical bar 3 x plus 1 close vertical bar plus open vertical bar 2 x minus 3 close vertical bar end cell equals 5 row cell 5 x minus 2 end cell equals 5 row cell 5 x end cell equals cell 5 plus 2 end cell row cell 5 x end cell equals 7 row x equals cell 7 over 5 end cell end table$

Diperoleh nilai $x equals 7 over 5$ , tetapi tidak memenuhi syarat $x greater or equal than 3 over 2$ , karena $7 over 5 less than 3 over 2$ .

Nilai $x$ pada $x plus 4$ untuk $negative 1 third less or equal than x less than 3 over 2$ .

Diperoleh nilai $x equals 1$ memenuhi syarat $negative 1 third less or equal than x less than 3 over 2$ .

Nilai $x$ pada $negative 5 x plus 2$ untuk $x less than negative 1 third$ .

$table attributes columnalign right center left columnspacing 2px end attributes row cell open vertical bar 3 x plus 1 close vertical bar plus open vertical bar 2 x minus 3 close vertical bar end cell equals 5 row cell negative 5 x plus 2 end cell equals 5 row cell negative 5 x end cell equals cell 5 minus 2 end cell row cell negative 5 x end cell equals 3 row x equals cell fraction numerator 3 over denominator negative 5 end fraction end cell row x equals cell negative 3 over 5 end cell end table$

Diperoleh nilai $x equals negative 3 over 5$ , memenuhi syarat $x less than negative 1 third$ .

Dengan demikian, diperoleh himpunan penyelesaiannya adalah $open curly brackets negative 3 over 5 comma space 1 close curly brackets$ .

Roboguru

Nilai yang memenuhi persamaan adalah ...

Pembahasan Soal:

Ingat!

Definisi dari suatu nilai mutlak adalah:

$open vertical bar x close vertical bar open curly brackets table row cell x comma space x greater or equal than 0 space end cell row blank row cell negative x comma space x less than 0 end cell end table close$

Maka:

$open vertical bar 2 x plus 4 close vertical bar open curly brackets table row cell 2 x plus 4 comma space 2 x plus 4 greater or equal than 0 end cell row cell 2 x greater or equal than negative 4 end cell row cell x greater or equal than negative 2 end cell row blank row cell negative left parenthesis 2 x plus 4 right parenthesis comma space 2 x plus 4 less than 0 end cell row cell 2 x less than negative 4 end cell row cell x less than negative 2 end cell end table close$

$open vertical bar 3 minus x close vertical bar open curly brackets table row cell 3 minus x comma space 3 minus x greater or equal than 0 end cell row cell negative x greater or equal than negative 3 end cell row cell x less or equal than 3 end cell row blank row cell negative left parenthesis 3 minus x right parenthesis comma space 3 minus x less than 0 end cell row cell negative x less than negative 3 end cell row cell x greater than 3 end cell end table close$

Sehingga daerah penyelesaian terbadi menjadi 3 yaitu:

Untuk setiap interval, maka:

Interval $x less than negative 2$

$table attributes columnalign right center left columnspacing 0px end attributes row cell negative open parentheses 2 x plus 4 close parentheses minus left parenthesis 3 minus x right parenthesis end cell equals cell negative 1 end cell row cell negative 2 x minus 4 minus 3 plus x end cell equals cell negative 1 end cell row cell negative x minus 7 end cell equals cell negative 1 end cell row cell negative x end cell equals cell negative 1 plus 7 end cell row cell negative x end cell equals 6 row x equals cell negative 6 end cell end table$

Karena $x equals negative 6$ berada pada interval $x less than negative 2$ , maka $x equals negative 6$ adalah penyelesaian.

Interval $negative 2 less or equal than x less or equal than 3$

$table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x plus 4 minus left parenthesis 3 minus x right parenthesis end cell equals cell negative 1 end cell row cell 2 x plus 4 minus 3 plus x end cell equals cell negative 1 end cell row cell 3 x plus 1 end cell equals cell negative 1 end cell row cell 3 x end cell equals cell negative 1 minus 1 end cell row cell 3 x end cell equals cell negative 2 end cell row x equals cell negative 2 over 3 end cell end table$

Karena $table attributes columnalign right center left columnspacing 0px end attributes row x equals cell negative 2 over 3 end cell end table$ berada pada interval $negative 2 less or equal than x less or equal than 3$ , maka $table attributes columnalign right center left columnspacing 0px end attributes row x equals cell negative 2 over 3 end cell end table$ merupakan penyelesaian.

Interval $x greater than 3$

$table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x plus 4 minus left parenthesis negative left parenthesis 3 minus x right parenthesis right parenthesis end cell equals cell negative 1 end cell row cell 2 x plus 4 plus 3 minus x end cell equals cell negative 1 end cell row cell x plus 7 end cell equals cell negative 1 end cell row x equals cell negative 1 minus 7 end cell row x equals cell negative 8 end cell end table$

Karena $x equals negative 8$ tidak berada pada interval $x greater than 3$ , maka $x equals negative 8$ bukan penyelesaian.

Dengan demikian, nilai $x$ yang memenuhi persamaan tersebut adalah $x equals negative 6 space atau space x equals negative 2 over 3$ .

Roboguru

Tentukan himpunan penyelesaian dari persamaan berikut!

Pembahasan Soal:

Dengan menggunakan definisi nilai mutlak diperoleh:

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x plus 4 close vertical bar end cell equals cell open curly brackets table attributes columnalign left end attributes row cell 2 x plus 4 comma space untuk space 2 x plus 4 greater or equal than 0 space left right double arrow x greater or equal than negative 2 end cell row cell negative open parentheses 2 x plus 4 close parentheses comma space untuk space 2 x plus 4 less than 0 space left right double arrow x less than negative 2 end cell end table close end cell row cell open vertical bar 3 minus x close vertical bar end cell equals cell open curly brackets table attributes columnalign left end attributes row cell 3 minus x comma space untuk space 3 minus x greater or equal than 0 space left right double arrow x less or equal than 3 end cell row cell negative open parentheses 3 minus x close parentheses comma space untuk space 3 minus x less than 0 space left right double arrow x greater than 3 end cell end table close end cell end table end style$

Pada interval $begin mathsize 14px style x less than negative 2 end style$ , diperoleh penyelesaian sebagai berikut.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative open parentheses 2 x plus 4 close parentheses minus open parentheses 3 minus x close parentheses end cell equals cell negative 1 end cell row cell negative 2 x minus 4 minus 3 plus x end cell equals cell negative 1 end cell row cell negative x minus 7 end cell equals cell negative 1 end cell row cell negative x end cell equals cell negative 1 plus 7 end cell row cell negative x end cell equals 6 row x equals cell negative 6 space open parentheses memenuhi close parentheses end cell end table end style$

Pada interval $begin mathsize 14px style negative 2 less or equal than x less or equal than 3 end style$ , diperoleh penyelesaian sebagai berikut.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 2 x plus 4 close parentheses minus open parentheses 3 minus x close parentheses end cell equals cell negative 1 end cell row cell 2 x plus 4 minus 3 plus x end cell equals cell negative 1 end cell row cell 3 x plus 1 end cell equals cell negative 1 end cell row cell 3 x end cell equals cell negative 1 minus 1 end cell row cell 3 x end cell equals cell negative 2 end cell row x equals cell negative 2 over 3 space open parentheses memenuhi close parentheses end cell end table end style$

Pada interval $begin mathsize 14px style x greater than 3 end style$ , diperoleh penyelesaian sebagai berikut.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 2 x plus 4 close parentheses minus open parentheses negative open parentheses 3 minus x close parentheses close parentheses end cell equals cell negative 1 end cell row cell 2 x plus 4 plus 3 minus x end cell equals cell negative 1 end cell row cell x plus 7 end cell equals cell negative 1 end cell row x equals cell negative 1 minus 7 end cell row x equals cell negative 8 space open parentheses tidak space memenuhi close parentheses end cell end table end style$

Jadi, himpunan penyelesaian dari persamaan tersebut adalah $begin mathsize 14px style HP equals open curly brackets negative 6 comma space minus 2 over 3 close curly brackets end style$ .

Roboguru

Jika penyelesaian adalah dan dengan , nilai dari adalah ...

Pembahasan Soal:

$open vertical bar x minus 3 close vertical bar equals open vertical bar 4 minus 3 x close vertical bar plus 1$

Menurut definisi

$open vertical bar x minus 3 close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell x minus 3 comma space x greater or equal than 3 end cell row cell negative x plus 3 comma x less than 3 end cell end table close$

$open vertical bar 4 minus 3 x close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell 4 minus 3 x comma space x greater or equal than 4 over 3 end cell row cell negative 4 plus 3 x comma x less than 4 over 3 end cell end table close$

Sehingga diperoleh

Untuk $x less than 4 over 3$ , maka

$table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x minus 3 close vertical bar end cell equals cell open vertical bar 4 minus 3 x close vertical bar plus 1 end cell row cell negative x plus 3 end cell equals cell negative 4 plus 3 x plus 1 end cell row cell negative x plus 3 x end cell equals cell 3 minus 4 plus 1 end cell row cell 2 x end cell equals 0 row x equals 0 end table$

Untuk $4 over 3 less or equal than x less than 3$ , maka

$table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x minus 3 close vertical bar end cell equals cell open vertical bar 4 minus 3 x close vertical bar plus 1 end cell row cell negative x plus 3 end cell equals cell 4 minus 3 x plus 1 end cell row cell 3 x minus x end cell equals cell 4 plus 1 minus 3 end cell row cell 2 x end cell equals 2 row x equals cell 1 open parentheses tidak space memenuhi close parentheses end cell end table$

Untuk $x greater or equal than 3$

$table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x minus 3 close vertical bar end cell equals cell open vertical bar 4 minus 3 x close vertical bar plus 1 end cell row cell x minus 3 end cell equals cell 4 minus 3 x plus 1 end cell row cell 3 x plus x end cell equals cell 4 plus 3 plus 1 end cell row cell 4 x end cell equals 8 row x equals 2 end table$

Sehingga diperoleh $HP equals open curly brackets 0 comma 2 close curly brackets$ , dengan $x subscript 1 equals 0$ dan $x subscript 2 equals 4$

Nilai dari $open parentheses x subscript 1 plus 2 x subscript 2 close parentheses$

$table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 plus 2 x subscript 2 end cell equals cell 0 plus 2 times 2 end cell row blank equals cell 0 plus 4 end cell row blank equals 4 end table$

Jadi, jawaban yang tepat adalah C

Roboguru