Pertanyaan

Tentukan himpunan penyelesaiannya! 4. ∣ 3 x + 1 ∣ + ∣ 2 x − 3 ∣ = 5

Tentukan himpunan penyelesaiannya!

4. $∣ 3 x + 1 ∣ + ∣ 2 x - 3 ∣ = 5$

G. Albiah

Master Teacher

Mahasiswa/Alumni Universitas Galuh Ciamis

Jawaban terverifikasi

Jawaban

himpunan penyelesaian dari adalah dan .

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

adalah

dan

Pembahasan

Ingat aturan nilai mutlak! Maka, a. b. Di dapatkan . Untuk Di dapatkan dan . Untuk Di dapatkan dan . Jadi, himpunan penyelesaian dari adalah dan .

Ingat aturan nilai mutlak!

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

Maka,

a. open vertical bar 3 x plus 1 close vertical bar

open vertical bar 3 x plus 1 close vertical bar equals 3 x plus 1

$table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x plus 1 end cell greater or equal than cell thin space 0 end cell row cell 3 x plus 1 minus 1 end cell greater or equal than cell thin space 0 minus 1 end cell row cell 3 x end cell greater or equal than cell thin space minus 1 end cell row cell fraction numerator 3 x over denominator 3 end fraction end cell greater or equal than cell fraction numerator negative 1 over denominator 3 end fraction end cell row x greater or equal than cell thin space minus 1 third end cell end table$

open vertical bar 3 x plus 1 close vertical bar equals negative open parentheses 3 x plus 1 close parentheses

$table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x plus 1 end cell less than 0 row cell 3 x plus 1 minus 1 end cell less than cell 0 minus 1 end cell row cell 3 x end cell less than cell negative 1 end cell row cell fraction numerator 3 x over denominator 3 end fraction end cell less than cell fraction numerator negative 1 over denominator 3 end fraction end cell row x less than cell negative 1 third end cell end table$

b. open vertical bar 2 x minus 3 close vertical bar

open vertical bar 2 x minus 3 close vertical bar equals 2 x minus 3

$table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus 3 end cell greater or equal than cell thin space 0 end cell row cell 2 x minus 3 plus 3 end cell greater or equal than cell thin space 0 plus 3 end cell row cell 2 x end cell greater or equal than cell thin space 3 end cell row cell fraction numerator 2 x over denominator 2 end fraction end cell greater or equal than cell 3 over 2 end cell row x greater or equal than cell 3 over 2 end cell end table$

open vertical bar 2 x minus 3 close vertical bar equals negative open parentheses 2 x minus 3 close parentheses

$table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus 3 end cell less than 0 row cell 2 x minus 3 plus 3 end cell less than cell 0 plus 3 end cell row cell 2 x end cell less than 3 row cell fraction numerator 2 x over denominator 2 end fraction end cell less than cell 3 over 2 end cell row x less than cell 3 over 2 end cell end table$

Di dapatkan x less than negative 1 third comma thin space minus 1 third less or equal than thin space x less than 3 over 2 comma thin space x greater or equal than 3 over 2 .

Untuk x less than negative 1 third

$table attributes columnalign right center left columnspacing 0px end attributes row cell negative open parentheses 3 x plus 1 close parentheses minus open parentheses 2 x minus 3 close parentheses end cell equals 5 row cell negative 3 x minus 1 minus 2 x plus 3 end cell equals 5 row cell negative 5 x plus 2 end cell equals 5 row cell negative 5 x plus 2 minus 2 end cell equals cell 5 minus 2 end cell row cell negative 5 x end cell equals 3 row cell fraction numerator negative 5 x over denominator negative 5 end fraction end cell equals cell fraction numerator 3 over denominator negative 5 end fraction end cell row x equals cell negative 3 over 5 end cell row blank blank blank end table$

Di dapatkan x equals negative 3 over 5 dan x less than negative 1 third .

Untuk negative 1 third less or equal than thin space x less than 3 over 2

Di dapatkan x equals 1 dan negative 1 third less or equal than thin space x less than 3 over 2 .

Jadi, 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 adalah x equals 1 dan x equals negative 3 over 5 .