Pertanyaan

Carilah himpunan penyelesaian dari setiap pertidaksamaan berikut. b. 2 − 2 ( 5 − 6 x ) + 3 ≤ − 2 7 + 14 x

Carilah himpunan penyelesaian dari setiap pertidaksamaan berikut.

b. $\frac{- 2 ( 5 - 6 x )}{2} + 3 \leq - \frac{7}{2} + 14 x$

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

himpunan penyelesaian dari adalah .

himpunan penyelesaian dari $fraction numerator negative 2 open parentheses 5 minus 6 x close parentheses over denominator 2 end fraction plus 3 less or equal than negative 7 over 2 plus 14 x$ adalah open curly brackets x vertical line space x greater or equal than 3 over 16 close curly brackets

open curly brackets x vertical line space x greater or equal than 3 over 16 close curly brackets

Pembahasan

b. Dengan menggunakan sifat-sifat dasar pertidaksamaan, penyelesaian dari pertidaksamaan di atas sebagai berikut. Dengan demikian, himpunan penyelesaian dari adalah .

b. Dengan menggunakan sifat-sifat dasar pertidaksamaan, penyelesaian dari pertidaksamaan di atas sebagai berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator negative 2 open parentheses 5 minus 6 x close parentheses over denominator 2 end fraction plus 3 end cell less or equal than cell negative 7 over 2 plus 14 x end cell row cell fraction numerator negative 10 plus 12 x over denominator 2 end fraction plus 3 end cell less or equal than cell negative 7 over 2 plus 14 x end cell row cell fraction numerator negative 10 plus 12 x over denominator 2 end fraction plus 3 minus 3 end cell less or equal than cell negative 7 over 2 plus 14 x minus 3 end cell row cell fraction numerator negative 10 plus 12 x over denominator 2 end fraction end cell less or equal than cell negative 7 over 2 minus 6 over 2 plus 14 x end cell row cell fraction numerator negative 10 plus 12 x over denominator 2 end fraction end cell less or equal than cell negative 13 over 2 plus 14 x end cell row cell fraction numerator negative 10 plus 12 x over denominator 2 end fraction minus 14 x end cell less or equal than cell negative 13 over 2 plus 14 x minus 14 x end cell row cell fraction numerator open parentheses negative 10 plus 12 x close parentheses over denominator 2 end fraction minus fraction numerator 14 x times 2 over denominator 2 end fraction end cell less or equal than cell negative 13 over 2 end cell row cell fraction numerator negative 10 plus 12 x minus 28 x over denominator 2 end fraction end cell less or equal than cell negative 13 over 2 end cell row cell fraction numerator negative 10 minus 16 x over denominator 2 end fraction end cell less or equal than cell negative 13 over 2 end cell row cell fraction numerator negative 10 minus 16 x over denominator 2 end fraction times 2 end cell less or equal than cell negative 13 over 2 times 2 end cell row cell negative 10 minus 16 x end cell less or equal than cell negative 13 end cell row cell negative 10 minus 16 x plus 10 end cell less or equal than cell negative 13 plus 10 end cell row cell negative 16 x end cell less or equal than cell negative 3 end cell row cell negative 16 x times open parentheses negative 1 over 16 close parentheses end cell greater or equal than cell negative 3 times open parentheses negative 1 over 16 close parentheses end cell row x greater or equal than cell 3 over 16 end cell end table$

Dengan demikian, himpunan penyelesaian dari $fraction numerator negative 2 open parentheses 5 minus 6 x close parentheses over denominator 2 end fraction plus 3 less or equal than negative 7 over 2 plus 14 x$ adalah open curly brackets x vertical line space x greater or equal than 3 over 16 close curly brackets .