Roboguru

Tentukan nilai  dari setiap pertidaksamaan berikut ini. d.   agar mempunyai penyelesaian .

Pertanyaan

Tentukan nilai a dari setiap pertidaksamaan berikut ini.

d.  1 minus fraction numerator 2 x plus a over denominator 3 end fraction greater than fraction numerator x plus 3 over denominator 2 end fraction agar mempunyai penyelesaian x less than negative 1 over 7.

Pembahasan Soal:

d.  Diketahui pertidaksamaan 1 minus fraction numerator 2 x plus a over denominator 3 end fraction greater than fraction numerator x plus 3 over denominator 2 end fraction, maka diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 minus fraction numerator 2 x plus a over denominator 3 end fraction end cell greater than cell fraction numerator x plus 3 over denominator 2 end fraction end cell row cell 6 over 6 minus fraction numerator 2 open parentheses 2 x plus a close parentheses over denominator 2 times 3 end fraction end cell greater than cell fraction numerator 3 open parentheses x plus 3 close parentheses over denominator 3 times 2 end fraction end cell row cell 6 over 6 minus fraction numerator 2 open parentheses 2 x plus a close parentheses over denominator 6 end fraction end cell greater than cell fraction numerator 3 open parentheses x plus 3 close parentheses over denominator 6 end fraction end cell row cell fraction numerator 6 minus 2 open parentheses 2 x plus a close parentheses over denominator 6 end fraction end cell greater than cell fraction numerator 3 open parentheses x plus 3 close parentheses over denominator 6 end fraction end cell row cell open parentheses fraction numerator 6 minus 2 open parentheses 2 x plus a close parentheses over denominator 6 end fraction close parentheses times 6 end cell greater than cell open parentheses fraction numerator 3 open parentheses x plus 3 close parentheses over denominator 6 end fraction close parentheses times 6 end cell row cell 6 minus 2 open parentheses 2 x plus a close parentheses end cell greater than cell 3 open parentheses x plus 3 close parentheses end cell row cell 6 minus 4 x minus 2 a end cell greater than cell 3 x plus 9 end cell row cell negative 4 x plus open parentheses 6 minus 2 a close parentheses minus open parentheses 6 minus 2 a close parentheses end cell greater than cell 3 x plus 9 minus open parentheses 6 minus 2 a close parentheses end cell row cell negative 4 x end cell greater than cell 3 x plus 9 minus 6 plus 2 a end cell row cell negative 4 x minus 3 x end cell greater than cell 3 x plus 5 plus 2 a minus 3 x end cell row cell negative 7 x end cell greater than cell 5 plus 2 a end cell row cell negative 7 x times open parentheses negative 1 over 7 close parentheses end cell less than cell open parentheses 5 plus 2 a close parentheses times open parentheses negative 1 over 7 close parentheses end cell row x less than cell negative 1 over 7 open parentheses 5 plus 2 a close parentheses end cell row blank blank blank end table

Kemudian, pertidaksamaan x less than negative 1 over 7 left parenthesis 5 plus 2 a right parenthesis agar mempunyai penyelesaian x less than negative 1 over 7, maka negative 1 over 7 open parentheses 5 plus 2 a close parentheses equals negative 1 over 7. Nilai a yang memenuhi sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell negative 1 over 7 open parentheses 5 plus 2 a close parentheses end cell equals cell negative 1 over 7 end cell row cell negative 1 over 7 open parentheses 5 plus 2 a close parentheses times open parentheses negative 7 close parentheses end cell equals cell negative 1 over 7 times open parentheses negative 7 close parentheses end cell row cell 5 plus 2 a end cell equals 1 row cell 5 plus 2 a minus 5 end cell equals cell 1 minus 5 end cell row cell 2 a end cell equals cell negative 4 end cell row cell 2 a times 1 half end cell equals cell negative 4 times 1 half end cell row a equals cell negative 2 end cell end table


Dengan demikian, nilai a yang memenuhi pertidaksamaan 1 minus fraction numerator 2 x plus a over denominator 3 end fraction greater than fraction numerator x plus 3 over denominator 2 end fraction agar mempunyai penyelesaian x less than negative 1 over 7 adalah a equals negative 2.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 10 Juni 2021

Ruangguru

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved