Pertanyaan

Solve the following sistem for a and b . b. { 2 a 1 − 2 b 1 = − 10 a 2 + b 3 = 5

Solve the following sistem for $a and b$ .

b. ${\frac{1}{2 a} - \frac{1}{2 b} = - 10 \frac{2}{a} + \frac{3}{b} = 5$

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

Klaim

S. Yoga

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Jawaban

penyelesaian dari setiap SPLDV tersebut adalah .

penyelesaian dari setiap SPLDV tersebut adalah open parentheses negative 1 over 11 comma thin space 1 over 9 close parentheses

Pembahasan

Diketahui: Mencari nilai dengan eliminasi : Substitusi ke persamaan pertama: Jadi,penyelesaian dari setiap SPLDV tersebut adalah .

Diketahui:

$open curly brackets table attributes columnalign left end attributes row cell fraction numerator 1 over denominator 2 a end fraction minus fraction numerator 1 over denominator 2 b end fraction equals negative 10 end cell row cell 2 over a plus 3 over b equals 5 end cell end table close$

Mencari nilai dengan eliminasi :

$table row cell fraction numerator 1 over denominator 2 a end fraction minus fraction numerator 1 over denominator 2 b end fraction equals negative 10 end cell cell open vertical bar cross times 4 close vertical bar end cell cell 2 over a minus 2 over b equals negative 40 end cell row cell 2 over a plus 3 over b equals 5 end cell cell open vertical bar cross times 1 close vertical bar end cell cell 2 over a plus 3 over b equals 5 space space space space space space space space minus end cell row blank blank cell fraction numerator negative 5 over denominator b end fraction equals negative 45 end cell row blank blank cell b equals 1 over 9 end cell end table$

Substitusi ke persamaan pertama:

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 1 over denominator 2 a end fraction minus fraction numerator 1 over denominator 2 b end fraction end cell equals cell negative 10 end cell row cell fraction numerator 1 over denominator 2 a end fraction minus fraction numerator 1 over denominator 2 open parentheses 1 over 9 close parentheses end fraction end cell equals cell negative 10 space open parentheses dikali space fraction numerator 4 a over denominator 9 end fraction close parentheses end cell row cell 2 over 9 minus 2 a end cell equals cell negative fraction numerator 40 a over denominator 9 end fraction end cell row cell negative 2 a end cell equals cell fraction numerator negative 40 a minus 2 over denominator 9 end fraction end cell row cell negative 18 a end cell equals cell negative 40 a minus 2 end cell row cell 22 a end cell equals cell negative 2 end cell row a equals cell negative 1 over 11 end cell end table$

Jadi, penyelesaian dari setiap SPLDV tersebut adalah open parentheses negative 1 over 11 comma thin space 1 over 9 close parentheses .