Pertanyaan

Solve the following system for x and y . in terms of a and b , where a  = b . { a x + b y = a 1 b 2 x + a 2 y = 1

Solve the following system for $x and y$ . in terms of $a and b$ , where $a \neq = b$ .

${a x + b y = \frac{1}{a} b^{2} x + a^{2} y = 1$

S. Yoga

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Jawaban

penyelesaian dari setiap SPLDV tersebut adalah .

penyelesaian dari setiap SPLDV tersebut adalah $open parentheses fraction numerator 1 over denominator a squared plus a b plus b squared end fraction comma space fraction numerator b plus a over denominator a open parentheses a squared plus a b plus b squared close parentheses end fraction close parentheses$ .

Pembahasan

Diketahui: Mencari nilai dengan eliminasi: Mencari nilai dengan eliminasi: Jadi,penyelesaian dari setiap SPLDV tersebut adalah .

Diketahui:

open curly brackets table attributes columnalign left end attributes row cell a x plus b y equals 1 over a end cell row cell b squared x plus a squared y equals 1 end cell end table close

Mencari nilai dengan eliminasi:

$table row cell a x plus b y equals 1 over a end cell cell open vertical bar cross times b squared close vertical bar end cell cell a b squared x plus b cubed y equals b squared over a end cell row cell b squared x plus a squared y equals 1 end cell cell open vertical bar cross times a space close vertical bar end cell cell a b squared x plus a cubed y equals a space space space space space minus end cell row blank blank cell open parentheses b cubed minus a cubed close parentheses y equals fraction numerator b squared minus a squared over denominator a end fraction end cell row blank blank cell y equals fraction numerator fraction numerator b squared minus a squared over denominator a end fraction over denominator b cubed minus a cubed end fraction end cell end table$

$table attributes columnalign right center left columnspacing 0px end attributes row y equals cell fraction numerator begin display style fraction numerator b squared minus a squared over denominator a end fraction end style over denominator b cubed minus a cubed end fraction end cell row blank equals cell fraction numerator left parenthesis b plus a right parenthesis left parenthesis b minus a right parenthesis over denominator a left parenthesis b minus a right parenthesis open parentheses b squared plus a b plus a squared close parentheses end fraction end cell row blank equals cell fraction numerator b plus a over denominator a open parentheses b squared plus a b plus a squared close parentheses end fraction end cell end table$

Mencari nilai dengan eliminasi:

$table row cell a x plus b y equals 1 over a end cell cell open vertical bar cross times a squared close vertical bar end cell cell a cubed x plus a squared b y equals a end cell row cell b squared x plus a squared y equals 1 end cell cell open vertical bar cross times b space close vertical bar end cell cell b cubed x plus a squared b y equals b space space space space space minus end cell row blank blank cell open parentheses b cubed minus a cubed close parentheses x equals a minus b end cell row blank blank cell x equals fraction numerator a minus b over denominator left parenthesis a minus b right parenthesis open parentheses a squared plus a b plus b squared close parentheses end fraction end cell row blank blank cell x equals fraction numerator 1 over denominator a squared plus a b plus b squared end fraction end cell end table$

Jadi, penyelesaian dari setiap SPLDV tersebut adalah $open parentheses fraction numerator 1 over denominator a squared plus a b plus b squared end fraction comma space fraction numerator b plus a over denominator a open parentheses a squared plus a b plus b squared close parentheses end fraction close parentheses$ .