Pertanyaan

Jika x dan y memenuhi y x + x y = 2 5 dan x - 3 y = 1 maka 5 x + 5 y = …

Jika x dan y memenuhi $\frac{x}{y} + \frac{y}{x} = \frac{5}{2}$ dan x - 3y = 1 maka 5x + 5y = …

-15 atau -3
-3 atau
-3 atau 15
3 atau
3 atau 15

M. Robo

Master Teacher

Jawaban terverifikasi

Pembahasan

x - 3 y = 1 x = 1 + 3 y …. (1) …. (2) Substitusi persamaan (1) ke persamaan (2) Untuk maka Untuk maka Jadi nilai dari: Atau

x - 3y = 1

x = 1 + 3y …. (1)

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight x over straight y plus straight y over straight x end cell equals cell 5 over 2 end cell row cell fraction numerator straight x squared plus straight y squared over denominator xy end fraction end cell equals cell 5 over 2 end cell end table end style$ …. (2)

Substitusi persamaan (1) ke persamaan (2)

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator left parenthesis 1 plus 3 straight y blank right parenthesis squared plus straight y squared over denominator left parenthesis 1 plus 3 straight y right parenthesis straight y end fraction end cell equals cell 5 over 2 end cell row cell fraction numerator left parenthesis 1 plus 3 straight y blank right parenthesis squared plus straight y squared over denominator straight y plus 3 straight y squared end fraction end cell equals cell 5 over 2 end cell row cell fraction numerator 1 plus 6 straight y plus 10 straight y squared over denominator straight y plus 3 straight y squared end fraction end cell equals cell 5 over 2 end cell row cell 2 plus 12 straight y plus 20 straight y squared end cell equals cell 5 straight y plus 15 straight y squared end cell row cell 5 straight y squared plus 7 straight y plus 2 end cell equals 0 row cell left parenthesis 5 straight y plus 2 right parenthesis left parenthesis straight y plus 1 right parenthesis end cell equals 0 row cell straight y subscript 1 end cell equals cell negative 2 over 5 space atau space straight y subscript 2 equals negative 1 end cell end table end style$

Untuk maka

Jadi nilai dari:

Atau