KUIS [tex]\displaystyle\rm\frac{a}{c}+\frac{a}{b}+1=-1\\\\\\\frac{b}{a}+\frac{b}{c}+1=-2[/tex] 2a + .... Question from @xcvi

July 2022 1 51 Report

KUIS

[tex]\displaystyle\rm\frac{a}{c}+\frac{a}{b}+1=-1\\\\\\\frac{b}{a}+\frac{b}{c}+1=-2[/tex]

2a + b + c = -18

b < a < c

a = .....

b = .....

c = .....

henriyulianto

a = –5, b = –10, c = 2.
 

Pembahasan

Diketahui

$\begin{cases}\dfrac{a}{c}+\dfrac{a}{b}+1=-1&(i)\\\\\dfrac{b}{a}+\dfrac{b}{c}+1=-2&(ii)\\\\2a+b+c=-18&(iii)\\\\b < a < c&(iv)\end{cases}$

Ditanyakan

Nilai a, b, dan c

PENYELESAIAN

$\begin{aligned}&(i):\frac{a}{c}+\frac{a}{b}+1=-1\\&{\Rightarrow\ }\frac{a}{c}+\frac{a}{b}=-2\quad\Bigr\}\times bc\\&{\therefore\ \:}ab+ac=-2bc\quad...(v)\\&\textsf{--------------------------------------}\end{aligned}$
$\begin{aligned}&(ii):\frac{b}{a}+\frac{b}{c}+1=-2\\&{\Rightarrow\ }\frac{b}{a}+\frac{b}{c}=-3\quad\Bigr\}\times ac\\&{\Rightarrow\ }bc+ab=-3ac\\&{\Rightarrow\ }ab+3ac=-bc\quad\bigr\}\times2\\&{\therefore\ \:}2ab+6ac=-2bc\quad...(vi)\\&\textsf{--------------------------------------}\end{aligned}$
$\begin{aligned}&(v)=(vi)\\&{\Rightarrow\ }ab+ac=2ab+6ac\\&{\Rightarrow\ }ab=-5ac\quad\bigr\}\div a\\&{\therefore\ \ }b=-5c\quad...(vii)\\&\textsf{--------------------------------------}\end{aligned}$
$\begin{aligned}&(vii)\to(v)\\&{\Rightarrow\ }a(-5c)+ac=-2(-5c)c\\&{\Rightarrow\ }{-}5ac+ac=10c^2\\&{\Rightarrow\ }{-}4ac=10c^2\quad\bigr\}\div(-2c)\\&{\therefore\ \ }2a=-5c\quad...(viii)\\&\textsf{--------------------------------------}\end{aligned}$
$\begin{aligned}&(vii,viii)\to(iii)\\&{\Rightarrow\ }{-}5c+(-5c)+c=-18\\&{\Rightarrow\ }{-}9c=-18\\&{\therefore\ \ }\boxed{\ c=\bf2\ }\quad...(ix)\\&\textsf{--------------------------------------}\end{aligned}$
$\begin{aligned}&(ix)\to(vii)\\&{\Rightarrow\ }b=-5(2)\\&{\therefore\ \ }\boxed{\ b=\bf{-}10\ }\quad...(x)\\&\textsf{--------------------------------------}\\&(ix)\to(viii)\\&{\Rightarrow\ }2a=-5(2)\\&{\therefore\ \ }\boxed{\ a=\bf{-}5\ }\quad...(xi)\\&\textsf{--------------------------------------}\\&(ix,x,xi)\to(iv)\\&{\Rightarrow\ }{-}10 < -5 < 2\\&{\therefore\ \ }(iv)\rm\ terpenuhi!\\&\textsf{--------------------------------------}\\&\blacksquare\end{aligned}$