Olimpiade Zone : Sistem Persamaan

Berikut ini adalah kumpulan soal mengenai Sistem Persamaan tingkat olimpiade. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

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No. 1

Diketahui $x$, $y$, $z$ dan $t$ adalah bilangan real tidak nol dan memenuhi persamaan

$x+y+z=t$
$\dfrac1x+\dfrac1y+\dfrac1z=\dfrac1t$
$x^3+y^3+z^3=1000^3$

Nilai dari $x+y+z+t$ adalah....

\(\eqalign{ \dfrac1x+\dfrac1y+\dfrac1z&=\dfrac1t\\[4pt] \dfrac{xy+yz+xz}{xyz}&=\dfrac1t\\[4pt] xy+yz+xz&=\dfrac{xyz}t}\)

\(\eqalign{ x^3+y^3+z^3&=(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz\\ 1000^3&=t^3-3\left(\cancel{t}\right)\left(\dfrac{xyz}{\cancel{t}}\right)+3xyz\\ 1000^3&=t^3-\cancel{3xyz}+\cancel{3xyz}\\ t&=10}\)

\(\eqalign{ x+y+z+t&=2t\\ &=2(1000)\\ &=\boxed{\boxed{2000}}}\)

Jadi, nilai dari $x+y+z+t$ adalah 2000.

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No. 2

Diketahui $(x,y)$ memenuhi dua persamaan :
$x^3-3x^2+5x+2016=2017$
$y^3-3y^2+5y+2017=2022$
Carilah nilai dari ${^{\frac1{16}}\negmedspace\log(x+y)}$

\(\eqalign{ x^3-3x^2+5x+2016&=2017\\ (x-1)^3+2x&=0\\ (x-1)^3+2(x-1)+2&=0 }\)
Misal $p=x-1$
$p^3+2p+2=0$

\(\eqalign{ y^3-3y^2+5y+2017&=2022\\ (y-1)^3+2(y-1)-2&=0 }\)
Misal $q=y-1$
$q^3+2q-2=0$

\(\eqalign{ p^3+2p+2&=0\\ q^3+2q-2&=0&\qquad+\\\hline p^3+q^3+2(p+q)&=0\\ (p+q)\left(p^2-pq+q^2\right)+2(p+q)&=0\\ (p+q)\left(p^2-pq+q^2+2\right)&=0\\ p+q&=0\\ x-1+y-1&=0\\ x+y&=2 }\)

Jadi, $^{\frac1{16}}\log(x+y)=2$.

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No. 3

$p$, $q$, dan $r$ adalah tiga bilangan real yang memenuhi persamaan
$p+q+r=9$
$pqr=10$
$r^2-p^2-q^2=13$
Nilai $r$ yang bulat yang memenuhi persamaan tersebut adalah ....

SUMBER

\(\eqalign{ r^2-p^2-q^2&=13\\ p^2+q^2&=r^2-13 }\)

\(\eqalign{ p+q+r&=9\\ (p+q+r)^2&=9^2\\ p^2+q^2+r^2+2(pq+qr+pr)&=81\\ r^2-13+r^2+2(pq+qr+pr)&=81\\ 2r^2+2(pq+qr+pr)&=94\\ pq+qr+pr&=47-r^2 }\)

$p$, $q$, dan $r$ merupakan akar-akar dari:

\(\eqalign{ x^3-9x^2+\left(47-r^2\right)x-10&=0\\ r^3-9r^2+\left(47-r^2\right)r-10&=0\\ r^3-9r^2+47r-r^3-10&=0\\ 9r^2-47r+10&=0\\ (9r-2)(r-5)&=0 }\)

$r=\dfrac29$ atau $r=5$

Jadi, nilai $r$ yang bulat yang memenuhi adalah 5.

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No. 4

Diketahui $a$, $b$, dan $c$ adalah tiga bilangan real yang memenuhi persamaan
$a^2-bc=7$
$b^2+ac=7$
$c^2+ab=7$
maka $a^2+b^2+c^2=$ ....

SUMBER

\(\eqalign{ a^2-bc&=b^2+ac\\ a^2-b^2&=ac+bc\\ (a+b)(a-b)&=(a+b)c\\ a-b&=c\\ a&=b+c }\)

\(\eqalign{ a^2-bc&=7\\ b^2+ac&=7\\ c^2+ab&=7&\qquad+\\\hline a^2+b^2+c^2+ab+ac-bc&=21\\ a^2+b^2+c^2+a(b+c)-bc&=21\\ a^2+b^2+c^2+a(a)-bc&=21\\ a^2+b^2+c^2+a^2-bc&=21\\ a^2+b^2+c^2+7&=21\\ a^2+b^2+c^2&=14 }\)

Jadi, $a^2+b^2+c^2=14$.

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No. 5

Diberikan sistem persamaan
$x+3y+2w=4444$,
$6x+2y=8888$,
$6z+3w=6666$,
Tentukan hasil dari $x+y+z+w$

$\begin{aligned} x+3y+2w&=4444\\ 6x+2y&=8888&\color{red}{:2}\\ 6z+3w&=6666&\color{red}{\times\dfrac23} \end{aligned}$

$\begin{aligned} x+3y+2w&=4444\\ 3x+y&=4444\\ 4z+2w&=4444&\color{red}{+}\\\hline 4x+4y+4z+4w&=3\cdot4444\\ x+y+z+w&=3\cdot1111\\ &=\boxed{\boxed{3333}} \end{aligned}$

Jadi, $x+y+z+w=3333$.

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No. 6

\begin{cases} a(b+c-5)=7\\ b(a+c-5)=7\\ a^2+b^2=50 \end{cases} Carilah nilai $(a,b,c)$

$\begin{aligned} a(b+c-5)&=7\\ ab+ac-5a&=7 \end{aligned}$

$\begin{aligned} b(a+c-5)&=7\\ ab+bc-5b&=7 \end{aligned}$

$\begin{aligned} ab+ac-5a&=7\\ ab+bc-5b&=7\qquad-\\\hline (a-b)c-5(a-b)&=0\\ (a-b)(c-5)&=0 \end{aligned}$

• Untuk $a=b$
  $\begin{aligned} a^2+b^2&=50\\ a^2+a^2&=50\\ 2a^2&=50\\ a^2&=25\\ a&=\pm5 \end{aligned}$
  
  • Untuk $a=b=5$
    $\begin{aligned} a(b+c-5)&=7\\ 5(5+c-5)&=7\\ 5c&=7\\ c&=\dfrac75 \end{aligned}$
    $(a,b,c)=\left(5,5,\dfrac75\right)$
    
  • Untuk $a=b=-5$
    $\begin{aligned} a(b+c-5)&=7\\ -5(-5+c-5)&=7\\ -5(c-10)&=7\\ -5c+50&=7\\ -5c&=-43\\ c&=\dfrac{43}5 \end{aligned}$
    $(a,b,c)=\left(-5,-5,\dfrac{43}5\right)$
    
• Untuk $c=5$
  $\begin{aligned} a(b+c-5)&=7\\ a(b+5-5)&=7\\ ab&=7\\ b&=\dfrac7a \end{aligned}$
  
  $\begin{aligned} a^2+b^2&=50\\ a^2+\left(\dfrac7a\right)^2&=50\\ a^2+\dfrac{49}{a^2}&=50\\ a^4+49&=50a^2\\ a^4-50a^2+49&=0\\ \left(a^2-1\right)\left(a^2-49\right)&=0 \end{aligned}$
  Untuk $a^2=1$ maka $a=\pm1$ dan $b=\pm7$
  $(a,b,c)=\{(-1,-7,5),(1,7,5)\}$
  Untuk $a^2=49$ maka $a=\pm7$ dan $b=\pm1$
  $(a,b,c)=\{(-7,-1,5),(7,1,5)\}$
  

Jadi,

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No. 7

Diketahui sistem persamaan ${a+3b+2d=8}$, ${6a+2b=8}$, dan ${6c+3d=12}$. Maka nilai ${a+b+c+d=}$...

SUMBER

$\begin{aligned} 6a+2b&=8\qquad\color{red}:2\\ 3a+b&=4 \end{aligned}$

$\begin{aligned} 6c+3d&=12\qquad\color{red}\times\dfrac23\\ 4c+2d&=8 \end{aligned}$

$\begin{array}{rcrcrcrcl} a&+&3b&+&&&2d&=&8\\ 3a&+&b&&&&&=&4\\ &&&&4c&+&2d&=&8\qquad\color{red}+\\\hline 4a&+&4b&+&4c&+&4d&=&20\\ a&+&b&+&c&+&d&=&\boxed{\boxed{5}} \end{array}$

Jadi, $a+b+c+d=5$.
JAWAB: D

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No. 8

Jika begin{cases}2a^2+2007a+3=0\3b^2+2007b+2=0\end{cases} dan ${ab\neq1}$, tentukan $\dfrac{a}b$

$\begin{aligned} 2a^2+2007a+3&=0\qquad&\color{red}{\times b}\\ 3b^2+2007b+2&=0\qquad&\color{red}{\times a} \end{aligned}$

$\begin{aligned} 2a^2b+2007ab+3b&=0\\ 3ab^2+2007ab+2a&=0\qquad&\color{red}{-}\\[-3pt]\hline\\[-12pt] 2a^2b-3ab^2+3b-2a&=0\\ ab(2a-3b)+3b-2a&=0\\ (ab-1)(2a-3b)&=0 \end{aligned}$

• $ab-1=0$
  $ab=1$ TM
• $2a-3b=0$
  $\begin{aligned} 2a&=3b\\ \dfrac{a}b&=\boxed{\boxed{\dfrac32}} \end{aligned}$

Jadi, $\dfrac{a}b=\dfrac32$.

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No. 9

Jika $$\{\begin{array}{l}2017a^2+2018a+2019=0\\ 2019b^2+2018b+2017=0\end{array}$$ dan ${ab\neq1}$, tentukan $\dfrac{a}b$

$\begin{aligned} 2017a^2+2018a+2019&=0\qquad&\color{red}{\times b}\\ 2019b^2+2018b+2017&=0\qquad&\color{red}{\times a} \end{aligned}$

$\begin{aligned} 2017a^2b+2018ab+2019b&=0\\ 2019ab^2+2018ab+2017a&=0\qquad&\color{red}{-}\\[-3pt]\hline\\[-12pt] 2017a^2b-2019ab^2+2019b-2017a&=0\\ ab(2017a-2019b)-(2017a-2019b)&=0\\ (ab-1)(2017a-2019b)&=0 \end{aligned}$

• $ab-1=0$
  $ab=1$ TM
• $2017a-2019b=0$
  $\begin{array}{rl}2017a& =2019b\\ {\displaystyle \frac{a}{b}}& =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{2019}{2017}}}}}}\end{array}$

Jadi, $\dfrac{a}b=\dfrac{2019}{2017}$.

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