Contoh Soal 4 Sistem Persamaan Linear Tiga Variabel

$\begin{array}{l}\\ 16.& \text{Diketahui suatu fungsi kuadrat}\\ & f(x)=a{x}^2+bx+c.\text{Jika fungsi}\\ & (-1,0),(1,4),\text{dan}(2,9),\text{maka}\\ & \text{fungsi yang dimaksud adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle f(x)=x^2-2x+3}\\ \text{b}.& f(x)=x^2+2x+3\\ \text{c}.& f(x)=x^2+2x-3\\ \text{d}.& f(x)=x^2-2x-3\\ \text{e}.& f(x)=x^2+2x+1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}(-1,0)\Rightarrow f(-1)=a-b+c=0....(1)\\ (1,4)\Rightarrow f(1)=a+b+c=4....(2)\\ (2,9)\Rightarrow f(2)=4a+2b+c=9....(3)\end{array}\\ & \text{Saat}(1)\mathrm{\&}(2),\text{didapatkan}\\ & b=2...............(4)\\ & \text{Saat}(1)\mathrm{\&}(3),\text{didapatkan}\\ & \begin{array}{l}\\ 4a+2b+c& =9& \\ a-b+c& =0& -\\ 3a+3b& =9& \\ a+b& =3& ...(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \{\begin{array}{ll}a& =1\\ c& =1\end{array}\\ & \text{Jadi},f(x)=a{x}^2+bx+c=x^2+2x+1\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \text{Diketahui persamaan}\{\begin{array}{ll}x-y& =2\\ kx+y& =3\end{array}\\ & \text{memiliki solusi}(x,y)\text{di kuadran I}\\ & \text{Jika dan hanya jika nilai}k\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle k=-1}\\ \text{b}.& k>-1\\ \text{c}.& k<{\displaystyle \frac{3}{2}}\\ \text{d}.& 0<k<{\displaystyle \frac{3}{2}}\\ \text{e}.& -1<k<{\displaystyle \frac{3}{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x-y=2....(1)\\ kx+y=3....(2)\end{array}\\ & \text{Dengan metode matriks didapatkan}\\ & x={\displaystyle \frac{\left|\begin{array}{cc}2& -1\\ 3& 1\end{array}\right|}{\left|\begin{array}{cc}1& -1\\ k& 1\end{array}\right|}={\displaystyle \frac{2-(-3)}{1+k}=\frac{5}{k+1}}}\\ & \text{Dengan cara yang sama pula}\\ & y={\displaystyle \frac{\left|\begin{array}{cc}1& 2\\ k& 3\end{array}\right|}{\left|\begin{array}{cc}1& -1\\ k& 1\end{array}\right|}={\displaystyle \frac{3-2k}{k+1}}}\\ & \text{Supaya memiliki solusi di kwadran I},\\ & \text{maka baik}x\text{maupun}y\\ & \text{haruslah positif, akibatnya}:\\ & k+1>0\Rightarrow k>-1\\ & \text{Sebagai akibat yang lain adalah}:\\ & 3-2k>0\Rightarrow k<{\displaystyle \frac{3}{2}}\\ & \text{Jadi},-1<k<{\displaystyle \frac{3}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \text{Diketahui sistem persamaan}\\ & y+{\displaystyle \frac{2}{x+z}=4}\\ & 5y+{\displaystyle \frac{18}{2x+y+z}=18}\\ & {\displaystyle \frac{8}{x+z}-\frac{6}{2x+y+z}=3}\\ & \text{Nilai}y+\sqrt{x^2-2xz+y^2}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 3}\\ \text{b}.& 5\\ \text{c}.& 7\\ \text{d}.& 9\\ \text{e}.& 11\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}y+{\displaystyle \frac{2}{x+z}=4}\\ 5y+{\displaystyle \frac{18}{2x+y+z}=18}\\ {\displaystyle \frac{8}{x+z}-\frac{6}{2x+y+z}=3}\end{array}\\ & \text{Jika disederhanakan beberapa bagian}\\ & \{\begin{array}{ll}y+2A& =4....(1)\\ 5y+18B& =18....(2)\\ 8A-6B& =3....(3)\end{array}\\ & \text{Saat}(1)+(2)\mathrm{\&}(3),\text{maka}\\ & \begin{array}{l}\\ y+2A& =4& |\times 5|& 5y+10A=20\\ 5y+3(8A-3)& =18& |\times 1|& 5y+24A=27& -\\ & & & -14A=-7\\ & & & A={\displaystyle \frac{1}{2}...(4)}\\ \text{maka}B={\displaystyle \frac{1}{6}\mathrm{\&}}& y=3& & \\ \text{akibatnya}\\ \{\begin{array}{ll}x& =1\\ z& =1\end{array}\end{array}\\ & \text{Jadi},y+\sqrt{x^2-2xz+z^2}=3+0=3\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Diberikan}a,b,\text{dan}c\text{adalah angka-angka}\\ & \text{dari bilangan 3 digit yang memenuhi}\\ & 49a+7b+c=286.\text{Nilai dari}a+b+c\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 16\\ \text{b}.& 17\\ \text{c}.& 18\\ \text{d}.& 19\\ \text{e}.& 20\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & 49a+7b+c=286\\ & \text{Nilai maksimum}a\text{adalah}5\\ & 49\times 5=245,\text{akibatnya}:\\ & 245+7b+c=286\Rightarrow 7b+c=286-245=41\\ & \text{Nilai maksimum}b\text{adalah}5\\ & 7\times 5=35,\text{akibatnya}:\\ & 35+c=41\Rightarrow c=41-35=6\\ & \text{Sehingga}a,b,\text{dan}c\text{adalah}5,5,\text{dan}6\\ & \text{Jadi},\text{nilai}a+b+c=5+5+6=16\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Diketahui sistem persamaan}\\ & (2x+3y{)}^{{.}^{\log (x-y+2z)}}=1\\ & 3^{2x+y+z}\times {27}^{3z+2y+x}=81\\ & 5x+3y+8z=2\\ & \text{Himpunan penyelesaian yang}\\ & \text{memenuhi adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left\{{\displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6}}\right\}\\ \text{b}.& \{-{\displaystyle \frac{17}{12},\frac{1}{2},\frac{7}{6}}\}\\ \text{c}.& \{-{\displaystyle \frac{17}{12},-\frac{1}{2},-\frac{7}{6}}\}\\ \text{d}.& \left\{{\displaystyle \frac{17}{12},\frac{1}{12},\frac{7}{6}}\right\}\\ \text{e}.& \{-{\displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6}}\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Untuk persamaan}(1)\\ & (2x+3y{)}^{{.}^{\log (x-y+2z)}}=(2x+3y{)}^0\\ & \iff (x-y+2z)={10}^0=1\\ & \text{Untuk persamaan}(2)\\ & 3^{2x+y+z}\times {27}^{3z+2y+x}=81\\ & \iff 3^{2x+y+z+3(3z+2y+x)}=3^4\\ & \iff 5x+7y+10z=4\\ & \text{Sehingga sistem persamaan akan terlihat}\\ & \{\begin{array}{c}x-y+2z=1....(1)\\ 5x+7y+10z=4....(2)\\ 5x+3y+8z=2....(3)\end{array}\\ & \text{Saat}(2)\mathrm{\&}(3),\text{maka}\\ & \begin{array}{l}\\ 5x+7y+10z& =4& \\ 5x+3y+8z& =2& -\\ 4y+2z& =2\\ 2y+z& =1...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ 5x-5y+10z& =5& \\ 5x+7y+10z& =4& -\\ -12y& =1\\ y& =-{\displaystyle \frac{1}{12}...(5)}\end{array}\\ & \text{Dari persamaan}(5)\text{disubstistusikan ke}(4)\\ & \begin{array}{rl}2y+z& =1\\ 2(-{\displaystyle \frac{1}{12}})+z& =1\\ z& =1+{\displaystyle \frac{1}{6}}\\ z& ={\displaystyle \frac{7}{6}}\end{array}\\ & \text{Cukup jelas juga}x=....\\ & \text{Jadi},\text{pilihannya adalah}e\end{array}\end{array}$

DAFTAR PUSTAKA

1. Bintari, N., Gunarto, D. 2007. Panduan Menguasai Soal-Soal Olimpiade MAtematika Nasional dan Internasional. Yogyakarta: INDONESIA CERDAS.
2. Kanginan, M. 2016. Matematika untuk SMA-MA/SMK-MAK Kelas X. Bandung: SRIKANDI EMPAT WIDYA UTAMA
3. Kurnianingsih, S. 2008. SPM Matematika SMA dan MA Program IPS Siap Tuntas Menghadapi Ujian. Jakarta: ESIS
4. Susianto, B. 2011. Soal dan Pembahasan Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Jakarta: GRASINDO
5. Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: TIGA SERANGKAI PUSTAKA MANDIRI