Contoh Soal 2 Sistem Persamaan Linear Tiga Variabel

$\begin{array}{l}\\ 6.& \text{Diketahui sistem persamaan berikut}\\ & \{\begin{array}{c}x+y-z=1\\ 2x-y+2z=9\\ x+3y-z=7\end{array}\\ & \text{Nilai}{\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{3}}\\ \text{b}.& {\displaystyle \frac{3}{4}}\\ \text{c}.& {\displaystyle \frac{13}{12}}\\ \text{d}.& {\displaystyle \frac{5}{4}}\\ \text{e}.& {\displaystyle \frac{7}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+y-z=1....(1)\\ 2x-y+2z=9....(2)\\ x+3y-z=7....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ x+y-z& =1& \\ 2x-y+2z& =9& +\\ 3x+z& =10& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ x+y-z& =1& |\times 3|& 3x+3y-3z& =3& \\ x+3y-z& =7& |\times 1|& x+3y-z& =7& -\\ & & & 2x-2z& =-4& \\ & & & x-z& =2& ....(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{l}\\ 3x+z& =10& \\ x-z& =-2& +\\ 4x& =8& \\ x& =2& .....(6)\\ \text{didapat pula}& z& =4......(7)\end{array}\\ & \text{Dari persamaan}(1)\mathrm{\&}(3)\text{didapatkan juga}\\ & \begin{array}{l}\\ x+y-z& =1& \\ x+3y-z& =-7& -\\ -2y& =-6& \\ y& =3& ....(8)\end{array}\\ & \text{Jadi},{\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}={\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}}}\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Diketahui sistem persamaan berikut}\\ & \{\begin{array}{c}x+y+z=5\\ x+y-4z=10\\ -2x+y+z=0\end{array}\\ & \text{Nilai dari}{\displaystyle \frac{xz}{y}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\frac{6}{13}}\\ \text{b}.& {\displaystyle -\frac{5}{13}}\\ \text{c}.& {\displaystyle -\frac{1}{13}}\\ \text{d}.& {\displaystyle \frac{1}{13}}\\ \text{e}.& {\displaystyle \frac{7}{13}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+y+z=5.....(1)\\ x+y-4z=10.....(2)\\ -2x+y+z=0.....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =5& \\ x+y-4z& =10& -\\ 5z& =-5& \\ z& =-1& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =5& \\ -2x+y+z& =0& -\\ 3x& =5& \\ x& ={\displaystyle \frac{5}{3}}& ....(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{rl}x+y+z& =5\\ {\displaystyle \frac{5}{3}+y-1}& =5\\ y& =5+1-{\displaystyle \frac{5}{3}=\frac{13}{3}}\end{array}\\ & \text{Jadi},{\displaystyle \frac{xz}{y}={\displaystyle \frac{\left({\displaystyle \frac{5}{3}}\right).(-1)}{{\displaystyle \frac{13}{3}}}=-{\displaystyle \frac{5}{13}}}}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Himpunan penyelesaian dari}\\ & \{\begin{array}{ll}{\displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}}& =8\\ {\displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}}& =10\\ {\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}}& =4\end{array}\\ & \text{adalah}\{(x,y,z)\},\text{maka}x+3z=....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& {\displaystyle \frac{1}{3}}\\ \text{c}.& 1\\ \text{d}.& 3\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{ll}{\displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}}& =8....(1)\\ {\displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}}& =10.....(2)\\ {\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}}& =4...........(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ {\displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}}& =8\\ {\displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}}& =10& -\\ -{\displaystyle \frac{1}{x}-\frac{1}{z}}& =-2\\ {\displaystyle \frac{1}{x}+\frac{1}{z}}& =2& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ {\displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}}& =8& |\times 2|& {\displaystyle \frac{4}{x}+\frac{4}{y}+\frac{8}{z}}& =16\\ {\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}}& =4& |\times 1|& {\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}}& =4& -\\ & & & {\displaystyle \frac{2}{x}+\frac{6}{z}}& =12\\ & & \iff & {\displaystyle \frac{1}{x}+\frac{3}{z}}& =6& ...(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{l}\\ {\displaystyle \frac{1}{x}+\frac{3}{z}}& =6\\ {\displaystyle \frac{1}{x}+\frac{1}{z}}& =2-\\ {\displaystyle \frac{2}{z}}& =4& \\ z& ={\displaystyle \frac{1}{2}......(6)}\\ x& =2-{\displaystyle \frac{1}{2}={\displaystyle \frac{3}{2}}}\end{array}\\ & \text{Jadi},x+3z={\displaystyle \frac{3}{2}+3.\frac{1}{2}=3}\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Diketahui tiga buah bilangan berturut-turut}\\ & a,b,\text{dan}c.\text{Rata-rata dari ke tiga bilangan}\\ & \text{itu adalah 12. Bilangan kedua sama dengan}\\ & \text{jumlah bilangan yang lain dikurangi 12}.\\ & \text{Jika bilangan ke tiga sama dengan jumlah}\\ & \text{bilangan yang lain, maka nilai}2a+b-c=....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle 42}\\ \text{b}.& -{\displaystyle 36}\\ \text{c}.& -{\displaystyle 18}\\ \text{d}.& -{\displaystyle 12}\\ \text{e}.& -{\displaystyle 6}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Model matematika dari persamaan di atas}\\ & \{\begin{array}{c}a+b+c=36....(1)\\ -a+b-x=12....(2)\\ a+b-c=0....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ a+b+c& =36& \\ -a+b-c& =12& +\\ 2b& =48& \\ b& =24& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ a+b+c& =36& \\ a+b-c& =0& -\\ 2c& =36& \\ c& =18& ....(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{rl}a+b+c& =36\\ a+24+18& =36\\ a& =36-42\\ & =-6\end{array}\\ & \text{Jadi},2a+b-c=2(-6)+24-18=-6\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Jumlah uang terdiri atas koin pecahan}Rp500,00\\ & Rp200,00danRp100,00\text{dengan nilai total}\\ & Rp100.000,00.\text{Jika nilai uang pecahan 500-an}\\ & \text{setengah dari nilai uang pecahan 200-an, tetapi}\\ & \text{tiga kali uang pecahan 100-an, maka banyak koin}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 460\\ \text{b}.& 440\\ \text{c}.& 420\\ \text{d}.& 380\\ \text{e}.& 350\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Model matematika dari kasus di atas}\\ & \{\begin{array}{c}A(500)+B(200)+C(100)=100.000....(1)\\ A(500)={\displaystyle \frac{1}{2}B(200)....(2)}\\ A(500)=3C(100)....(3)\end{array}\\ & \text{Dari persamaan}(2)\text{didapatkan}\\ & 2A(500)=B(200)\\ & \text{Dari persamaan}(3)\text{akan didapatkan}\\ & {\displaystyle \frac{1}{3}A(500)=C(100)}\\ & \text{Dari persamaan}(1)\text{maka},\\ & A(500)+B(200)+C(100)=100.000\\ & A(500)+2A(500)+{\displaystyle \frac{1}{3}A(500)=100.000}\\ & {\displaystyle \frac{10}{3}A(500)=100.000\iff A(500)=30.000}\\ & \text{maka akan didapatkan}\\ & B(200)=2(30.000)=60.000\\ & C(100)={\displaystyle \frac{1}{3}(30.000)=10.000}\\ & \{\begin{array}{ll}A(500)& =30.000\Rightarrow A={\displaystyle \frac{30.000}{500}=60}\\ B(200)& =60.000\Rightarrow B={\displaystyle \frac{60.000}{200}=300}\\ C(100)& =10.000\Rightarrow C={\displaystyle \frac{10.000}{100}=100}\end{array}\\ & \text{Jadi},A+B+C=60+300+100=460\end{array}\end{array}$