$\begin{array}{ll}\\ 11.&\textrm{Suatu bilangan terdiri atas 3 angka. Jumlah}\\ &\textrm{ketiga angka tersebut adalah 9. Angka kedua}\\ &\textrm{dikurangi angka pertama dan angka ketiga }\\ &\textrm{sama dengan 1. Dua kali angka pertama sama}\\ &\textrm{dengan jumlah angka kedua dan angka ketiga.}\\ &\textrm{Angka puluhan pada bilangan tersebut adalah}\\ &....\\ &\begin{array}{llll}\\ \textrm{a}.&3\\ \textrm{b}.&4\\ \color{red}\textrm{c}.&5\\ \textrm{d}.&6\\ \textrm{e}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Model matematikanya}\\ &\left\{\begin{matrix} A+B+C=9\: \: \qquad....(1)\\ 2B-A-C=1\qquad\: ....(2)\\ 2A=B+C\: \: \: \: \qquad\quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\color{blue}\begin{array}{llll}\\ \displaystyle A+B+C&=9\\ \displaystyle -A+B-C&=1&+\\\hline \qquad2B&=10\\ \: \: \: \: \qquad\qquad B&=5&...(4)\\ \end{array}\\ &\color{blue}\textrm{Jadi},\: \textrm{bilangan kedua adalah}\: =\: B=5 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 12.&(\textbf{SIMAK UI 2010})\\ &\textrm{Jika}\: \: x+y+2z=K,\: x+2y+z=K,\\ &2x+y+z=K\: \: \textrm{dengan}\: \: K\neq 0,\: \textrm{maka}\\ &x^{2}+y^{2}+z^{2}\: \: \textrm{bila dinyatakan dalam}\: \: K\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{16}K^{2}\\ \color{red}\textrm{b}.&\displaystyle \frac{3}{16}K^{2}\\ \textrm{c}.&\displaystyle \frac{4}{17}K^{2}\\ \textrm{d}.&\displaystyle \frac{3}{8}K^{2}\\ \textrm{e}.&\displaystyle \frac{2}{3}K^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y+2z=K\: \: \qquad....(1)\\ x+2y+z=K\qquad\: ....(2)\\ 2x+y+z=K\: \: \qquad ....(3)\end{matrix}\right.\\ &\color{black}\textrm{maka}\\ &\color{red}\left\{\begin{matrix} z+(x+y+z)=K\: \: \qquad....(1)\\ y+(x+y+z)=K\qquad\: ....(2)\\ x+(x+y+z)=K\: \: \qquad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2)+(3),\: \textrm{maka}\\ &\color{blue}\begin{array}{llll}\\ \displaystyle x+y+2z&=K\\ \displaystyle x+2y+z&=K&\\ \displaystyle 2x+y+z&=K&+\\\hline 4x+4y+4z&=3K\\ x+y+z&=\displaystyle \frac{3}{4}K&...(4)\\ \end{array}\\ &\textrm{Saat}\: \: (4)\: \: \textrm{disubstitusikan ke}\: \: (1),(2),\: \textrm{dan}\: (3)\\ &\textrm{Jelas bahwa akan didapatkan}\\ &x=y=z=\displaystyle \frac{1}{4}K\\ &\color{blue}\textrm{Jadi},\: \: x^{2}+y^{2}+y^{2}=3\left ( \displaystyle \frac{1}{4}K \right )^{2}=\displaystyle \frac{3}{16}K^{2} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 13.&\textrm{Diketahui}\: \: 0,15252525252...=\displaystyle \frac{p}{2q+r}\\ &\textrm{Jika jumlah}\: \: p\: \: \textrm{dan}\: \: q=\textrm{3 kali}\: \: r,\: \textrm{maka}\\ &\textrm{masing-masing harga}\: \: p,q, \: \textrm{dan}\: \: r=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 152,2819,2584\\ \textrm{b}.&\displaystyle 252,\displaystyle \frac{5638}{7},\frac{8102}{21}\\ \color{red}\textrm{c}.&\displaystyle 151,\frac{2819}{7},\frac{1292}{7}\\ \textrm{d}.&\displaystyle 151,\displaystyle \frac{2819}{7},\frac{2584}{7}\\ \textrm{e}.&\displaystyle 152,\frac{2819}{14},\frac{1292}{7} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Dari soal diketahui}\\ &\color{red}\begin{cases} 0,1\overline{5252}& =\displaystyle \frac{p}{2q+r}\: .....(1)\\ \quad p+q & =3r\: ............(2) \end{cases}\\ &\color{black}\textrm{dan}\\ &\color{blue}\begin{array}{llll}\\ \displaystyle \qquad \color{red}x&=0,15252525252...\\ \displaystyle 1000\color{red}x&=152,5252525252...\\ \displaystyle \quad10\color{red}x&=\: \: \: \: \: 1,5252525252...&-\\\hline \: \: 990\color{red}x&=151\\ \qquad \color{red}x&=\displaystyle \frac{151}{990},\: \: \color{black}\textrm{maka}\\ \displaystyle \frac{p}{2q+r}&=\displaystyle \frac{151}{990}\\ &\begin{cases} p &=151 \: \: .......(3)\\ 2p+r &=990 \: \: .......(4) \end{cases} \end{array}\\ &\textrm{Dari}\: \: (3)\: \textrm{diperoleh}:q=3r-p=3r-151\: ....\color{blue}(5)\\ &\textrm{Dari}\: \: (5)\: \: \textrm{disubstitusikan ke}\: \: (4)\\ &\begin{aligned}2q+r&=990\\ 2(3r-151)+r&=990\\ 6r-302+r&=990\\ 7r&=990+302=1292\\ r&=\displaystyle \frac{1292}{7}\: .....\color{blue}(6) \end{aligned}\\ &\textrm{Dari}\: \: (3)\&(6)\: \: \textrm{disubstitusikan ke}\: \: (2)\\ &\color{purple}\begin{aligned}p+q&=3r\\ 151+q&=3\left ( \displaystyle \frac{1292}{7} \right )\\ q&=\displaystyle \frac{3876}{7}-151\\ &=\displaystyle \frac{3876-1057}{7}\\ &=\displaystyle \frac{2819}{7}\: .....\color{blue}(7) \end{aligned}\\ &\color{blue}\textrm{Jadi},\: \: p,q,r\: \: \textrm{adalah}\: :\: \displaystyle 151,\frac{2819}{7},\frac{1292}{7} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 14.&\textrm{Perhatikanlah sistem persamaan berikut}\\ &\begin{cases} 3x+2y-5z & =3 \\ 2x-6y+kz & =9 \\ 5x-4y-z & =5 \end{cases}\\ &\textrm{agar sistem persamaan ini tidak}\\ &\textrm{memiliki penyelesaian, maka nilai}\: \: k=....\\ &\begin{array}{llll}\\ \textrm{a}.&-4\\ \textrm{b}.&2\\ \textrm{c}.&3\\ \color{red}\textrm{d}.&4\\ \textrm{e}.&6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Agar sistem persamaan}\\ &\color{red}\begin{cases} 3x+2y-5z & =3 \\ 2x-6y+kz & =9 \\ 5x-4y-z & =5 \end{cases}\\ &\color{black}\textrm{tidak berpenyelesaian, maka}\\ &\color{black}\textrm{ingat penyelesaian metode matrik}\\ &\color{black}\textrm{buatlah penyebutnya}=0,\: \: \textrm{yaitu}:\\ &\color{blue}\begin{vmatrix} 3 & 2 & -5\\ 2 & -6 & k\\ 5 & -4 & -1 \end{vmatrix}=0\\ &\textrm{Selanjutnya}\\ &3\begin{vmatrix} -6 & k\\ -4 & -1 \end{vmatrix}-2\begin{vmatrix} 2 & k\\ 5 & -1 \end{vmatrix}-5\begin{vmatrix} 2 & -6\\ 5 & -4 \end{vmatrix}=0\\ &3(6+4k)-2(-2-5k)-5(-8+30)=0\\ &18+12k+4+10k+40-150=0\\ &22x=88\\ &\quad \color{blue}x=4 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 15.&\textrm{Diketahui}\\ &\begin{pmatrix} \displaystyle \frac{1}{5} & \displaystyle \frac{1}{5} & \displaystyle \frac{1}{5}\\ \displaystyle \frac{1}{5} & \displaystyle \frac{1}{5} & -\displaystyle \frac{4}{5}\\ -\displaystyle \frac{2}{5} & \displaystyle \frac{1}{10} & \displaystyle \frac{1}{10} \end{pmatrix}\begin{pmatrix} x\\ y\\ z \end{pmatrix}=\begin{pmatrix} 1\\ 2\\ 0 \end{pmatrix}\\ &\textrm{Nilai}\: \: x,y,\: \: \textrm{dan}\: \: z\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{5},\frac{4}{5},-\frac{1}{10}\\ \textrm{b}.&-1,5,1\\ \color{red}\textrm{c}.&1,5,-1\\ \textrm{d}.&-1,1,5\\ \textrm{e}.&5,1,-1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\color{red}\left\{\begin{matrix} \displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z=1\quad \quad....(1)\\ \displaystyle \frac{1}{5}x+\frac{1}{5}y-\frac{4}{5}z=2\qquad\: ....(2)\\ -\displaystyle \frac{2}{5}x+\frac{1}{10}y+\frac{1}{10}z=0\: \quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\color{blue}\begin{array}{llll}\\ \displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z&=1&\\ \displaystyle \frac{1}{5}x+\frac{1}{5}y-\frac{4}{5}z&=2&-\\\hline \quad\qquad \qquad \displaystyle \frac{5}{5}z&=-1&\\ \: \: \: \quad\qquad \qquad \displaystyle z&=-1&...(4) \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\color{blue}\begin{array}{lllllll}\\ \displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z&=1&\left | \times 1 \right |&\displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z=1\\ -\displaystyle \frac{2}{5}x+\frac{1}{10}y+\frac{1}{10}z&=0&\left | \times 2 \right |&-\displaystyle \frac{4}{5}+\frac{1}{5}y+\frac{1}{5}z=0&-\\\hline &&&\: \: \: \: \displaystyle \frac{5}{5}x\qquad\qquad\: =1&\\ &&&\: \: \: \quad x=-1\: ........(5) \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{akan didapatkan}\\ &y=5\\ &\color{blue}\textrm{Jadi},\: \: (x,y,z)=(1,5,-1) \end{aligned} \end{array}$
Tidak ada komentar:
Posting Komentar
Informasi