Latihan Soal 8 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll}\\ 66.&\textrm{Suatu unit pekerjaan dapat diselesaikan oleh A}\\ &\textrm{B, dan C bersama-sama dalam 2 jam saja.}\\ &\textrm{Jika pekerjaan itu dapat diselesaikan oleh A dan}\\ &\textrm{B bersama-sama dalam 2 jam 24 menit, dan oleh}\\ &\textrm{B dan C bersama-sama dalam waktu 3 jam,}\\ &\textrm{maka sistem persamaan berikut yang memenuhi}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\begin{cases} A+B+C&=2 \\ A+B & =\displaystyle \frac{12}{5} \\ B+C &=3 \end{cases}\\ \textrm{b}.&\begin{cases} A+B+C&=\displaystyle \frac{1}{2} \\ A+B & =\displaystyle \frac{5}{12} \\ B+C &=\displaystyle \frac{1}{3} \end{cases}\\ \textrm{c}.&\begin{cases} \displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}&=2 \\ \displaystyle \frac{1}{A}+\frac{1}{B}& =\displaystyle \frac{12}{5} \\ \displaystyle \frac{1}{B}+\frac{1}{C}&=3 \end{cases}\\ \color{red}\textrm{d}.&\begin{cases} \displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}&=\displaystyle \frac{1}{2} \\ \displaystyle \frac{1}{A}+\frac{1}{B}& =\displaystyle \frac{5}{12} \\ \displaystyle \frac{1}{B}+\frac{1}{C}&=\displaystyle \frac{1}{3} \end{cases}\\ \textrm{e}.&\begin{cases} \displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}&=2 \\ \displaystyle \frac{1}{A}+\frac{1}{B}-\frac{1}{C}& =\displaystyle \frac{12}{5} \\ \displaystyle -\frac{1}{A}+\frac{1}{B}+\frac{1}{C}&=3 \end{cases} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\textrm{Per}&\textrm{hatikan bahwa}:\color{red}\textrm{Waktu penyelesaian}\\ \color{red}\textrm{sua}&\color{red}\textrm{tu pekerjaan adalah termasuk}\\ \color{red}\textrm{per}&\color{red}\textrm{bandingan berbalik nilai},\: \color{blue}\textrm{maka}\\ \bullet \: \: \: &A,B,\: \textrm{dan}\: C \: \textrm{dalam 2 jam, artinya}:\\ &\color{black}\displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{2},\: \color{blue}\textrm{demikian juga}\\ \bullet \: \: \: &A\: \textrm{dan}\: B\: \textrm{bersama-sama selesai dalam}\\ &\textrm{2 jam 24 menit atau}\: \displaystyle \frac{12}{5}\: \textrm{jam}:\\ &\color{black}\displaystyle \frac{1}{A}+\frac{1}{B}=\frac{5}{12}\\ \bullet \: \: \: &B\: \textrm{dan}\: C\: \textrm{selesai dalam 3 jam}:\\ &\color{black}\displaystyle \frac{1}{B}+\frac{1}{C}=\frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 67.&\textrm{Himpunan penyelesaian dari}\\ &\left\{\begin{matrix} x+y+4z=15\quad\\ x-y+z=2\qquad\\ x+2y-3z=-4 \end{matrix}\right.\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ (-1,1,3) \right \}\\ \color{red}\textrm{b}.&\left \{ (1,2,3) \right \}\\ \textrm{c}.&\left \{ (-2,1,1) \right \}\\ \textrm{d}.&\left \{ (3,2,-1) \right \}\\ \textrm{e}.&\left \{ (1,-2,3) \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\textrm{Semunya dikerjakan dengan metode}\\ &\color{blue}\textrm{matriks}\: (\color{black}\textbf{Cara Cramer})\\ &\begin{aligned} \color{blue}x&=\displaystyle \frac{\begin{vmatrix} 15 & 1 & 4\\ 2& -1 & 1\\ -4& 2 & -3 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{15\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 2 & 1\\ -4 & -3 \end{vmatrix}+4\begin{vmatrix} 2 & -1\\ -4 & 2 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{15(3-2)-1(-6+4)+4(4-4)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{15(1)-1(-2)+4(0)}{1(1)-1(-4)+4(3)}=\frac{17}{17}=1 \\ \color{blue}y&=\displaystyle \frac{\begin{vmatrix} 1 & 15 & 4\\ 1& 2 & 1\\ 1& -4 & -3 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{1\begin{vmatrix} 2 & 1\\ -4 & -3 \end{vmatrix}-15\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & 2\\ 1 & -4 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{1(-6+4)-15(-3-1)+4(-4-2)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{1(-2)-15(-4)+4(-6)}{1(1)-1(-4)+4(3)}=\frac{34}{17}=2\\ \color{blue}z&=\displaystyle \frac{\begin{vmatrix} 1 & 1 & 15\\ 1& -1 & 2\\ 1& 2 & -4 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{1\begin{vmatrix} -1 & 2\\ 2 & -4 \end{vmatrix}-1\begin{vmatrix} 1 & 2\\ 1 & -4 \end{vmatrix}+15\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{1(4-4)-1(-4-2)+15(2+1)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{1(0)-1(-6)+15(3)}{1(1)-1(-4)+4(3)}=\frac{51}{17}=3 \end{aligned} \end{array}$

$.\quad\quad \color{blue}\textrm{Cara di atas}$  full matriks-Cramer

$\begin{array}{ll}\\ 68.&\textrm{Hasil dari}\: \: xyz\: \: \textrm{yang memenuhi}\\ &\left\{\begin{matrix} x+y+z=2\quad\\ x-y+z=-2\: \\ x-y-z=2\quad \end{matrix}\right.\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-8\\ \textrm{b}.&-4\\ \textrm{c}.&2\\ \textrm{d}.&4\\ \textrm{e}.&8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y+z=2\quad.....(1)\\ x-y+z=-2\: .....(2)\\ x-y-z=2\quad .....(3) \end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llc}\\ x+y+z&=2&\\ x-y+z&=-2&-\\\hline \: \, \quad2y&=4&\\ \qquad\quad y&=2&....(4) \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{lcc}\\ x+y+z&=2&\\ x-y-z&=2&+\\\hline 2x&=4&\\ \qquad\quad x&=2&....(5) \end{array} \\ &\textrm{Persamaan}\: \: (4)\&(5)\: \: \textrm{ke}\: \: (1)\\ &\color{red}\begin{aligned}x+y+z&=2\\ (2)+(2)+z&=2\\ z&=-2 \end{aligned}\\ &\textrm{Jadi},\: \: xyz=(2).(2).(-2)=\color{red}-8 \end{aligned} \end{array}$

$.\quad\: \:  \color{black}\textrm{Cara di atas}$  full eliminasi-substitusi

$\begin{array}{ll}\\ 69.&\textrm{Diketahui sistem persamaan berikut}\\ &\left\{\begin{matrix} x+y+z=-6\quad\\ x-2y+z=3\quad\: \\ -2x+y+z=9\quad \end{matrix}\right.\\ &\textrm{Nilai}\: \: xyz=....\\ &\begin{array}{llll}\\ \textrm{a}.&-30\\ \textrm{b}.&-15\\ \textrm{c}.&5\\ \color{red}\textrm{d}.&30\\ \textrm{e}.&35 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y+z=-6\quad ....(1)\\ x-2y+z=3\quad\: ....(2)\\ -2x+y+z=9\quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llc}\\ x+y+z&=-6&\\ x-2y+z&=3&-\\\hline \: \: \: \quad 3y&=-9&\\ \qquad\quad y&=-3&....(4) \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{llc}\\ x+y+z&=-6&\\ -2x+y+z&=9&-\\\hline 3x&=-15&\\ \qquad\quad x&=-5&....(5) \end{array} \\ &\textrm{Persamaan}\: \: (4)\&(5)\: \: \textrm{ke}\: \: (1)\\ &\color{red}\begin{aligned}x+y+z&=2\\ (-5)+(-3)+z&=-6\\ z&=2 \end{aligned}\\ &\textrm{Jadi},\: \: xyz=(-5).(-3).(2)=\color{red}30 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 70.&\textrm{Diketahui sistem persamaan berikut}\\ &\left\{\begin{matrix} x+2y+z=4\: \: \qquad\\ 3x+y+2z=-5\quad\: \\ x-2y+2z=-6\quad \end{matrix}\right.\\ &\textrm{Nilai}\: \: xyz=....\\ &\begin{array}{llll}\\ \textrm{a}.&-96\\ \color{red}\textrm{b}.&-24\\ \textrm{c}.&24\\ \textrm{d}.&32\\ \textrm{e}.&96 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+2y+z=4\: \qquad.......(1)\\ 3x+y+2z=-5\quad\: ......(2)\\ x-2y+2z=-6\quad .......(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llclll}\\ x+2y+z&=4&\left | \times 1 \right |&\: \: x+2y+z&=4\\ 3x+y+2z&=-5&\left | \times 2 \right |&6x+2y+4z&=-10&-\\\hline &&&-5x\: \: \quad-3z&=14&...(4)\\ \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{lll}\\ x+2y+z&=4&\\ x-2y+2z&=-6&+\\\hline 2x\: \: \: \, \quad +3z&=-2&...(5)\\ \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\begin{array}{lll}\\ -5x-3z&=14&\\ 2x+3z&=-2&+\\\hline -3x&=12&\\ \qquad\quad x&=-4&.....(6)\\ \color{red}\textrm{didapat pula}&z&=2......(7) \end{array}\\ &\textrm{Dari persamaan}\: \: (6)\&(7)\: \: \textrm{didapatkan}\\ &\color{red}\begin{aligned}x+2y+z&=4\\ (-4)+2y+2&=4\\ y&=3 \end{aligned}\\ &\textrm{Jadi},\: \: xyz=(-4).(3).(2)=\color{red}-24 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 71.&\textrm{Diketahui sistem persamaan berikut}\\ &\left\{\begin{matrix} x+y-z=1\: \: \qquad\\ 2x-y+2z=9\quad\: \\ x+3y-z=7\: \: \: \: \quad \end{matrix}\right.\\ &\textrm{Nilai}\: \: \displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{3}\\ \textrm{b}.&\displaystyle \frac{3}{4}\\ \color{red}\textrm{c}.&\displaystyle \frac{13}{12}\\ \textrm{d}.&\displaystyle \frac{5}{4}\\ \textrm{e}.&\displaystyle \frac{7}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y-z=1\: \: \qquad....(1)\\ 2x-y+2z=9\quad\: ....(2)\\ x+3y-z=7\: \: \: \: \quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llll}\\ x+y-z&=1&\\ 2x-y+2z&=9&+\\\hline 3x\: \: \qquad+z&=10&...(4)\\ \end{array} \\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{llllll}\\ x+y-z&=1&\left | \times 3 \right |&3x+3y-3z&=3&\\ x+3y-z&=7&\left | \times 1 \right |&\quad x+3y-z&=7&-\\\hline &&&2x\quad \: \: \quad-2z&=-4&\\ &&&\: \: x\quad \: \: \: \: \quad-z&=2&....(5)\\ \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\begin{array}{lll}\\ 3x+z&=10&\\ x-z&=-2&+\\\hline 4x&=8&\\ \qquad\quad x&=2&.....(6)\\ \color{red}\textrm{didapat pula}&z&=4......(7) \end{array} \\ &\textrm{Dari persamaan}\: \: (1)\&(3)\: \: \textrm{didapatkan juga}\\ &\begin{array}{lll}\\ x+y-z&=1&\\ x+3y-z&=-7&-\\\hline \quad -2y&=-6&\\ \qquad\qquad y&=3&....(8) \end{array}\\ &\textrm{Jadi},\: \: \displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\color{red}\frac{13}{12} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 72.&\textrm{Diketahui sistem persamaan berikut}\\ &\left\{\begin{matrix} x+y+z=5\: \: \: \: \quad\\ x+y-4z=10\quad \\ -2x+y+z=0 \quad \end{matrix}\right.\\ &\textrm{Nilai dari}\: \: \displaystyle \frac{xz}{y}\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -\frac{6}{13}\\ \color{red}\textrm{b}.&\displaystyle -\frac{5}{13}\\ \textrm{c}.&\displaystyle -\frac{1}{13}\\ \textrm{d}.&\displaystyle \frac{1}{13}\\ \textrm{e}.&\displaystyle \frac{7}{13} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y+z=5\: \: \: \: \quad.....(1)\\ x+y-4z=10\quad .....(2)\\ -2x+y+z=0 \quad .....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llll}\\ x+y+z&=5&\\ x+y-4z&=10&-\\\hline \: \: \qquad \: \: \: \: \: 5z&=-5&\\ \: \: \qquad\quad \: \: \: z&=-1&...(4)\\ \end{array} \\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{llllll}\\ x+y+z&=5&\\ -2x+y+z&=0&-\\\hline 3x\quad \: \quad&=5&\\ \: \: \quad \: \: \: \: \quad x&=\displaystyle \frac{5}{3}&....(5)\\ \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\color{blue}\begin{aligned}x+y+z&=5\\ \displaystyle \frac{5}{3}+y-1&=5\\ y&=5+1-\displaystyle \frac{5}{3}=\frac{13}{3} \end{aligned}\\ &\textrm{Jadi},\: \: \displaystyle \frac{xz}{y}=\displaystyle \frac{\left ( \displaystyle \frac{5}{3} \right ).(-1)}{\displaystyle \frac{13}{3}}=\color{red}-\displaystyle \frac{5}{13} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 73.&\textrm{Himpunan penyelesaian dari}\\ &\begin{cases} \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{cases}\\ &\textrm{adalah}\: \: \left \{ (x,y,z) \right \},\: \textrm{maka}\: \: x+3z=....\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&\displaystyle \frac{1}{3}\\ \textrm{c}.&1\\ \color{red}\textrm{d}.&3\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\begin{cases} \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{cases}\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llll}\\ \displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}&=8\\ \displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}&=10&-\\\hline -\displaystyle \frac{1}{x}\: \: \: \: \: -\frac{1}{z}&=-2\\ \displaystyle \frac{1}{x}\: \: \: \: \: \: \: \: +\frac{1}{z}&=2&...(4)\\ \end{array} \\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{lllllll}\\ \displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}&=8&\left | \times 2 \right |&\displaystyle \frac{4}{x}+\frac{4}{y}+\frac{8}{z}&=16\\ \displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}&=4&\left | \times 1 \right |&\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}&=4&-\\\hline &&&\displaystyle \frac{2}{x}\: \: \: \: \: +\frac{6}{z}&=12\\ &&\Leftrightarrow &\displaystyle \frac{1}{x}\: \: \: \: \: +\frac{3}{z}&=6&...(5)\\ \end{array}\\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\begin{array}{llll}\\ \displaystyle \frac{1}{x}+\frac{3}{z}&=6\\ \displaystyle \frac{1}{x}+\frac{1}{z}&=2\: \: \: -\\\hline \qquad\displaystyle \frac{2}{z}&=4&\\ \qquad z&=\displaystyle \frac{1}{2}\: \: ......(6)\\ \qquad x&=2-\displaystyle \frac{1}{2}=\displaystyle \frac{3}{2} \end{array}\\ &\textrm{Jadi},\: \: x+3z=\displaystyle \frac{3}{2}+3.\frac{1}{2}=\color{red}3 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 74.&\textrm{Diketahui tiga buah bilangan berturut-turut}\\ &a,\: b,\: \textrm{dan}\: c.\: \textrm{Rata-rata dari ke tiga bilangan}\\ &\textrm{itu adalah 12. Bilangan kedua sama dengan}\\ &\textrm{jumlah bilangan yang lain dikurangi 12}.\\ &\textrm{Jika bilangan ke tiga sama dengan jumlah}\\ &\textrm{bilangan yang lain, maka nilai}\: \: 2a+b-c=....\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle 42\\ \textrm{b}.&-\displaystyle 36\\ \textrm{c}.&-\displaystyle 18\\ \textrm{d}.&-\displaystyle 12\\ \color{red}\textrm{e}.&-\displaystyle 6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\textrm{Model matematika dari persamaan di atas}\\ &\left\{\begin{matrix} a+b+c=36\: \: \qquad....(1)\\ -a+b-x=12\quad\: ....(2)\\ a+b-c=0\: \: \: \: \: \: \: \quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\begin{array}{llll}\\ a+b+c&=36&\\ -a+b-c&=12&+\\\hline \: \: \: \: \: \: \: \: \: \: \: 2b&=48&\\ \: \: \qquad\quad \: \: \: b&=24&...(4)\\ \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\begin{array}{llllll}\\ a+b+c&=36&\\ a+b-c&=0&-\\\hline \quad\qquad2c &=36&\\ \: \: \quad \: \: \: \: \quad c&=18&....(5)\\ \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\color{blue}\begin{aligned}a+b+c&=36\\ a+24+18&=36\\ a&=36-42\\ &=-6 \end{aligned} \\ &\textrm{Jadi},\: \: 2a+b-c=2(-6)+24-18=\color{red}-6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 75.&\textrm{Jumlah uang terdiri atas koin pecahan}\: \: Rp500,00\\ &Rp200,00\: \: dan\: \: Rp100,00\: \: \textrm{dengan nilai total}\\ &Rp100.000,00.\: \textrm{Jika nilai uang pecahan 500-an}\\ &\textrm{setengah dari nilai uang pecahan 200-an, tetapi}\\ &\textrm{tiga kali uang pecahan 100-an, maka banyak koin}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&460\\ \textrm{b}.&440\\ \textrm{c}.&420\\ \textrm{d}.&380\\ \textrm{e}.&350 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Model matematika dari kasus di atas}\\ &\left\{\begin{matrix} A(500)+B(200)+C(100)=100.000\: ....(1)\\ A(500)=\displaystyle \frac{1}{2}B(200)\qquad\qquad\qquad\qquad\: ....(2)\\ A(500)=3C(100)\qquad\qquad\qquad\: \: \, \: \: \: \: \quad ....(3)\end{matrix}\right.\\ &\textrm{Dari persamaan}\: \: (2)\: \textrm{didapatkan}\\ &2A(500)=B(200)\\ &\textrm{Dari persamaan}\: \: (3)\: \textrm{akan didapatkan}\\ &\displaystyle \frac{1}{3}A(500)=C(100)\\ &\textrm{Dari persamaan}\: \: (1)\: \: \textrm{maka},\\ &A(500)+B(200)+C(100)=100.000\\ &A(500)+2A(500)+\displaystyle \frac{1}{3}A(500)=100.000\\ &\displaystyle \frac{10}{3}A(500)=100.000\Leftrightarrow A(500)=30.000\\ &\textrm{maka akan didapatkan}\\ &B(200)=2(30.000)=60.000\\ &C(100)=\displaystyle \frac{1}{3}(30.000)=10.000\\ &\color{red}\begin{cases} A(500) &=30.000\Rightarrow \color{black}A=\displaystyle \frac{30.000}{500}=60 \\ B(200) &=60.000\Rightarrow \color{black}B=\displaystyle \frac{60.000}{200}=300 \\ C(100) &=10.000\Rightarrow \color{black}C=\displaystyle \frac{10.000}{100}=100 \end{cases}\\ &\textrm{Jadi},\: \: A+B+C=60+300+100=\color{red}460 \end{aligned} \end{array}$.


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