Blog Competition 2019 #2019GANTIBIMBEL

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Kelas 10 - BAB 3 Sistem persamaan Linear

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A. Sistem Persamaan Linear Dua Variabel (SPLDV)
A. 1  Bentuk Umum
\begin{array}{ll}\\ \textbf{\underline{Bentuk Umum}}&:\\\\ &\begin{cases} a_{1}x+b_{1}y &=c_{1} \\\\ a_{2}x+b_{2}y &=c_{2} \end{cases}\\\\ \qquad\qquad \textit{\textbf{dengan}}&\bullet \quad a_{1},\: a_{2},\: b_{1},\: b_{2},\: c_{1}\: \: \textrm{dan}\: \: c_{2}\: \: \textrm{bilangan real}.\\ &\bullet \quad \textrm{Jika}\: \: \begin{cases} c_{1}=c_{2}=0 & ,\: \textrm{maka dikatakan SPLDV }\textit{homogen}\\ c_{1}\neq 0\: \: \: \textrm{atau}& \: c_{2} \neq 0 \: \: ,\: \textrm{maka dikatakan SPLDV }\textit{tidak homogen} \end{cases} \end{array}.
Perhatikan bahwa
\begin{array}{ll}\\ \textrm{Jika}\: \: x=x_{0}\: \: \textrm{dan}\: \: y=y_{0}&\textrm{memenuhi}\\\\ &\begin{cases} a_{1}x+b_{1}y &=c_{1} \\\\ a_{2}x+b_{2}y &=c_{2} \end{cases}\\\\ \qquad \textrm{maka akan berlaku}&\textbf{hubungan}\\\\ &\begin{cases} a_{1}x_{0}+b_{1}y_{0} &=c_{1} \\\\ a_{2}x_{0}+b_{2}y_{0} &=c_{2} \end{cases}\\\\ \textrm{sehingga pasangan terurut}&\left ( x_{0},y_{0} \right )\: \: \textrm{merupakan penyelesaian SPLDV tersebut}. \end{array}.
A. 2  Penyelesaian SPLDV 
Ada banyak cara penyelesaian SPLDV, yaitu:
  • metode grafik
  • cara substitusi
  • cara eliminasi
  • cara gabungan eliminasi-substitusi
  • cara determinan
\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{Dengan}\: &\: \underline{\textbf{determinan}}\: \: (D)\\\\ \textrm{misal}&\: \begin{cases} ax+by=c \\ px+qy=r \end{cases}\\ D&=\begin{vmatrix} a & b\\ p & q \end{vmatrix}=aq-bp\\ & \end{aligned}}\\\hline D_{x}=\begin{vmatrix} c&b\\ r&q \end{vmatrix}&D_{y}=\begin{vmatrix} a & c\\ p & r \end{vmatrix}\\\hline \begin{aligned}&\\ x&=\displaystyle \frac{D_{x}}{D}\\ &=\displaystyle \frac{cq-br}{aq-bp}\\ & \end{aligned}&\begin{aligned}&\\ y&=\displaystyle \frac{D_{y}}{D}\\ &=\displaystyle \frac{ar-cp}{aq-bp}\\ & \end{aligned}\\\hline \end{array}.
A. 3  Kemungkinan Banyak Solusi untuk SPLDV
\begin{array}{|c|c|c|c|}\hline \textbf{Bentuk}&\multicolumn{3}{|c|}{\textbf{Banyak Solusi Penyelesaian SPLDV}}\\\hline &\textrm{Tunggal}&\textrm{Tak berhingga}&\textrm{Tidak memiliki}\\\cline{2-4} \begin{cases} a_{1}x+b_{1}y=c_{1} \\ a_{2}x+b_{2}y=c_{2} \end{cases}&\begin{matrix} a_{1}b_{2}-a_{2}b_{1}\neq 0\\ \textrm{atau}\\ a_{1}b_{2}\neq a_{2}b_{1} \end{matrix}&\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}&\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}\\ &&&\\\hline \end{array}.