گام به گام تمرین صفحه 112 درس 6 ریاضی نهم (خط و معادله های خطی)

1

\(\begin{array}{l}\left\{ \begin{array}{l}2(x - y) + 3y = 4\\\\3x - 2(2x - y) = 7\end{array} \right. \Rightarrow \left\{ \begin{array}{l}2x - 2y + 3y = 4\\\\3x - 4x + 2y = 7\end{array} \right.\\\\\left\{ \begin{array}{l}2x + y = 4\\\\ - x + 2y = 7\end{array} \right. \Rightarrow \begin{array}{*{20}{c}}{}\\{}\\{2 \times }\end{array}\left\{ \begin{array}{l}2x + y = 4\\\\ - x + 2y = 7\end{array} \right.\\\\\left\{ \begin{array}{l}2x + y = 4\\\\ - 2x + 4y = 14\end{array} \right.\\\,\,\overline {5y = 18 \Rightarrow y = \frac{{18}}{5}} \\\\ - x + 2 \times \frac{{18}}{5} = 7 \Rightarrow - x = - \frac{1}{5}\\\\x = \frac{1}{5}\end{array}\)

2

\(\begin{array}{l}\left\{ \begin{array}{l}\frac{{x - 1}}{2} - \frac{{y - 1}}{3} = \frac{1}{6}\\\\x + y = 4\end{array} \right. \Rightarrow \begin{array}{*{20}{c}}{6 \times }\\{}\\{}\end{array}\left\{ \begin{array}{l}\frac{{x - 1}}{2} - \frac{{y - 1}}{3} = \frac{1}{6}\\\\x + y = 4\end{array} \right.\\\\\left\{ \begin{array}{l}3(x - 1) - 2(y - 1) = 1\\\\x + y = 4\end{array} \right. \Rightarrow \left\{ \begin{array}{l}3x - 3 - 2y + 2 = 1\\\\x + y = 4\end{array} \right.\\\\\left\{ \begin{array}{l}3x - 2y = 2\\\\x + y = 4\end{array} \right. \Rightarrow \begin{array}{*{20}{c}}{}\\{}\\{2 \times }\end{array}\left\{ \begin{array}{l}3x - 2y = 2\\\\x + y = 4\end{array} \right.\\\\\left\{ \begin{array}{l}3x - 2y = 2\\\\2x + 2y = 8\end{array} \right.\\\,\,\,\,\,\,\overline {5x = 10 \Rightarrow x = 2} \\\\2 + y = 4 \Rightarrow y = 2\end{array}\)

دو عدد توان دار با پایه های مختلف در صورتی با هم برابر می شوند که توان آن ها برابر صفر باشد. پس \(2x - y - 2 = 0\) و \(x + y - 1 = 0\) خواهد بود.

\(\begin{array}{l}\left\{ \begin{array}{l}x + y - 1 = 0\\\\2x - y - 2 = 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x + y = 1\\\\2x - y = 2\end{array} \right.\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline {3x = 3 \Rightarrow x = 1} \\\\x + y = 1 \Rightarrow 1 + y = 1 \Rightarrow y = 0\end{array}\)

ابتدا نقطه تقاطع دو خط را بدست می آوریم:

\(\begin{array}{l}\left\{ \begin{array}{l}x - y = 1\\\\x + y = 1\end{array} \right.\\\,\,\,\,\,\,\overline {2x = 2 \Rightarrow x = 1} \\\\x + y = 1 \Rightarrow 1 + y = 1 \Rightarrow y = 0\end{array}\)

مختصات نقطۀ تقاطع، \(\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\) می باشد. حال به کمک این نقطه، معادلۀ خط مورد نظر را بدست می آوریم:

\(\begin{array}{l}y = - \frac{2}{3}x + b\,\,\,\mathop \Rightarrow \limits^{\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]} 0 = - \frac{2}{3}(1) + b\\\\ - \frac{2}{3} + b = 0 \Rightarrow b = \frac{2}{3}\\\\y = - \frac{2}{3}x + \frac{2}{3}\end{array}\)

هر سه خط یک جواب مشترک دارند. همه از نقطه \(\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\) می گذرد.

تعداد گاوها را x و تعداد شترمرغ ها را y در نظر می گیریم:

 \(\begin{array}{l}\left\{ \begin{array}{l}x + y = 20\\\\4x + 2y = 56\end{array} \right. \Rightarrow \begin{array}{*{20}{c}}{ - 2 \times }\\{}\\{}\end{array}\left\{ \begin{array}{l}x + y = 20\\\\4x + 2y = 56\end{array} \right.\\\\\left\{ \begin{array}{l} - 2x - 2y = - 40\\\\4x + 2y = 56\end{array} \right.\\\,\,\,\,\,\,\overline {2x = 16 \Rightarrow x = 8} \\\\8 + y = 20 \Rightarrow y = 12\end{array}\)

در این مزرعه 8 رأس گاو و 12 رأس شترمرغ وجود دارد.