اكتب: مجموعة حل للنظام: $\begin{array}{r}5x-6y=0\\ x+2y=4\end{array}\}$

$$\begin{gathered} \begin{array}{rl}& 1\times [5x-6y=0]...\left(1\right)\\ & 5\times [x+2y=4]...\left(2\right)\end{array} \\ \begin{array}{rl}& 5x-6y=0...\left(1\right)\\ & \mp 5x\mp 10y=\mp 20...\left(2\right)\end{array}\mathrm{بالطرح} \\ -16y=-20\Rightarrow \frac{-16y}{-16}=\frac{-20}{-16}\Rightarrow y=\frac{5}{4} \end{gathered}$$

نعوض في المعادلة (1)

$\begin{array}{c}5x-6y=0\Rightarrow 5x-6\left(\frac{5}{4}\right)=0\Rightarrow 5x-\frac{30}{4}=0\Rightarrow 5x=\frac{30}{4}\Rightarrow x=\frac{30}{20}=\frac{3}{2}\\ S=\left\{(\frac{3}{2},\frac{5}{4})\right\}\end{array}$

فكر واكتب

تحدٍ: جد مجموعة الحل للنظام:

$\begin{array}{l}\frac{2}{6}x-\frac{1}{3}y=1\\ \frac{1}{2}x+\frac{1}{2}y=3\end{array}\}$

$$\begin{gathered} \begin{array}{rl}& [\frac{2}{6}x-\frac{1}{3}y=1]\times 6\dots \dots \left(1\right)\\ & [\frac{1}{2}x+\frac{1}{2}y=3]\times 2\dots \dots \left(2\right)\\ & [\frac{12}{6}x-\frac{6}{3}y=6]\dots \dots \left(1\right)\\ & [\frac{2}{2}x+\frac{2}{2}y=6]\dots \dots \left(2\right)\\ & 1\times [2x-2y=6]\dots \dots \left(1\right)\end{array} \\ \begin{array}{rl}& 2\times [x+y=6]....\left(2\right)\\ & 2x-2y=6....\left(1\right)\\ & 2x+2y=12...\left(2\right) \mathrm{بالجمع}\\ & 4x=18\Rightarrow \frac{4x}{4}=\frac{18}{4}\Rightarrow x=\frac{9}{2} \left(2 \mathrm{معادلة} \mathrm{في} \mathrm{نعوض}\right)\\ & x+y=6\Rightarrow \frac{9}{2}+y=6\Rightarrow y=6-\frac{9}{2}=\frac{12-9}{2}=\frac{3}{2}\\ & S=\left\{(\frac{2}{2},\frac{3}{2})\right\}\end{array} \end{gathered}$$

أصحح الخطأ: قال أحمد إن مجموعة حل للنظام:

$\begin{array}{l}2x+3y=6\\ 3x+2y=1\end{array}\}$ هي المجموعة $\left\{(\frac{5}{16},\frac{5}{9})\right\}$

اكتشف خطأ أحمد وصححه

$\begin{array}{rl}& 2x+3y=6...\left(1\right)\\ & 3x+2y=1...\left(2\right)\end{array}$

• نضرب المعادلة (2) في 3
• نضرب المعادلة (2) في 2

$$\begin{gathered} \begin{array}{c}6x+9x=18...\left(1\right)\\ \mp 6x\mp 4y=\mp 2...\left(2\right)\end{array}\mathrm{بالطرح} \\ 5y=16\Rightarrow y=\frac{16}{5} \end{gathered}$$

نعوض عن قيمة y بالمعادلة (1) لإيجاد قيمة x

$2x+3\left(\frac{16}{5}\right)=6\Rightarrow 2x=3\Rightarrow -\frac{48}{5}+6\Rightarrow 2x=-\frac{18}{5}\Rightarrow x=\cdot \frac{18}{10}=\cdot \frac{9}{5}$

مجموعة الحل للنظام $S=\left\{(-\frac{9}{5},\frac{16}{5})\right\}$

اكتب: مجموعة حل للنظام: $\begin{array}{r}5x-6y=0\\ x+2y=4\end{array}\}$

$$\begin{gathered} \begin{array}{rl}& 1\times [5x-6y=0]...\left(1\right)\\ & 5\times [x+2y=4]...\left(2\right)\end{array} \\ \begin{array}{rl}& 5x-6y=0...\left(1\right)\\ & \mp 5x\mp 10y=\mp 20...\left(2\right)\end{array}\mathrm{بالطرح} \\ -16y=-20\Rightarrow \frac{-16y}{-16}=\frac{-20}{-16}\Rightarrow y=\frac{5}{4} \end{gathered}$$

نعوض في المعادلة (1)

$\begin{array}{c}5x-6y=0\Rightarrow 5x-6\left(\frac{5}{4}\right)=0\Rightarrow 5x-\frac{30}{4}=0\Rightarrow 5x=\frac{30}{4}\Rightarrow x=\frac{30}{20}=\frac{3}{2}\\ S=\left\{(\frac{3}{2},\frac{5}{4})\right\}\end{array}$