جد مجموعة الحل للنظام باستعمال الحذف لكل مما يأتي:

$\begin{array}{l}6y-2x-8=0\\ y+x-12=0\end{array}\}$

• $\left\{\right(8,-4\left)\right\}$
• $\mathbf{\left\{}\mathbf{\right(}\mathbf{8}\mathbf{,}\mathbf{4}\mathbf{\left)}\mathbf{\right\}}$
• $\left\{\right(-8,4\left)\right\}$
• $\left\{\right(-8,-4\left)\right\}$

[3-1] حل نظام من معادلتين خطيتين بمتغيرين

اختر الإجابة الصحيحة لكل مما يأتي:

جد مجموعة حل للنظام بيانياً:

$\begin{array}{l}y=4x-6\\ y=x\end{array}\}$

• $\left\{\right(-2,-2\left)\right\}$
• $\left\{\right(-2,2\left)\right\}$
• $\left\{\right(2,-2\left)\right\}$
• د- $\mathbf{\left\{}\mathbf{\right(}\mathbf{2}\mathbf{,}\mathbf{2}\mathbf{\left)}\mathbf{\right\}}$

$\begin{array}{l}y=x-3\\ y=3-x\end{array}\}$

• $\left\{\right(-3,0\left)\right\}$
• $\mathbf{\left\{}\mathbf{\right(}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{\left)}\mathbf{\right\}}$
• $\left\{\right(0,-3\left)\right\}$
• $\left\{\right(0,3\left)\right\}$

جد مجموعة الحل للنظام باستعمال التعويض لكل مما يأتي:

$\begin{array}{c}3x+4y=26\\ 5x-2y=0\end{array}\}$

• $\mathbf{\left\{}\mathbf{\right(}\mathbf{2}\mathbf{,}\mathbf{5}\mathbf{\left)}\mathbf{\right\}}$
• $\left\{\right(-2,-5\left)\right\}$
• $\left\{\right(2,-5\left)\right\}$
• $\left\{\right(-2,5\left)\right\}$

$\begin{array}{c}y=6x+12\\ 3y=2x-8\end{array}\}$

• $\left\{(\frac{-11}{4},\frac{9}{2})\right\}$
• $\left\{(\frac{11}{4},\frac{-9}{2})\right\}$
• $\mathbf{\left\{}\mathbf{(}\frac{\mathbf{-}\mathbf{11}}{\mathbf{4}}\mathbf{,}\frac{\mathbf{-}\mathbf{9}}{\mathbf{2}}\mathbf{)}\mathbf{\right\}}$
• $\left\{(\frac{11}{4},\frac{9}{2})\right\}$

$\begin{array}{l}\frac{3x}{4}-\frac{y}{2}=4\\ \frac{y}{2}-\frac{x}{4}=2\end{array}\}$

• $\left\{\right(12,-10\left)\right\}$
• $\left\{\right(-12,-10\left)\right\}$
• $\mathbf{\left\{}\mathbf{\right(}\mathbf{12}\mathbf{,}\mathbf{10}\mathbf{\left)}\mathbf{\right\}}$
• $\left\{\right(-12,10\left)\right\}$

جد مجموعة الحل للنظام باستعمال الحذف لكل مما يأتي:

$\begin{array}{c}7x-4y=12\\ 3x-y=5\end{array}\}$

• $\left\{(-\frac{8}{5},\frac{1}{5})\right\}$
• $\left\{(-\frac{8}{5},-\frac{1}{5})\right\}$
• $\left\{(\frac{8}{5},\frac{1}{5})\right\}$
• $\mathbf{\left\{}\mathbf{(}\frac{\mathbf{8}}{\mathbf{5}}\mathbf{,}\mathbf{-}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{)}\mathbf{\right\}}$

$\begin{array}{l}6y-2x-8=0\\ y+x-12=0\end{array}\}$

• $\left\{\right(8,-4\left)\right\}$
• $\mathbf{\left\{}\mathbf{\right(}\mathbf{8}\mathbf{,}\mathbf{4}\mathbf{\left)}\mathbf{\right\}}$
• $\left\{\right(-8,4\left)\right\}$
• $\left\{\right(-8,-4\left)\right\}$

$\begin{array}{l}\frac{2}{3}x-\frac{1}{6}y=2\frac{1}{3}\\ \frac{1}{4}x-\frac{1}{2}y=3\frac{1}{2}\end{array}\}$

• $\left\{\right(-2,-6\left)\right\}$
• $\left\{\right(-2,6\left)\right\}$
• $\mathbf{\left\{}\mathbf{\right(}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{6}\mathbf{\left)}\mathbf{\right\}}$
• $\left\{\right(2,6\left)\right\}$