Đáp án:
13) \(\left\{ \begin{array}{l}
y = - 3\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\left\{ \begin{array}{l}
3x + y = 5\\
x - 2y = - 3
\end{array} \right. \to \left\{ \begin{array}{l}
6x + 2y = 10\\
x - 2y = - 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
7x = 7\\
y = 5 - 3x
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1\\
y = 2
\end{array} \right.\\
2)\left\{ \begin{array}{l}
2x + y = 1\\
3x + 4y = - 1
\end{array} \right. \to \left\{ \begin{array}{l}
8x + 4y = 4\\
3x + 4y = - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
5x = 5\\
y = 1 - 2x
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1\\
y = - 1
\end{array} \right.\\
3)\left\{ \begin{array}{l}
2x + 3y = 2\\
3x - 3y = \dfrac{1}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
5x = \dfrac{5}{2}\\
y = x - \dfrac{1}{6}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
y = \dfrac{1}{3}
\end{array} \right.\\
4)DK:x \ne 0;y \ne 0\\
\left\{ \begin{array}{l}
x = y - 1\\
\dfrac{2}{{y - 1}} + \dfrac{3}{y} = 2\left( 1 \right)
\end{array} \right.\\
\left( 1 \right) \to 2y + 3y - 3 = 2\left( {{y^2} - y} \right)\\
\to 2{y^2} - 7y + 3 = 0\\
\to \left[ \begin{array}{l}
y = 3\\
y = \dfrac{1}{2}
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = - \dfrac{1}{2}
\end{array} \right.\\
5)\left\{ \begin{array}{l}
6x + 3y = 15\\
x - 3y = - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
7x = 14\\
y = 5 - 2x
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 2\\
y = 1
\end{array} \right.\\
6)\left\{ \begin{array}{l}
2x - 2 + y = 3\\
x - 3y = - 8
\end{array} \right. \to \left\{ \begin{array}{l}
2x + y = 5\\
x - 3y = - 8
\end{array} \right.\\
\to \left\{ \begin{array}{l}
6x + 3y = 15\\
x - 3y = - 8
\end{array} \right.\\
\to \left\{ \begin{array}{l}
7x = 7\\
y = 5 - 2x
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1\\
y = 3
\end{array} \right.\\
7)\left\{ \begin{array}{l}
8x + 2y = 10\\
3x - 2y = - 12
\end{array} \right.\\
\to \left\{ \begin{array}{l}
11x = - 2\\
y = 5 - 4x
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = - \dfrac{2}{{11}}\\
y = \dfrac{{63}}{{11}}
\end{array} \right.\\
8)\left\{ \begin{array}{l}
4x + 7y = 18\\
21x - 7y = 7
\end{array} \right.\\
\to \left\{ \begin{array}{l}
25x = 25\\
y = 3x - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1\\
y = 2
\end{array} \right.\\
9)\left\{ \begin{array}{l}
9x + 6y = 18\\
2x - 6y = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
11x = 22\\
y = \dfrac{{x - 2}}{3}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 2\\
y = 0
\end{array} \right.\\
10)\left\{ \begin{array}{l}
y = x - 4\\
x = - \dfrac{3}{2}
\end{array} \right. \to \left\{ \begin{array}{l}
x = - \dfrac{3}{2}\\
y = - \dfrac{{11}}{2}
\end{array} \right.\\
11)\left\{ \begin{array}{l}
6x + 2y = 18\\
x - 2y = - 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
7x = 14\\
y = 9 - 3x
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 2\\
y = 3
\end{array} \right.\\
12)\left\{ \begin{array}{l}
3x - 2y = 1\\
4x + 2y = - 8
\end{array} \right.\\
\to \left\{ \begin{array}{l}
7x = - 7\\
y = - 4 - 2x
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = - 1\\
y = - 2
\end{array} \right.\\
13)\left\{ \begin{array}{l}
- 2x + 6y = - 20\\
2x + y = - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
7y = - 21\\
x = 3y + 10
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = - 3\\
x = 1
\end{array} \right.\\
15)DK:x \ne 0;y \ne 0\\
\left\{ \begin{array}{l}
6x + 6y = 5xy\\
4y - 3x = xy
\end{array} \right.\\
\to \left\{ \begin{array}{l}
6x + 6y = 5xy\\
- 15x + 20y = 5xy
\end{array} \right.\\
\to \left\{ \begin{array}{l}
21x - 14y = 0\\
4y - 3x = xy
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{2}{3}y\\
- 3.\dfrac{2}{3}y + 4y = \dfrac{2}{3}y.y
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{2}{3}y\\
\dfrac{2}{3}{y^2} - 2y = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = 0\left( l \right)\\
y = 3
\end{array} \right. \to x = 2
\end{array}\)
