1,{x+y=1. 2, {x-y=-2 3,{x+y=5. 4, {x+y=-1
{2x-3y=12. {2x+y=5 {3x-y=3. {2x+y=0
5,{x-y=1
{3x-4y=2. Giải các hpt sau bằng phương pháp thế.
1,{x+y=1. 2, {x-y=-2 3,{x+y=5. 4, {x+y=-1
{2x-3y=12. {2x+y=5 {3x-y=3. {2x+y=0
5,{x-y=1
{3x-4y=2. Giải các hpt sau bằng phương pháp thế.
Đáp án:
5) \(\left\{ \begin{array}{l}
y = 1\\
x = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\left\{ \begin{array}{l}
x + y = 1\\
2x – 3y = 12
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1 – y\\
2\left( {1 – y} \right) – 3y = 12
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1 – y\\
– 5y = 10
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 3\\
y = – 2
\end{array} \right.\\
2)\left\{ \begin{array}{l}
x – y = – 1\\
2x + y = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = y – 1\\
2\left( {y – 1} \right) + y = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = y – 1\\
3y = 7
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{7}{3}\\
x = \dfrac{4}{3}
\end{array} \right.\\
3)\left\{ \begin{array}{l}
x + y = 5\\
3x – y = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 3x – 3\\
x + 3x – 3 = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
4x = 8\\
y = 3x – 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 2\\
y = 3
\end{array} \right.\\
4)\left\{ \begin{array}{l}
x + y = – 1\\
2x + y = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = – 1 – x\\
2x – 1 – x = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1\\
y = – 2
\end{array} \right.\\
5)\left\{ \begin{array}{l}
x – y = 1\\
3x – 4y = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = y + 1\\
3\left( {y + 1} \right) – 4y = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = y + 1\\
– y = – 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 1\\
x = 2
\end{array} \right.
\end{array}\)