Gibbs Vega
11/27/2023 · Middle School
\( \left[ \begin{array} { l l } { 1 } & { 0 } \\ { 3 } & { 1 } \end{array} \right] \times x + \left[ \begin{array} { c } { 5 } \\ { 10 } \end{array} \right] = \left[ \begin{array} { l } { 10 } \\ { 28 } \end{array} \right] \)
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Данное уравнение представляет собой систему линейных уравнений. Чтобы найти \( x \), нам нужно решить эту систему.
Давайте сначала разберемся с уравнением:
\[
\begin{bmatrix}
1 & 0 \\
3 & 1
\end{bmatrix} \times x + \begin{bmatrix}
5 \\
10
\end{bmatrix} = \begin{bmatrix}
10 \\
28
\end{bmatrix}
\]
Это уравнение можно записать как систему двух линейных уравнений:
1. \(1x_1 + 0x_2 + 5 = 10\)
2. \(3x_1 + 1x_2 + 10 = 28\)
Теперь давайте решим эту систему уравнений.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
\(\left\{ \begin{array}{l}1\times x_{1}+0\times x_{2}+5=10\\3x_{1}+1\times x_{2}+10=28\end{array}\right.\)
- step1: Calculate:
\(\left\{ \begin{array}{l}x_{1}+5=10\\3x_{1}+x_{2}+10=28\end{array}\right.\)
- step2: Solve the equation:
\(\left\{ \begin{array}{l}x_{1}=5\\3x_{1}+x_{2}+10=28\end{array}\right.\)
- step3: Substitute the value of \(x_{1}:\)
\(3\times 5+x_{2}+10=28\)
- step4: Simplify:
\(25+x_{2}=28\)
- step5: Move the constant to the right side:
\(x_{2}=28-25\)
- step6: Subtract the numbers:
\(x_{2}=3\)
- step7: Calculate:
\(\left\{ \begin{array}{l}x_{1}=5\\x_{2}=3\end{array}\right.\)
- step8: Check the solution:
\(\left\{ \begin{array}{l}x_{1}=5\\x_{2}=3\end{array}\right.\)
- step9: Rewrite:
\(\left(x_{1},x_{2}\right) = \left(5,3\right)\)
Решение системы уравнений:
\[
\begin{bmatrix}
1 & 0 \\
3 & 1
\end{bmatrix} \times x + \begin{bmatrix}
5 \\
10
\end{bmatrix} = \begin{bmatrix}
10 \\
28
\end{bmatrix}
\]
Дает нам \( x = \begin{bmatrix} 5 \\ 3 \end{bmatrix} \).
Следовательно, \( x = \begin{bmatrix} 5 \\ 3 \end{bmatrix} \).
Quick Answer
\( x = \begin{bmatrix} 5 \\ 3 \end{bmatrix} \).
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