Respuesta :

Answer:

x=4 and y=5

Step-by-step explanation:

The given system of equations are

[tex]2x+6y=38[/tex]

[tex]5x-y=15[/tex]

The matrix form is

[tex]\begin{bmatrix}2&6\\ \:5&-1\end{bmatrix}\begin{bmatrix}x\\ \:y\end{bmatrix}=\begin{bmatrix}38\\ \:15\end{bmatrix}[/tex]

Let as assume

[tex]A=\begin{bmatrix}2&6\\ \:5&-1\end{bmatrix}[/tex]

[tex]X=\begin{bmatrix}x\\ \:y\end{bmatrix}[/tex]

[tex]B=\begin{bmatrix}38\\ \:15\end{bmatrix}[/tex]

then,

[tex]AX=B[/tex]

[tex]X=A^{-1}B[/tex]

We know that,

[tex]\begin{bmatrix}a\:&\:b\:\\ c\:&\:d\:\end{bmatrix}^{-1}=\frac{1}{\det \begin{bmatrix}a\:&\:b\:\\ c\:&\:d\:\end{bmatrix}}\begin{bmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{bmatrix}[/tex]

[tex]A^{-1}=\frac{1}{\det \begin{bmatrix}2&6\\ 5&-1\end{bmatrix}}\begin{bmatrix}-1&-6\\ -5&2\end{bmatrix}[/tex]

[tex]A^{-1}=\frac{1}{-32}\begin{bmatrix}-1&-6\\ -5&2\end{bmatrix}[/tex]

[tex]X=\frac{1}{-32}\begin{bmatrix}-1&-6\\ -5&2\end{bmatrix}\begin{bmatrix}38\\ \:15\end{bmatrix}[/tex]

[tex]X=\frac{1}{-32}\begin{bmatrix}\left(-1\right)\cdot \:38+\left(-6\right)\cdot \:15\\ \left(-5\right)\cdot \:38+2\cdot \:15\end{bmatrix}[/tex]

[tex]X=\frac{1}{-32}\begin{bmatrix}-128\\ -160\end{bmatrix}[/tex]

[tex]\begin{bmatrix}x\\ \:y\end{bmatrix}=\begin{bmatrix}4\\ 5\end{bmatrix}[/tex]

Therefore, the value of x is 4 and value of y is 5.

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