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.