[tex]\huge{x=5,y=3}[/tex]
Step-by-step explanation:
Given system of equations, [tex]4x\textrm{ }-\textrm{ }2y\textrm{ }=\textrm{ }14\textrm{ , }x\textrm{ }-\textrm{ }6y\textrm{ }=\textrm{ }-13[/tex]
The system can be equivalenty represented as a matrix product as [tex]\left[\begin{array}{cc}4&-2\\1&-6\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]\textrm{ = }\left[\begin{array}{c}14\\-13\end{array}\right][/tex]
This can be simplifies, [tex]Ax=B[/tex]⇒[tex]x=A^{-1} B[/tex]
For a matrix [tex]A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right][/tex] of size [tex]2[/tex] × [tex]2[/tex], [tex]A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d&-b\\-c&a\end{array}\right][/tex].
So [tex]\left[\begin{array}{cc}4&-2\\1&-6\end{array}\right]^{-1}=\frac{1}{-22}\left[\begin{array}{cc}-6&2\\-1&4\end{array}\right][/tex]
[tex]\left[\begin{array}{c}x\\y\end{array}\right]=\frac{1}{-22}\left[\begin{array}{cc}-6&2\\-1&4\end{array}\right]\left[\begin{array}{c}14\\-13\end{array}\right]=\left[\begin{array}{c}5\\3\end{array}\right][/tex]
∴[tex]x=5,y=3[/tex]