Given
[tex]\begin{bmatrix}3 & {-1} \\ {1} & {2}\end{bmatrix}\begin{pmatrix}x \\ \:y\end{pmatrix}=\begin{pmatrix}8 \\ \:5\end{pmatrix}[/tex]
Explanation
From the above, the equation can be rewritten as
[tex]\begin{gathered} 3\times1-1\times y=8 \\ 1\times x+2\times y=5 \\ \begin{pmatrix}3x-y \\ x+2y\end{pmatrix}=\begin{pmatrix}8 \\ 5\end{pmatrix} \\ isolate\text{ x in eqn 1} \\ \mathrm{Substitute\:}x=\frac{8+y}{3} \\ \begin{bmatrix}\frac{8+y}{3}+2y=5\end{bmatrix} \\ 8+y+6y=15 \\ 7y=15-8 \\ 7y=7 \\ y=\frac{7}{7} \\ y=1 \\ \mathrm{For\:}x=\frac{8+y}{3}\text{ substitute y=1} \\ x=\frac{8+1}{3} \\ x=\frac{9}{3} \\ x=3 \end{gathered}[/tex]
Answer: Option A
