Question D.4: Solve for x and y: x = 3 - 4y, 20y=-1 + 3x.

In this case, the equations must first be placed in the format of Eqs. (D.1a) and (D.1b):

\begin{matrix} Col. 1  Col. 2  Col. 3\\\hline \\ a_{1} x+b_{1} y=c_{1} \\a_{2} x+b_{2} y=c_{2}\end{matrix}                                      \begin{matrix} (D.1a) \\ (D.1b) \end{matrix}

\begin{aligned}x+4 y &=3 \\-3 x+20 y &=-1 \\\hline\end{aligned}.

x=\frac{\left|\begin{array}{rr}3 & 4 \\-1 & 20\end{array}\right|}{\left|\begin{array}{rr}1 & 4 \\-3 & 20\end{array}\right|}=\frac{(3)(20)-(-1)(4)}{(1)(20)-(-3)(4)}.

=\frac{60+4}{20+12}=\frac{64}{32}=2.

y=\frac{\left|\begin{array}{rr}1 & 3 \\-3 & -1\end{array}\right|}{32}=\frac{(1)(-1)-(-3)(3)}{32}.

=\frac{-1+9}{32}=\frac{8}{32}=\frac{1}{4}.