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At this point this is mostly LaTeX practice, but why not?!?...

Related video is at:
http://www.khanacademy.org/math/algebra/solving-linear-equations/v/order-of-operations-example

1b.
\begin{eqnarray*}
ans & = & 2 + 7 \times 11 - 12 \div 3 \\
ans & = & 2 + (7 \times 11) - (12 \div 3) \\
ans & = & 2 + 77 - 4 \\
ans & = & 79 - 4 \\
ans & = & 75
\end{eqnarray*}

1d.
\begin{eqnarray*}
ans & = & \frac{2 \times (3 + (2 - 1))}{4 - (6 + 2)} - (3 - 5) \\
ans & = & \frac{2 \times (3 + 1)}{4 - (6 + 2)} - (3 - 5) \\
ans & = & \frac{2 \times 4}{4 - (6 + 2)} - (3 - 5) \\
ans & = & \frac{8}{4 - 8} - (3 - 5) \\
ans & = & \frac{8}{-4} - (3 - 5) \\
ans & = & -2 - (3 - 5) \\
ans & = & -2 - -2 \\
ans & = & -2 + 2 \\
ans & = & 0
\end{eqnarray*}

2b.
\begin{eqnarray*}
ans & = & 2y^2 \mbox{ when } x = 1 \mbox{ and } y = 5 \\
ans & = & 2 \times (5^2) \\
ans & = & 2 \times 25 \\
ans & = & 50
\end{eqnarray*}

2d.
\begin{eqnarray*}
ans & = & (y^2 - x)^2 \mbox{ when } x = 2 \mbox{ and } y = 1 \\
ans & = & (1^2 - 2)^2 \\
ans & = & (1 - 2)^2 \\
ans & = & (-1)^2 \\
ans & = & 1
\end{eqnarray*}

3b.
\begin{eqnarray*}
ans & = & \frac{z^2}{x + y} + \frac{x^2}{x - y} \mbox{ when } x = 1, y = -2, \mbox{ and } z = 4 \\
ans & = & \frac{4^2}{1 + -2} + \frac{1^2}{1 - -2} \\
ans & = & \frac{16}{-1} + \frac{1}{3} \\
ans & = & -16 + \frac{1}{3} \\
ans & = & \frac{-48}{3} + \frac{1}{3} \\
ans & = & \frac{-47}{3}
\end{eqnarray*}

3d.
\begin{eqnarray*}
\end{eqnarray*}

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