Section 2.2 Examples
To help make these expectations clear, I’ve produced a number of examples. In each of these examples, Solution 1 is technically correct but very poorly presented. I would not give full marks for this solution. Solution 2 contains the same technical details, but is much more clearly presented and includes references. Solution 2 would be given full marks.
Solve this system of equations.
\begin{align*}
x + y \amp = 1 \\
x + z \amp = 0 \\
y + z \amp = 2
\end{align*}
Solution 1.
Solution 2.
\begin{align*}
x + y \amp = 1 \amp \amp \amp \amp\\
x \amp 1 - y \amp \amp \amp \amp\\
1 - y + z \amp = 0 \amp \amp \amp \amp\\
y \amp = 1 + z \amp \amp \amp \amp\\
1 + z + z \amp = 2 \amp \amp \amp \amp\\
2z \amp = 1 \amp \amp \amp \amp\\
z \amp = \frac{1}{2}
\amp y \amp = \frac{3}{2}
\amp x \amp = \frac{-1}{2}
\end{align*}
I can solve for \(x\) in the first equation.
\begin{equation*}
x = 1 - y
\end{equation*}
Then I put this expression for \(x\) into the second equation.
\begin{equation*}
1 - y + z = 0
\end{equation*}
Now I can solve for \(y\) in this second equation.
\begin{equation*}
y = 1 + z
\end{equation*}
I can put this expression for \(y\) into the third equation.
\begin{equation*}
1 + z + z = 2
\end{equation*}
Now \(z\) is the only variable, so I can simply solve for it.
\begin{equation*}
z = \frac{1}{2}
\end{equation*}
I can use this value for \(z\) to determine \(y\) using a previous expression.
\begin{equation*}
y = 1 + z = \frac{3}{2}
\end{equation*}
Finally, I can use this value for \(y\) to determine \(x\) using the first equation.
\begin{equation*}
x = 1 - y = \frac{-1}{2}
\end{equation*}
That gives me the full solution.
\begin{align*}
\amp x = \frac{-1}{2} \amp \amp y = \frac{3}{2} \amp \amp
z = \frac{1}{2}
\end{align*}
Example 2.2.2.
Solution 1.
Solution 2.
\begin{align*}
9 \amp = 3^2 \\
3 \times 4 \amp = 12 \\
12 \amp \rightarrow 1200\\
\frac{1200}{15} \amp = 80
\end{align*}
I have been given the area, but I need to know the perimeter. This is a square, so the area is the square of the side length. Since the area is \(9m^2\text{,}\) the side length must be \(3m\text{,}\) since \(3^3 = 9\text{.}\) There are four sides, each with length \(3m\text{,}\) so the total perimeter is \(12m\text{.}\) The planks are in \(cm\text{,}\) so I need to write \(12m\) as \(1200cm\text{.}\) Assume that the fence is built of planks that touch without any gaps, I can just divide the perimeter by the plank width.
\begin{equation*}
\frac{1200}{15} = 80
\end{equation*}
I need 80 planks. It may not be necessary in the solution, but notice, also, that this gives 20 planks for each side, divided evenly, so I don’t have to worry about using part of a plank for one side and part for another side.
Example 2.2.3.
What are the roots of \(4x^2 - 7x - 23 = 0 \text{?}\)
Solution 1.
Solution 2.
\begin{align*}
x \amp = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\
x \amp = \frac{-7 \pm \sqrt{7^2 - 4(4)(23)}}{2(4)} \\
x \amp = \frac{7 \pm \sqrt{417}}{8} \\
x \amp \doteq -1.68, 3.43
\end{align*}
\begin{align*}
x \amp = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\
x \amp = \frac{-7 \pm \sqrt{7^2 - 4(4)(23)}}{2(4)} \\
x \amp = \frac{7 \pm \sqrt{417}}{8}
\end{align*}
These are the exact solutions. Approximate values, by calculator, are \(x \doteq -1.68, 3.43\text{.}\)
OR
Approximate values using Wolfram Alpha: \(x \doteq 1.68,
3.43\text{.}\)
OR
Approximate values using Symbolab: \(x \doteq 01.68, 3.43\text{.}\)
OR
Approximate values using Desmos: \(x \doteq -1.68, 3.43\text{.}\)
Example 2.2.4.
Find the approximate values of \(\sin (5)\text{,}\) \(\ln
(23)\text{,}\) \(\sqrt{41}\text{,}\) and \(\frac{75}{342}\text{.}\)
Solution 1.
Solution 2.
\begin{align*}
\sin (5) \amp \doteq 0.087 \\
\ln (23) \amp \doteq 3.135\\
\sqrt{41} \amp \doteq 6.403 \\
\frac{75}{342} \amp \doteq 0.219
\end{align*}
Throughout this assignment, use of Wolfram Alpha for calcaultions is indicated by (WA).
\begin{align*}
\sin (5) \amp \doteq 0.087 \amp \amp (\text{WA})\\
\ln (23) \amp \doteq 3.135 \amp \amp (\text{WA})\\
\sqrt{41} \amp \doteq 6.403 \amp \amp (\text{WA})\\
\frac{75}{342} \amp \doteq 0.219 \amp \amp (\text{WA})
\end{align*}
