已知变量 x 和 y 满足关系式 $y = px + \frac{q}{x}$，其中 p 和 q 为常数。

1. Calculate the values of $X = x^2$ and $Y = xy$ for each pair of $(x, y)$.

For $x=1, y=8.55$: $x^2 = 1, xy = 8.55$
For $x=2, y=6.20$: $x^2 = 4, xy = 12.40$
For $x=3, y=5.32$: $x^2 = 9, xy = 15.96$
For $x=4, y=5.72$: $x^2 = 16, xy = 22.88$
For $x=5, y=6.00$: $x^2 = 25, xy = 30.00$

2. Plot the points and draw the line of best fit.

Plot the points $(1, 8.55)$, $(4, 12.40)$, $(9, 15.96)$, $(16, 22.88)$, and $(25, 30.00)$ on the grid with $x^2$ on the horizontal axis and $xy$ on the vertical axis. Draw a straight line of best fit passing as close as possible to all plotted points.

1. Convert the given non-linear equation into a linear form $Y = mX + c$.

Given $y = px + \frac{q}{x}$.
Multiply throughout by $x$:
$xy = px^2 + q$
This is in the form $Y = mX + c$, where $Y = xy$, $X = x^2$, gradient $m = p$, and vertical intercept $c = q$.

2. Determine the value of $p$ by finding the gradient of the line of best fit.

Using two points on the line of best fit, for example $(4, 11.8)$ and $(25, 30.3)$ (values may vary slightly based on the drawn line):
$p = \text{gradient} = \frac{30.3 - 11.8}{25 - 4} = \frac{18.5}{21} \approx 0.88$

3. Determine the value of $q$ by finding the vertical intercept of the line of best fit.

From the graph, the line intersects the $xy$-axis (where $x^2 = 0$) at approximately $8.25$.
$q = \text{vertical intercept} \approx 8.25$

1. Relate the given equation to the variables plotted on the graph

The given equation is $px + \frac{q}{x} = \frac{20}{x}$.
Since $y = px + \frac{q}{x}$, this is equivalent to $y = \frac{20}{x}$.
Multiplying by $x$ gives $xy = 20$.

2. Use the graph to find the value of $x^2$ when $xy = 20$ and solve for $x$

On the graph of $xy$ against $x^2$, locate $xy = 20$ on the vertical axis.
Read the corresponding value on the horizontal axis ($x^2$).
From the line of best fit, when $xy = 20$, $x^2 \approx 13.1$.
$x = \sqrt{13.1} \approx 3.62$ (taking the positive root).

1. Rearrange the given equation to isolate $x$ terms on one side to prepare for division

$y = px + \frac{q}{x}$
$xy = px^2 + q$
$px^2 = xy - q$
$x^2 = \frac{1}{p}xy - \frac{q}{p}$

2. Divide by $xy$ to obtain the required variables $\frac{x}{y}$ and $\frac{1}{xy}$

Divide by $xy$:
$\frac{x^2}{xy} = \frac{1}{p} - \frac{q}{p}\left(\frac{1}{xy}\right)$
$\frac{x}{y} = -\frac{q}{p}\left(\frac{1}{xy}\right) + \frac{1}{p}$

3. Identify the gradient and vertical intercept

Comparing with $Y = mX + c$ where $Y = \frac{x}{y}$ and $X = \frac{1}{xy}$:
Gradient $m = -\frac{q}{p}$
Vertical intercept $c = \frac{1}{p}$