下表显示了两个变量 $x$ 和 $y$ 的实验值。

$x$	2	4	6	8
$y$	2.25	0.81	0.47	0.33

1. Calculate the values of $X = \frac{1}{x}$ and $Y = xy$ for each data point.

For $x=2, y=2.25$: $X = \frac{1}{2} = 0.5$, $Y = 2(2.25) = 4.5$
For $x=4, y=0.81$: $X = \frac{1}{4} = 0.25$, $Y = 4(0.81) = 3.24$
For $x=6, y=0.47$: $X = \frac{1}{6} \approx 0.167$, $Y = 6(0.47) = 2.82$
For $x=8, y=0.33$: $X = \frac{1}{8} = 0.125$, $Y = 8(0.33) = 2.64$

2. Plot the points $(X, Y)$ on the graph paper and draw the line of best fit.

Plot the points $(0.5, 4.5)$, $(0.25, 3.24)$, $(0.167, 2.82)$, and $(0.125, 2.64)$ on a grid with $X$-axis representing $\frac{1}{x}$ and $Y$-axis representing $xy$. Draw a straight line passing through or close to these points.

1. Rearrange the given equation to match the vertical axis variable of the graph.

The given equation is $x = \frac{3}{y}$.
Rearranging this gives $xy = 3$.

2. Use the graph to find the corresponding value of the horizontal axis variable.

On the graph of $xy$ against $\frac{1}{x}$, locate the point where the vertical coordinate is $xy = 3$.
Reading the corresponding horizontal coordinate from the straight line graph gives:
$\frac{1}{x} \approx 0.198$

3. Calculate the values of $x$ and $y$.

$x = \frac{1}{0.198} \approx 5.05$
Substitute $x$ back into the equation to find $y$:
$y = \frac{3}{5.05} \approx 0.594$
(Note: Acceptable range depends on the student's plotted graph, typically $x \approx 5.0$ to $5.1$ and $y \approx 0.59$ to $0.60$)

1. Determine the gradient and $Y$-intercept of the straight line graph of $xy$ against $\frac{1}{x}$

From the table, the coordinates $(X, Y) = (\frac{1}{x}, xy)$ are:
$x=2 \implies X=0.5, Y=4.5$
$x=4 \implies X=0.25, Y=3.24$
$x=6 \implies X=0.167, Y=2.82$
$x=8 \implies X=0.125, Y=2.64$

Gradient $m = \frac{4.5 - 3.24}{0.5 - 0.25} = \frac{1.26}{0.25} = 5.04 \approx 5$ (or using points $(0.5, 4.5)$ and $(0.125, 2.64)$, $m = \frac{4.5 - 2.64}{0.5 - 0.125} = \frac{1.86}{0.375} = 4.96 \approx 5$).
Let's assume the exact gradient is $5$.
$Y$-intercept $c$: $4.5 = 5(0.5) + c \implies c = 2$.

2. Form the linear equation relating $Y$ and $X$

The equation of the line is $Y = mX + c$.
Substituting $m = 5$ and $c = 2$, we get:
$Y = 5X + 2$

3. Convert the linear equation back to the original variables $x$ and $y$

Substitute $Y = xy$ and $X = \frac{1}{x}$:
$xy = 5\left(\frac{1}{x}\right) + 2$
$xy = \frac{5}{x} + 2$
Divide by $x$:
$y = \frac{5}{x^2} + \frac{2}{x}$