The diagram shows part of a straight line obtained by plotting $x^2 y$ against $x$.

1. Calculate the gradient of the straight line

Gradient $m = \frac{7 - 3}{6 - 0} = \frac{4}{6} = \frac{2}{3}$

2. Identify the vertical intercept

The line intersects the vertical axis at $(0, 3)$, so the vertical intercept $c = 3$.

3. Form the equation of the straight line

The vertical axis is $x^2 y$ and the horizontal axis is $x$.
$x^2 y = \frac{2}{3}x + 3$

4. Rearrange to express y in terms of x

$y = \frac{\frac{2}{3}x + 3}{x^2}$
$y = \frac{2}{3x} + \frac{3}{x^2}$

1. Substitute the given expression for $y$ into the linear equation from part (i)

From part (i), the equation of the line is $x^2 y = \frac{2}{3}x + 3$.
Substitute $y = \frac{5}{x^2}$:
$x^2 \left(\frac{5}{x^2}\right) = \frac{2}{3}x + 3$
$5 = \frac{2}{3}x + 3$

2. Solve the resulting linear equation for $x$

$\frac{2}{3}x = 5 - 3$
$\frac{2}{3}x = 2$
$x = 3$

1. Recall the equation from part (i) and manipulate it to form a new linear relationship

From part (i), the equation is $x^2 y = \frac{2}{3}x + 3$.
Divide the entire equation by $x$:
$xy = \frac{2}{3} + \frac{3}{x}$
$xy = 3\left(\frac{1}{x}\right) + \frac{2}{3}$

2. Identify the new variables for the vertical and horizontal axes

Comparing $xy = 3\left(\frac{1}{x}\right) + \frac{2}{3}$ with the linear form $Y = mX + c$:
Vertical axis ($Y$) = $xy$
Horizontal axis ($X$) = $\frac{1}{x}$

3. State the gradient and $y$-intercept of the new equation

Gradient ($m$) = $3$
$y$-intercept ($c$) = $\frac{2}{3}$