The diagram shows part of the curve $y = \frac{15}{(0.5x+1)}$ and a straight line $y = 1.25x - 10$. The straight line intersects the $y$-axis at $P$, the $x$-axis at $Q$ and the curve at $R$.

1. Find the coordinates of $P$ by setting $x = 0$

For the $y$-intercept $P$, substitute $x = 0$ into the equation of the line:
$y = 1.25(0) - 10$
$y = -10$
So, $P$ is $(0, -10)$.

2. Find the coordinates of $Q$ by setting $y = 0$

For the $x$-intercept $Q$, substitute $y = 0$ into the equation of the line:
$0 = 1.25x - 10$
$1.25x = 10$
$x = \frac{10}{1.25} = 8$
So, $Q$ is $(8, 0)$.

1. Equate the equations of the curve and the line to find the intersection points

$\frac{15}{0.5x+1} = 1.25x - 10$

2. Rearrange into a standard quadratic equation

$15 = (1.25x - 10)(0.5x + 1)$
$15 = 0.625x^2 + 1.25x - 5x - 10$
$0.625x^2 - 3.75x - 25 = 0$
Multiply by 8 to remove decimals:
$5x^2 - 30x - 200 = 0$
$x^2 - 6x - 40 = 0$

3. Solve the quadratic equation for $x$

$(x - 10)(x + 4) = 0$
$x = 10 \text{ or } x = -4$

4. Determine the valid coordinates of $R$

From the diagram, $R$ is in the first quadrant, so $x > 0$. Therefore, $x = 10$.
Substitute $x = 10$ into the equation of the line:
$y = 1.25(10) - 10 = 12.5 - 10 = 2.5$
The coordinates of $R$ are $(10, 2.5)$.

1. Find the $x$-coordinate of the intersection point $R$ by equating the equations of the curve and the line.

$\frac{15}{0.5x+1} = 1.25x - 10$
$15 = (1.25x - 10)(0.5x + 1)$
$15 = 0.625x^2 + 1.25x - 5x - 10$
$0.625x^2 - 3.75x - 25 = 0$
Multiply by 8:
$5x^2 - 30x - 200 = 0$
$x^2 - 6x - 40 = 0$
$(x - 10)(x + 4) = 0$
Since $R$ is in the first quadrant, $x = 10$.

2. Set up the definite integral for the area bounded by the curve, the line, and the $y$-axis.

$\text{Area} = \int_{0}^{10} \left( \frac{15}{0.5x+1} - (1.25x - 10) \right) dx$

3. Integrate the expression.

$\int \left( \frac{15}{0.5x+1} - 1.25x + 10 \right) dx = \left[ \frac{15 \ln(0.5x+1)}{0.5} - \frac{1.25x^2}{2} + 10x \right]_0^{10}$
$= \left[ 30 \ln(0.5x+1) - 0.625x^2 + 10x \right]_0^{10}$

4. Evaluate the definite integral to find the area.

Substitute $x = 10$:
$30 \ln(0.5(10)+1) - 0.625(10)^2 + 10(10) = 30 \ln(6) - 62.5 + 100 = 30 \ln 6 + 37.5$
Substitute $x = 0$:
$30 \ln(1) - 0 + 0 = 0$
$\text{Area} = 30 \ln 6 + 37.5 \approx 91.2527 \approx 91.3 \text{ units}^2$