The problem states:

"If $y + 1$ varies directly as $x$.
(i) Find an expression for $y$ in terms of $x$.
(ii) It is given that $y = 12$ when $x = 3$. Find $y$ when $x = 8$."

Solution:

Part (i):

Since $y + 1$ varies directly as $x$, we can write: $y + 1 = kx$ where $k$ is a constant of proportionality.

To express $y$ in terms of $x$, we rearrange the equation: $y = kx - 1$

Part (ii):

We are given that $y = 12$ when $x = 3$. We use this information to find the constant $k$.

$12 + 1 = k \cdot 3$ $13 = 3k$ $k = \frac{13}{3}$

Now, we need to find $y$ when $x = 8$.

Using the equation $y = kx - 1$: $y = \left(\frac{13}{3}\right) \cdot 8 - 1$ $y = \frac{104}{3} - 1$ $y = \frac{104}{3} - \frac{3}{3}$ $y = \frac{104 - 3}{3}$ $y = \frac{101}{3}$ $y \approx 33.67$

So, $y = \frac{101}{3}$ when $x = 8$.

Summary:

1. The expression for $y$ in terms of $x$ is $y = \frac{13}{3}x - 1$.
2. When $x = 8$, $y = \frac{101}{3}$.

Would you like more details or have any questions? Here are some related questions for further exploration:

1. What does it mean for two quantities to vary directly?
2. How do you determine the constant of proportionality in direct variation problems?
3. What are some examples of direct variation in real-life situations?
4. Can you explain the steps to rearrange an equation to solve for a specific variable?
5. How can you verify the solution to a direct variation problem?
6. What is the significance of the constant $k$ in the context of this problem?
7. How would the expression for $y$ change if $y + 1$ varied inversely with $x$?
8. How can you graph the relationship between $y$ and $x$ for this problem?

Tip: When dealing with direct variation, always check if the relationship holds true for different given values to ensure consistency.