Let the original number of people who intended to take the trip be $x$.

Step 1: Write the expression for the original cost per person

If all $x$ people went, the cost per person would be:

$\frac{1050}{x}$

Step 2: Write the expression for the new cost per person

When 8 people drop out, the number of people becomes $x - 8$. The new cost per person is:

$\frac{1050}{x - 8}$

Step 3: Set up the equation based on the problem

The problem states that the new cost per person is $4 more than the original cost. Therefore, we can set up the equation:

$\frac{1050}{x - 8} = \frac{1050}{x} + 4$

Step 4: Solve the equation

1. Subtract $\frac{1050}{x}$ from both sides:

$\frac{1050}{x - 8} - \frac{1050}{x} = 4$

2. Factor out $1050$ on the left-hand side:

$1050 \left( \frac{1}{x - 8} - \frac{1}{x} \right) = 4$

3. Combine the fractions inside the parentheses:

$1050 \left( \frac{x - (x - 8)}{x(x - 8)} \right) = 4$

$1050 \left( \frac{8}{x(x - 8)} \right) = 4$

4. Divide both sides by 1050:

$\frac{8}{x(x - 8)} = \frac{4}{1050}$

5. Cross-multiply:

$8 \times 1050 = 4 \times x(x - 8)$

$8400 = 4x(x - 8)$

6. Divide both sides by 4:

$2100 = x(x - 8)$

7. Expand the right-hand side:

$2100 = x^2 - 8x$

8. Rearrange the equation:

$x^2 - 8x - 2100 = 0$

Step 5: Solve the quadratic equation

Use the quadratic formula:

$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-2100)}}{2(1)}$

$x = \frac{8 \pm \sqrt{64 + 8400}}{2}$

$x = \frac{8 \pm \sqrt{8464}}{2}$

$x = \frac{8 \pm 92}{2}$

So, $x = \frac{8 + 92}{2} = 50$ or $x = \frac{8 - 92}{2} = -42$.

Since the number of people must be positive, we have $x = 50$.

Final Answer:

The original number of people who intended to take the trip was 50.

Would you like more details or have any questions?

Here are 5 related questions to think about:

1. What would the new cost per person be after 8 people dropped out?
2. How would the total cost change if the bus charter was $1200 instead?
3. What would happen to the price difference if 5 people dropped out instead of 8?
4. How can we set up a similar problem involving a different number of people and total cost?
5. What are other real-world applications of such algebraic problems?

Tip: When dealing with cost per person problems, setting up expressions based on the number of people helps organize the equation structure.