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The problem states:

"A car park has 53 parking bays. Each bay can hold either a car or a bike. One day, it is noted that every bay is occupied and the total number of vehicles is 126. Calculate how many cars and how many bikes there are in the car park using algebra (not trial and error)."

Let's solve this using algebra.

### Step 1: Define the variables
- Let \( c \) represent the number of cars.
- Let \( b \) represent the number of bikes.

### Step 2: Set up the equations
We have two key pieces of information:
1. The total number of parking bays (spaces) is 53. This gives us the equation:
   \[
   c + b = 53
   \]
2. The total number of vehicles is 126. Since a car takes 1 space and a bike takes 1 space but counts as more vehicles, we'll set this up:
   \[
   4c + 1b = 126
   \]

### Step 3: Solve the system of equations

We already have \( c + b = 53 \). We can rearrange this equation for \( b \):
\[
b = 53 - c
\]

Now, substitute \( b = 53 - c \) into the second equation:
\[
4c + (53 - c) = 126
\]
Simplify this equation:
\[
4c + 53 - c = 126
\]
\[
3c + 53 = 126
\]
Subtract 53 from both sides:
\[
3c = 73
\]
Now, divide by 3:
\[
c = \frac{73}{3} = 24.33
\]

However, since \( c \) must be an integer (since you can't have a fraction of a car), let's recheck the equation setup for possible errors. 

Looking at the description of the problem, we see that there is a mistake in the algebraic setup of vehicle counting.

Given the total vehicle count, it's more likely that the cars and bikes refer to different vehicles. Let's interpret it as:

- 1 car (1 space, 1 vehicle).
- 1 bike (1 space, 2 vehicles) 

So let's adjust the equation:
If each bike counts as 2 vehicles, we have:
\[
c + 2b = 126
\]

Now solving the system:
1. \( c + b = 53 \)
2. \( c + 2b = 126 \)

Subtract the first equation from the second:
\[
(c + 2b) - (c + b) = 126 - 53
\]
\[
b = 73
\]

Substitute \( b = 73 \) into the first equation:
\[
c + 73 = 53
\]
\[
c = 53 - 73 = -20
\]

Since this leads to an impossibility with the provided vehicle numbers, the initial equation interpretations need revising; the numbers provided might be incorrectly copied or intended for another setup. 

The given problem might be clearer with a different ratio or setting, such as multiple bikes occupying different spaces.

However, given the problem setup, and assuming some error, the initial ratio is likely one of the simpler relationships corrected by the method.

Would you like to look into possible other interpretations, or is the problem clarified by your input or alternative situation? 

### Related Questions:
1. How would the scenario change if each bike was replaced by a motorbike taking more space?
2. What if the space count was increased to accommodate bigger vehicles like trucks?
3. How does vehicle arrangement change when vehicle type space usage is different?
4. What would happen in a situation with uneven parking spaces?
5. Could a mixed vehicle type approach (cars, bikes, vans) affect the problem setup?

### Tip:
Always double-check problem statements for clarity and correctness of numbers when setting up equations. Sometimes the smallest details can affect outcomes significantly!

A car park has 53 parking bays. Each bay can hold either a car or a bike. One day, it is noted that every bay is occupied and the total number of vehicles is 126. Calculate how many cars and how many bikes there are in the car park using algebra, not trial and error.

Learn how to solve a system of equations involving parking bays with 53 spaces and 126 vehicles, using algebra to find the number of cars and bikes. This problem covers key algebraic concepts for Grades 8-10.

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