<Question>
I have a bag with yellow and blue marbles in it. At the moment, the ratio of blue marbles to yellow marbles is 4:3. If I add 5 blue marbles and remove 3 yellow marbles, the ratio will be 7:3. How many blue marbles were in the bag before I added more?
</Question>

<think>
<Plan_1>
Okay, let's try to solve this marble ratio problem. Hmm, so the problem says that initially, the ratio of blue marbles to yellow marbles is 4:3. Then, if I add 5 blue marbles and remove 3 yellow marbles, the new ratio becomes 7:3. I need to find out how many blue marbles there were originally.

Let me start by understanding the problem step by step.
First, let's denote the number of blue marbles as B and the number of yellow marbles as Y. The initial ratio is 4:3, which means B/Y = 4/3. So, I can write that as an equation: B/Y = 4/3. From this, I can express B in terms of Y. Let's see, cross-multiplying gives 3B = 4Y, so B = (4/3)Y. That might be useful later.

Then, after adding 5 blue marbles and removing 3 yellow marbles, the new ratio becomes 7:3. So, the new number of blue marbles is B + 5, and the new number of yellow marbles is Y - 3. The ratio of these two is (B + 5)/(Y - 3) = 7/3. So that's another equation: (B + 5)/(Y - 3) = 7/3.

Now, I have two equations:

1. B = (4/3)Y
2. (B + 5)/(Y - 3) = 7/3

I can substitute the first equation into the second one to solve for Y, and then find B.

Let me try substituting B = (4/3)Y into the second equation.

<prompt_1.1> Substitute B = (4/3)Y into the second equation and simplify. </prompt_1.1>
-----

Alternatively, maybe I can approach this problem using variables based on the ratio. Since the initial ratio is 4:3, let's say there are 4k blue marbles and 3k yellow marbles, where k is a positive integer. This is another way to represent the ratio.

<prompt_1.2> Use an alternative method by setting up variables based on the ratio and solve for k. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
So, replacing B in the second equation:

[(4/3)Y + 5] / [Y - 3] = 7/3

Now, let's solve this equation for Y. To eliminate the fractions, maybe multiply both sides by 3 to make it easier. Let's see:

Multiply both sides by 3:

3 * [(4/3)Y + 5] / [Y - 3] = 3 * (7/3)

Simplifying the right side: 7

Left side: [3*(4/3)Y + 3*5] / [Y - 3] = [4Y + 15]/[Y - 3]

So now, the equation is:

(4Y + 15)/(Y - 3) = 7

Now, multiply both sides by (Y - 3) to get rid of the denominator:

4Y + 15 = 7(Y - 3)

Expand the right side:

4Y + 15 = 7Y - 21

Now, subtract 4Y from both sides:

15 = 3Y - 21

Then, add 21 to both sides:

36 = 3Y

Divide both sides by 3:

Y = 12

So, the original number of yellow marbles is 12. Then, from the first equation, B = (4/3)Y = (4/3)*12 = 16.
</execution_1.1>

<execution_1.2>
Then, after adding 5 blue marbles, the number of blue marbles becomes 4k + 5. The number of yellow marbles becomes 3k - 3 after removing 3. The new ratio is given as 7:3. Therefore:

(4k + 5)/(3k - 3) = 7/3

Cross-multiplying:

3*(4k + 5) = 7*(3k - 3)

12k + 15 = 21k - 21

Bring variables to one side and constants to the other:

15 + 21 = 21k - 12k

36 = 9k

So, k = 4.

Therefore, the original number of blue marbles is 4k = 4*4 = 16, and yellow marbles is 3k = 12. This also gives the same answer as before.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me check if this makes sense.
Original marbles: 16 blue and 12 yellow. The ratio 16:12 simplifies to 4:3, which is correct. Then, adding 5 blue marbles makes it 21 blue, and removing 3 yellow marbles makes it 9 yellow. The new ratio is 21:9, which simplifies to 7:3.
-----

Based on execution_1.2:
But let me verify once more to be sure.

<prompt_2.1> Verify the solution using both methods again and conclude. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
Original ratio: 16:12 = 4:3. After adding 5 blue and removing 3 yellow: 16+5=21, 12-3=9. 21:9 = 7:3. Correct.
</execution_2.1>
</Execution_2>

<Final_answer>
Therefore, the number of blue marbles in the bag before adding more was $\boxed{16}$.
</Final_answer>
</think>