<Question>
Hillary has eleven coins, all dimes and nickels. In total, the coins are worth 75 cents. How many nickels does she have?
</Question>

<think>
<Plan_1>
Okay, so Hillary has 11 coins, and they're all dimes and nickels. The total value is 75 cents. I need to find out how many nickels she has. Hmm, let's think about how to approach this.

<prompt_1.1> Set up the equations based on the problem statement. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, let me recall that dimes are worth 10 cents each and nickels are 5 cents each. So, if I let the number of dimes be, say, D, and the number of nickels be N, then I can set up some equations. Since all the coins are dimes and nickels, the total number of coins should be D + N = 11. That's one equation. Then, the total value would be 10D + 5N = 75 cents. So, two equations here. I can solve this system of equations to find N, the number of nickels.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me write them down again:

1. D + N = 11
2. 10D + 5N = 75

I can simplify the second equation. Maybe divide everything by 5 to make the numbers smaller.
That gives 2D + N = 15. Wait, 75 divided by 5 is 15, right? So the second equation becomes 2D + N = 15.

Now, I have:

1. D + N = 11
2. 2D + N = 15

If I subtract the first equation from the second, that might eliminate N. Let's try:

<prompt_2.1> Subtract the first equation from the second equation to find the number of dimes. </prompt_2.1>
-----

Alternatively, maybe I can use substitution. From the first equation, D = 11 - N. Then plug that into the second equation.

<prompt_2.2> Use substitution to verify the number of nickels. </prompt_2.2>

Another way to think about this might be using mental math. If all 11 coins were dimes, then the total would be 110 cents.

<prompt_2.3> Use a mental math approach to verify the solution. </prompt_2.3>

Or maybe trial and error. Suppose there are 7 nickels. Then dimes would be 11 -7 =4.

<prompt_2.4> Use trial and error to verify the solution. </prompt_2.4>

Wait, just to make sure there's no mistake in the algebra. Let's write the equations again:

<prompt_2.5> Re-evaluate the algebraic solution to ensure correctness. </prompt_2.5>

Alternatively, matrix methods or Cramer's rule, but that's overkill for such simple equations. But since substitution and elimination both give the same answer, it's solid.

Another angle: suppose we think in terms of coins.

<prompt_2.6> Explore different methods to confirm the solution. </prompt_2.6>

Let me just check once more with different numbers. Suppose someone claims there are 5 nickels.

<prompt_2.7> Perform a final verification of the solution using different numbers. </prompt_2.7>
</Plan_2>

<Execution_2>
<execution_2.1>
(2D + N) - (D + N) = 15 - 11

That simplifies to D = 4. So, D = 4. Then, substituting back into the first equation, 4 + N = 11, so N = 7.
</execution_2.1>

<execution_2.2>
2D + N = 15
D + N = 11
Subtract the second from the first: (2D + N) - (D + N) = 15 - 11
That gives D = 4. Then, substitute back into D + N = 11: 4 + N = 11, so N = 7. Same result.
</execution_2.2>

<execution_2.3>
But she only has 75 cents, which is 35 cents less. Each nickel is 5 cents, so replacing a dime with a nickel reduces the total by 5 cents. To reduce by 35 cents, she needs 35 /5 =7 nickels. That's the same answer. So that's another way to think about it without algebra. Quick check: 7 nickels is 35 cents, 4 dimes is 40 cents. 35 +40 =75. 7+4=11 coins.
</execution_2.3>

<execution_2.4>
Each nickel is 5 cents less than a dime. So, replacing a dime with a nickel reduces the total by 5 cents. The difference between 40 and 110 is 70 cents. So, each nickel replaces a dime, reducing the total by 5 cents. So, 70 divided by 5 is 14. Wait, but that would mean 14 nickels, which is impossible because there are only 11 coins. Hmm, maybe that approach is wrong. Let me see. Wait, maybe not. Let's check:

If all 11 coins were dimes, that's 110 cents. But she has 75 cents, which is 35 cents less. Each nickel instead of a dime reduces the total by 5 cents. So, 35 divided by 5 is 7. So, 7 nickels. That makes sense.
</execution_2.4>

<execution_2.5>
D + N = 11

10D +5N =75

Express D in terms of N: D =11 -N. Substitute into the value equation:

10*(11 - N) +5N =75

110 -10N +5N =75

110 -5N =75

Subtract 75 from both sides:

35 -5N =0 → 35=5N → N=7.
</execution_2.5>

<execution_2.6>
If all coins were dimes, 11 coins would be 110 cents. But the total is 75 cents, which is 35 cents less. Each nickel is worth 5 cents, and each dime is 10. The difference between a dime and a nickel is 5 cents. So, if each nickel takes away 5 cents from the total, how many nickels would account for the 35 cent difference? Wait, maybe this approach is similar to the first one. Let me see.

Wait, the total difference is 110 - 75 = 35 cents. Each nickel instead of a dime reduces the total by 5 cents. So, number of nickels would be 35 / 5 = 7. That's another way to get the same answer. So, that's a quicker way, maybe. So, if you think in terms of replacing dimes with nickels, each replacement reduces the total by 5 cents. So, to reduce from 110 to 75, you need to replace 7 dimes with nickels. Hence, 7 nickels. That's a neat method. So, same answer.
</execution_2.6>

<execution_2.7>
Then dimes would be 6. 5*5 +6*10=25 +60=85, which is too much. So 5 is too low. 7 nickels gives 5*7=35 and 4*10=40, total 75. Perfect. So yes, 7 nickels.
</execution_2.7>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, so she has 4 dimes and 7 nickels. Let's check that. 4 dimes would be 4*10=40 cents.

<prompt_3.1> Verify the solution by checking the total value and number of coins. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
7 nickels would be 7*5=35 cents. 40 +35=75, which matches. And 4+7=11 coins. That seems right.
</execution_3.1>
</Execution_3>

<Final_answer>
Therefore, Hillary has $\boxed{7}$ nickels.
</Final_answer>
</think>