Explanation:
Okay, so I need to find the inverse of 17 modulo 83. That means I need to find some number x such that when I multiply 17 by x and then divide by 83, the remainder is 1. In other words, 17x ≡ 1 mod 83. Hmm, how do I approach this?

The problem also mentions that 17 times 5 is 85, which is 83 + 2. Maybe that's a hint? Let me think. If 17*5 = 85, then 17*5 ≡ 2 mod 83 because 85 - 83 = 2. So, 17*5 ≡ 2 mod 83. Maybe I can use this to find the inverse?

Let me recall that to find the inverse of a number modulo another number, one method is the Extended Euclidean Algorithm. But maybe there's a simpler way here since they gave me that 17*5 = 85, which is 2 mod 83. Let me try to use that.

So, if 17*5 ≡ 2 mod 83, then maybe I can manipulate this equation to get 17 times something equals 1 mod 83. Let me see. If I can find a multiple of 2 that gives me 1 mod 83, then I can combine that with the previous equation.

Wait, since 17*5 ≡ 2 mod 83, then multiplying both sides by the inverse of 2 mod 83 would give me 17*(5 * inverse of 2) ≡ 1 mod 83. So, if I can find the inverse of 2 mod 83, then I can multiply it by 5 to get the inverse of 17. That seems like a plan.

But how do I find the inverse of 2 mod 83? Well, the inverse of 2 is a number y such that 2y ≡ 1 mod 83. Let me think. Since 2*42 = 84, which is 1 mod 83 because 84 - 83 = 1. So, 2*42 ≡ 1 mod 83. Therefore, the inverse of 2 is 42. So, then the inverse of 17 would be 5*42 mod 83. Let me compute that.

5*42 is 210. Now, divide 210 by 83 to find the remainder. Let me see, 83*2 = 166. 210 - 166 = 44. So, 210 mod 83 is 44. Therefore, 17^{-1} ≡ 44 mod 83? Let me check if that's correct.

Let me verify by multiplying 17 and 44 and see if it's congruent to 1 mod 83. 17*44. Let me compute that. 17*40 = 680, and 17*4 = 68, so total is 680 + 68 = 748. Now, divide 748 by 83 to find the remainder. Let me see how many times 83 goes into 748.

First, 83*9 = 747. Because 80*9=720, 3*9=27, so 720+27=747. Then 748 - 747 = 1. Therefore, 748 ≡ 1 mod 83. Perfect! So 17*44 ≡ 1 mod 83. Therefore, 44 is indeed the inverse of 17 modulo 83. So the answer is 44?

Wait, but let me make sure I didn't make any mistakes in my steps. Let me go through again.

Starting with 17*5 = 85. 85 mod 83 is 2. So 17*5 ≡ 2 mod 83. Then, since 2*42 = 84 ≡ 1 mod 83, so multiplying both sides of 17*5 ≡ 2 by 42 gives 17*5*42 ≡ 2*42 ≡ 1 mod 83. Therefore, 17*(5*42) ≡ 1 mod 83. So 5*42 is 210, which mod 83 is 210 - 2*83 = 210 - 166 = 44. So yes, 44. Then verifying 17*44 = 748, which is 83*9 + 1, so 748 mod 83 is 1. Correct.

Alternatively, maybe I can use the Extended Euclidean Algorithm to find the inverse. Let me try that method as a check.

The Extended Euclidean Algorithm finds integers x and y such that ax + by = gcd(a, b). Since 17 and 83 are coprime (because 83 is prime and 17 is less than 83 and not a multiple), their gcd is 1. So, we can find x such that 17x + 83y = 1. Then x mod 83 is the inverse.

Let me perform the algorithm step by step.

First, divide 83 by 17:

83 = 17*4 + 15. Because 17*4 = 68, 83 - 68 = 15.

Then divide 17 by 15:

17 = 15*1 + 2.

Then divide 15 by 2:

15 = 2*7 + 1.

Then divide 2 by 1:

2 = 1*2 + 0.

So the gcd is 1, as expected. Now, working backwards to express 1 as a combination of 17 and 83.

Starting from the second last equation:

1 = 15 - 2*7.

But 2 is from the previous equation: 2 = 17 - 15*1. Substitute that into the above:

1 = 15 - (17 - 15*1)*7 = 15 - 17*7 + 15*7 = 15*8 - 17*7.

Now, 15 is from the first equation: 15 = 83 - 17*4. Substitute that into the equation:

1 = (83 - 17*4)*8 - 17*7 = 83*8 - 17*32 - 17*7 = 83*8 - 17*(32 + 7) = 83*8 - 17*39.

Therefore, rearranged: -17*39 + 83*8 = 1. Which means that -39*17 ≡ 1 mod 83. Therefore, the inverse of 17 mod 83 is -39. But we need to express this as a positive residue between 0 and 82. So, -39 mod 83 is equal to 83 - 39 = 44. Which matches the previous result! So that's reassuring. So the inverse is indeed 44.

Therefore, both methods give me 44. So I think that's the correct answer. Let me just check once more with another method, maybe trial and error? Although that might take longer, but just to be thorough.

Alternatively, since 17*44 ≡ 1 mod 83, as we saw, and 44 is between 0 and 82, that's the answer. But maybe I can think of another way. For example, since 17*5 = 85 ≡ 2 mod 83, then 17*5*42 ≡ 2*42 ≡ 84 ≡ 1 mod 83. So 17*(5*42) ≡ 1 mod 83. Which again gives 5*42 = 210 ≡ 44 mod 83. So same result.

Alternatively, maybe using Fermat's little theorem? Since 83 is prime, then for any a not divisible by 83, a^{82} ≡ 1 mod 83. Therefore, the inverse of a is a^{81} mod 83. So, 17^{-1} ≡ 17^{81} mod 83. But computing 17^81 mod 83 sounds tedious, but maybe we can find a pattern or use exponentiation by squaring? Although that might be more work than necessary, but let me try.

But given that we already have two methods giving 44, maybe it's safe. However, just for practice, let me try to compute 17^k mod 83 for some exponents to see if we can find a pattern or something.

But honestly, given that two different methods have given me 44, and the verification by multiplying 17*44 gives 748 ≡ 1 mod 83, I think that's solid. So maybe I can just go with 44.

But just to make sure, let me check another multiplication. Let me compute 17*44 again. 17*40 is 680, 17*4 is 68, so 680 + 68 = 748. Then divide 748 by 83. Let me do that division again. 83*9 is 747, as before. So 748 - 747 = 1. So yes, 748 mod 83 is 1. Perfect. So that's correct.

Alternatively, maybe I can check with another number. Suppose I thought the inverse was something else, like 21. Let me check 17*21. 17*20=340, 17*1=17, total 357. 357 divided by 83. 83*4=332, 357-332=25. So 25 mod 83. Not 1. How about 17* something else? Let me try 17*  something. Maybe 17*  5 is 85 ≡ 2, 17*  5* something. Wait, we already did that.

Alternatively, maybe I can think of 17x ≡ 1 mod 83. So 17x - 1 = 83k for some integer k. So 17x - 83k = 1. Which is the same as the equation from the Extended Euclidean Algorithm. Which we solved as x = -39, which is 44 mod 83. So that's consistent.

Alternatively, maybe I can use the fact that 17*5 = 85 ≡ 2 mod 83, so 17*5 ≡ 2. Then, if I can find a multiple of 2 that is 1 mod 83, which we did as 2*42=84≡1. So combining these, 17*5*42 ≡ 1. So 17*(5*42) ≡ 1. Which is 17*210 ≡ 1. Then 210 mod 83. Let me compute 83*2=166, 210-166=44. So 210≡44. Therefore, 17*44≡1. Yep.

So all methods lead to 44. Therefore, I think that's the correct answer. I don't see any mistakes in my reasoning. So I think the inverse of 17 modulo 83 is 44.

**Final Answer**
The inverse of $17$ modulo $83$ is $\boxed{44}$.

To find the modular inverse of $ 17 \mod 83 $, we are looking for an integer $ x $ such that:

$$
17x \equiv 1 \pmod{83}
$$

This means that $ x $ is the number that, when multiplied by 17, gives a product that is congruent to 1 modulo 83.

---

### Step 1: Use the Given Hint

We are told that:

$$
17 \cdot 5 = 85 \equiv 2 \pmod{83}
$$

So, we now have:

$$
17 \cdot 5 \equiv 2 \pmod{83}
$$

This is helpful because we can now try to find the inverse of 2 modulo 83, and then multiply that by 5 to get the inverse of 17.

---

### Step 2: Find the Inverse of 2 Modulo 83

We want a number $ y $ such that:

$$
2y \equiv 1 \pmod{83}
$$

Try $ y = 42 $:

$$
2 \cdot 42 = 84 \equiv 1 \pmod{83}
$$

So, the inverse of 2 modulo 83 is 42.

---

### Step 3: Multiply to Find the Inverse of 17

Now, recall:

$$
17 \cdot 5 \equiv 2 \pmod{83}
$$

Multiply both sides by the inverse of 2 (which is 42):

$$
17 \cdot 5 \cdot 42 \equiv 2 \cdot 42 \equiv 1 \pmod{83}
$$

So:

$$
17 \cdot (5 \cdot 42) \equiv 1 \pmod{83}
$$

Compute $ 5 \cdot 42 = 210 $, and then reduce modulo 83:

$$
210 \mod 83 = 210 - 2 \cdot 83 = 210 - 166 = 44
$$

Thus:

$$
17 \cdot 44 \equiv 1 \pmod{83}
$$

---

### Step 4: Final Verification

To confirm, compute:

$$
17 \cdot 44 = 748
$$

Now divide 748 by 83:

$$
83 \cdot 9 = 747 \Rightarrow 748 - 747 = 1
$$

So:

$$
748 \equiv 1 \pmod{83}
$$

This confirms that $ 44 $ is indeed the modular inverse of 17 modulo 83.

---

### Final Answer

$$
\boxed{44}
$$

[Assistant]
Explanation:
Okay, so I need to find the inverse of 17 modulo 83. That means I need to find some number x such that when I multiply 17 by x and then divide by 83, the remainder is 1. In other words, 17x ≡ 1 mod 83. Hmm, how do I approach this?

The problem also mentions that 17 times 5 is 85, which is 83 + 2. Maybe that's a hint? Let me think. If 17*5 = 85, then 17*5 ≡ 2 mod 83 because 85 - 83 = 2. So, 17*5 ≡ 2 mod 83. Maybe I can use this to find the inverse?

Let me recall that to find the inverse of a number modulo another number, one method is the Extended Euclidean Algorithm. But maybe there's a simpler way here since they gave me that 17*5 = 85, which is 2 mod 83. Let me try to use that.

So, if 17*5 ≡ 2 mod 83, then maybe I can manipulate this equation to get 17 times something equals 1 mod 83. Let me see. If I can find a multiple of 2 that gives me 1 mod 83, then I can combine that with the previous equation.

Wait, since 17*5 ≡ 2 mod 83, then multiplying both sides by the inverse of 2 mod 83 would give me 17*(5 * inverse of 2) ≡ 1 mod 83. So, if I can find the inverse of 2 mod 83, then I can multiply it by 5 to get the inverse of 17. That seems like a plan.

But how do I find the inverse of 2 mod 83? Well, the inverse of 2 is a number y such that 2y ≡ 1 mod 83. Let me think. Since 2*42 = 84, which is 1 mod 83 because 84 - 83 = 1. So, 2*42 ≡ 1 mod 83. Therefore, the inverse of 2 is 42. So, then the inverse of 17 would be 5*42 mod 83. Let me compute that.

5*42 is 210. Now, divide 210 by 83 to find the remainder. Let me see, 83*2 = 166. 210 - 166 = 44. So, 210 mod 83 is 44. Therefore, 17^{-1} ≡ 44 mod 83? Let me check if that's correct.

Let me verify by multiplying 17 and 44 and see if it's congruent to 1 mod 83. 17*44. Let me compute that. 17*40 = 680, and 17*4 = 68, so total is 680 + 68 = 748. Now, divide 748 by 83 to find the remainder. Let me see how many times 83 goes into 748.

First, 83*9 = 747. Because 80*9=720, 3*9=27, so 720+27=747. Then 748 - 747 = 1. Therefore, 748 ≡ 1 mod 83. Perfect! So 17*44 ≡ 1 mod 83. Therefore, 44 is indeed the inverse of 17 modulo 83. So the answer is 44?

Wait, but let me make sure I didn't make any mistakes in my steps. Let me go through again.

Starting with 17*5 = 85. 85 mod 83 is 2. So 17*5 ≡ 2 mod 83. Then, since 2*42 = 84 ≡ 1 mod 83, so multiplying both sides of 17*5 ≡ 2 by 42 gives 17*5*42 ≡ 2*42 ≡ 1 mod 83. Therefore, 17*(5*42) ≡ 1 mod 83. So 5*42 is 210, which mod 83 is 210 - 2*83 = 210 - 166 = 44. So yes, 44. Then verifying 17*44 = 748, which is 83*9 + 1, so 748 mod 83 is 1. Correct.

Alternatively, maybe I can use the Extended Euclidean Algorithm to find the inverse. Let me try that method as a check.

The Extended Euclidean Algorithm finds integers x and y such that ax + by = gcd(a, b). Since 17 and 83 are coprime (because 83 is prime and 17 is less than 83 and not a multiple), their gcd is 1. So, we can find x such that 17x + 83y = 1. Then x mod 83 is the inverse.

Let me perform the algorithm step by step.

First, divide 83 by 17:

83 = 17*4 + 15. Because 17*4 = 68, 83 - 68 = 15.

Then divide 17 by 15:

17 = 15*1 + 2.

Then divide 15 by 2:

15 = 2*7 + 1.

Then divide 2 by 1:

2 = 1*2 + 0.

So the gcd is 1, as expected. Now, working backwards to express 1 as a combination of 17 and 83.

Starting from the second last equation:

1 = 15 - 2*7.

But 2 is from the previous equation: 2 = 17 - 15*1. Substitute that into the above:

1 = 15 - (17 - 15*1)*7 = 15 - 17*7 + 15*7 = 15*8 - 17*7.

Now, 15 is from the first equation: 15 = 83 - 17*4. Substitute that into the equation:

1 = (83 - 17*4)*8 - 17*7 = 83*8 - 17*32 - 17*7 = 83*8 - 17*(32 + 7) = 83*8 - 17*39.

Therefore, rearranged: -17*39 + 83*8 = 1. Which means that -39*17 ≡ 1 mod 83. Therefore, the inverse of 17 mod 83 is -39. But we need to express this as a positive residue between 0 and 82. So, -39 mod 83 is equal to 83 - 39 = 44. Which matches the previous result! So that's reassuring. So the inverse is indeed 44.

Therefore, both methods give me 44. So I think that's the correct answer. Let me just check once more with another method, maybe trial and error? Although that might take longer, but just to be thorough.

Alternatively, since 17*44 ≡ 1 mod 83, as we saw, and 44 is between 0 and 82, that's the answer. But maybe I can think of another way. For example, since 17*5 = 85 ≡ 2 mod 83, then 17*5*42 ≡ 2*42 ≡ 84 ≡ 1 mod 83. So 17*(5*42) ≡ 1 mod 83. Which again gives 5*42 = 210 ≡ 44 mod 83. So same result.

Alternatively, maybe using Fermat's little theorem? Since 83 is prime, then for any a not divisible by 83, a^{82} ≡ 1 mod 83. Therefore, the inverse of a is a^{81} mod 83. So, 17^{-1} ≡ 17^{81} mod 83. But computing 17^81 mod 83 sounds tedious, but maybe we can find a pattern or use exponentiation by squaring? Although that might be more work than necessary, but let me try.

But given that we already have two methods giving 44, maybe it's safe. However, just for practice, let me try to compute 17^k mod 83 for some exponents to see if we can find a pattern or something.

But honestly, given that two different methods have given me 44, and the verification by multiplying 17*44 gives 748 ≡ 1 mod 83, I think that's solid. So maybe I can just go with 44.

But just to make sure, let me check another multiplication. Let me compute 17*44 again. 17*40 is 680, 17*4 is 68, so 680 + 68 = 748. Then divide 748 by 83. Let me do that division again. 83*9 is 747, as before. So 748 - 747 = 1. So yes, 748 mod 83 is 1. Perfect. So that's correct.

Alternatively, maybe I can check with another number. Suppose I thought the inverse was something else, like 21. Let me check 17*21. 17*20=340, 17*1=17, total 357. 357 divided by 83. 83*4=332, 357-332=25. So 25 mod 83. Not 1. How about 17* something else? Let me try 17*  something. Maybe 17*  5 is 85 ≡ 2, 17*  5* something. Wait, we already did that.

Alternatively, maybe I can think of 17x ≡ 1 mod 83. So 17x - 1 = 83k for some integer k. So 17x - 83k = 1. Which is the same as the equation from the Extended Euclidean Algorithm. Which we solved as x = -39, which is 44 mod 83. So that's consistent.

Alternatively, maybe I can use the fact that 17*5 = 85 ≡ 2 mod 83, so 17*5 ≡ 2. Then, if I can find a multiple of 2 that is 1 mod 83, which we did as 2*42=84≡1. So combining these, 17*5*42 ≡ 1. So 17*(5*42) ≡ 1. Which is 17*210 ≡ 1. Then 210 mod 83. Let me compute 83*2=166, 210-166=44. So 210≡44. Therefore, 17*44≡1. Yep.

So all methods lead to 44. Therefore, I think that's the correct answer. I don't see any mistakes in my reasoning. So I think the inverse of 17 modulo 83 is 44.

**Final Answer**
The inverse of $17$ modulo $83$ is $\boxed{44}$.

To find the modular inverse of $ 17 \mod 83 $, we are looking for an integer $ x $ such that:

$$
17x \equiv 1 \pmod{83}
$$

This means that $ x $ is the number that, when multiplied by 17, gives a product that is congruent to 1 modulo 83.

---

### Step 1: Use the Given Hint

We are told that:

$$
17 \cdot 5 = 85 \equiv 2 \pmod{83}
$$

So, we now have:

$$
17 \cdot 5 \equiv 2 \pmod{83}
$$

This is helpful because we can now try to find the inverse of 2 modulo 83, and then multiply that by 5 to get the inverse of 17.

---

### Step 2: Find the Inverse of 2 Modulo 83

We want a number $ y $ such that:

$$
2y \equiv 1 \pmod{83}
$$

Try $ y = 42 $:

$$
2 \cdot 42 = 84 \equiv 1 \pmod{83}
$$

So, the inverse of 2 modulo 83 is 42.

---

### Step 3: Multiply to Find the Inverse of 17

Now, recall:

$$
17 \cdot 5 \equiv 2 \pmod{83}
$$

Multiply both sides by the inverse of 2 (which is 42):

$$
17 \cdot 5 \cdot 42 \equiv 2 \cdot 42 \equiv 1 \pmod{83}
$$

So:

$$
17 \cdot (5 \cdot 42) \equiv 1 \pmod{83}
$$

Compute $ 5 \cdot 42 = 210 $, and then reduce modulo 83:

$$
210 \mod 83 = 210 - 2 \cdot 83 = 210 - 166 = 44
$$

Thus:

$$
17 \cdot 44 \equiv 1 \pmod{83}
$$

---

### Step 4: Final Verification

To confirm, compute:

$$
17 \cdot 44 = 748
$$

Now divide 748 by 83:

$$
83 \cdot 9 = 747 \Rightarrow 748 - 747 = 1
$$

So:

$$
748 \equiv 1 \pmod{83}
$$

This confirms that $ 44 $ is indeed the modular inverse of 17 modulo 83.

---

### Final Answer

$$
\boxed{44}
$$

Solution:
We may begin by noting that $17\cdot 5=85\equiv 2\pmod{83}$. However, we are looking for $n$ such that $17\cdot n\equiv 1\pmod{83}$.

Note that $2\cdot 42=84\equiv 1\pmod{83}$. Therefore, \begin{align*}
17\cdot 5\cdot 42 &\equiv 2\cdot 42 \\
&\equiv 1\pmod{83},
\end{align*}which tells us that $17$ and $5\cdot 42$ are each other's inverses modulo $83$. We can evaluate $5\cdot 42=210$, but this is not in the range $0$ to $82$, so we take its residue $\pmod{83}$, which is $44$.

Therefore, $17^{-1}\equiv \boxed{44}\pmod{83}$.

We may check our answer: $17\cdot 44 = 748 = 9\cdot 83+1 \equiv 1\pmod{83}$, so our answer is correct.


Explanation:
Okay, so I need to find the average of the two smallest positive integer solutions to the congruence 14u ≡ 46 mod 100. Hmm, let me think about how to approach this.

First, I remember that solving a linear congruence like ax ≡ b mod m involves finding integers x that satisfy the equation. The general method is to find the modular inverse of a modulo m, provided that a and m are coprime. But wait, in this case, a is 14 and m is 100. Are they coprime? Let me check their greatest common divisor (gcd).

The gcd of 14 and 100. Let's see, factors of 14 are 1, 2, 7, 14. Factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, 100. The common factors are 1 and 2, so gcd(14, 100) is 2. Since the gcd is not 1, the congruence 14u ≡ 46 mod 100 might not have solutions unless 2 divides 46. Let me check that.

Yes, 46 divided by 2 is 23, which is an integer. So, since the gcd divides 46, there are solutions. The number of solutions modulo 100 should be equal to the gcd, which is 2. Therefore, there are two solutions modulo 100. But the question asks for the two smallest positive integer solutions. So, I need to find all solutions and pick the two smallest ones, then average them.

Alright, so how do I solve 14u ≡ 46 mod 100? Let me recall the steps. First, divide the entire congruence by the gcd, which is 2. That gives me:

(14/2)u ≡ 46/2 mod 100/2

Which simplifies to:

7u ≡ 23 mod 50

Now, since 7 and 50 are coprime (gcd(7,50)=1), there is a unique solution modulo 50. So, I need to find the inverse of 7 modulo 50.

To find the inverse of 7 mod 50, I need an integer x such that 7x ≡ 1 mod 50. Let me try to find that.

Let me use the extended Euclidean algorithm. Let me compute gcd(7,50) and express it as a linear combination of 7 and 50.

First, divide 50 by 7: 50 = 7*7 + 1. Then, divide 7 by 1: 7 = 1*7 + 0. So, gcd is 1. Now, backtracking:

1 = 50 - 7*7

Therefore, 1 = (-7)*7 + 1*50. So, the inverse of 7 mod 50 is -7. But -7 mod 50 is 43, since 50 - 7 = 43. Let me check: 7*43 = 301. Now, divide 301 by 50: 50*6 = 300, so 301 - 300 = 1. Yes, 7*43 ≡ 1 mod 50. Perfect.

So, the inverse of 7 mod 50 is 43. Therefore, multiplying both sides of the congruence 7u ≡ 23 mod 50 by 43 gives:

u ≡ 23*43 mod 50

Let me compute 23*43. Let me do 20*43 + 3*43. 20*43 is 860, 3*43 is 129. Adding them together: 860 + 129 = 989. Now, compute 989 mod 50.

Divide 989 by 50. 50*19 = 950. 989 - 950 = 39. So, 989 mod 50 is 39. Therefore, u ≡ 39 mod 50.

So, the solutions to the original congruence 14u ≡ 46 mod 100 are all integers congruent to 39 mod 50. But wait, since we divided the original congruence by 2, the solutions modulo 100 would be... Let me think.

When we have a congruence ax ≡ b mod m, and we divide by d = gcd(a, m), we get (a/d)x ≡ (b/d) mod (m/d). The solutions to the original congruence are the solutions to this new congruence plus multiples of m/d. Wait, so in this case, the solutions to 7u ≡ 23 mod 50 are u ≡ 39 mod 50. Therefore, the solutions to the original congruence 14u ≡ 46 mod 100 are u ≡ 39 mod 50 and u ≡ 39 + 50 mod 100? Wait, no. Wait, when you divide the modulus by d, the solutions are congruent modulo m/d. Wait, maybe I need to think again.

Let me recall that if we have ax ≡ b mod m, and d = gcd(a, m), then the solutions are given by x ≡ x0 + (m/d)*k, where k is an integer, and x0 is a particular solution.

In our case, after dividing by d=2, we had 7u ≡ 23 mod 50, which has a unique solution modulo 50, which is u ≡ 39 mod 50. Therefore, the solutions to the original congruence 14u ≡ 46 mod 100 are u ≡ 39 mod 50 and u ≡ 39 + 50 mod 100? Wait, no. Wait, the modulus for the original congruence is 100, and since we divided by d=2, the solutions are spaced by m/d = 100/2 = 50. Therefore, the solutions are u ≡ 39 mod 50. But since the original modulus is 100, the solutions are u ≡ 39 mod 50 and u ≡ 39 + 50 mod 100? Wait, no. Wait, if the solutions are u ≡ 39 mod 50, then modulo 100, the solutions would be 39 and 39 + 50 = 89. Because 50 is half of 100. Therefore, the solutions are 39 and 89 mod 100.

Wait, let me check that. Let me take u = 39. Then 14*39 = 546. Let me compute 546 mod 100. 546 divided by 100 is 5*100=500, so 546 - 500 = 46. So 546 ≡ 46 mod 100. Correct.

Now, take u = 39 + 50 = 89. 14*89. Let me compute 14*90 = 1260, so 14*89 = 1260 - 14 = 1246. 1246 mod 100 is 46. Correct. So, 89 is also a solution.

Therefore, the solutions are 39, 89, 139, 189, etc. So the two smallest positive integer solutions are 39 and 89. Therefore, the average is (39 + 89)/2 = 128/2 = 64.

Wait, but let me make sure there are no smaller solutions. For example, is there a solution less than 39? Let me check.

Suppose u is a positive integer less than 39. Let me check if there's another solution. Since the solutions are congruent to 39 mod 50, the next solution before 39 would be 39 - 50 = -11, which is negative. So, the smallest positive solution is 39, then 89. Therefore, the two smallest positive solutions are 39 and 89.

Therefore, their average is (39 + 89)/2 = 64.

But let me verify once again. Maybe I made a mistake in solving the congruence. Let me check my steps again.

Original congruence: 14u ≡ 46 mod 100.

Divide both sides by 2: 7u ≡ 23 mod 50.

Find inverse of 7 mod 50. As before, 7*43 = 301 ≡ 1 mod 50. So, multiplying both sides by 43: u ≡ 23*43 mod 50.

23*43: Let me compute 23*40 = 920, 23*3 = 69, so total is 920 + 69 = 989. 989 divided by 50: 50*19 = 950, 989 - 950 = 39. So, u ≡ 39 mod 50.

Therefore, solutions are u = 39 + 50k, where k is integer. Therefore, the smallest positive solutions are when k=0: 39, k=1: 89, k=2: 139, etc.

Therefore, the two smallest positive solutions are 39 and 89.

Average is (39 + 89)/2 = 128/2 = 64.

But let me check if there's another solution between 1 and 39. For example, maybe there's another solution? Let me check.

Suppose u is between 1 and 39. Let me check if 14u mod 100 equals 46.

Alternatively, since we know that the solutions are 39 mod 50, so the solutions are numbers congruent to 39 mod 50. So, numbers like 39, 89, 139,... So, there are no solutions between 1 and 39 except 39 itself. Therefore, 39 is the smallest positive solution, then 89.

Alternatively, maybe I can solve the original congruence another way. Let me try solving 14u ≡ 46 mod 100.

Let me write this as 14u - 46 = 100k for some integer k. Therefore, 14u - 100k = 46.

This is a linear Diophantine equation. Let me try solving it.

First, check if gcd(14, 100) divides 46. As before, gcd(14, 100) is 2, and 2 divides 46, so solutions exist.

Let me divide the equation by 2: 7u - 50k = 23.

So, 7u - 50k = 23.

We can write this as 7u = 50k + 23.

Looking for integer solutions u, k.

Let me solve for u: u = (50k + 23)/7.

We need (50k + 23) divisible by 7.

So, 50k + 23 ≡ 0 mod 7.

Compute 50 mod 7: 7*7=49, so 50 ≡ 1 mod 7.

Similarly, 23 mod 7: 7*3=21, so 23 ≡ 2 mod 7.

Therefore, the equation becomes 1*k + 2 ≡ 0 mod 7 => k + 2 ≡ 0 mod 7 => k ≡ -2 mod 7 => k ≡ 5 mod 7.

Therefore, k = 5 + 7m, where m is an integer.

Therefore, substituting back into u:

u = (50*(5 + 7m) + 23)/7 = (250 + 350m + 23)/7 = (273 + 350m)/7 = 273/7 + 350m/7 = 39 + 50m.

Therefore, u = 39 + 50m.

Therefore, the solutions are u = 39 + 50m, where m is integer.

Therefore, the smallest positive solutions are when m=0: 39, m=1: 89, m=2: 139, etc.

Therefore, the two smallest positive solutions are 39 and 89.

Therefore, their average is (39 + 89)/2 = 64.

So, that seems consistent with my previous result.

Alternatively, maybe I can check with another method. Let me try to solve 14u ≡ 46 mod 100.

Let me note that 14u ≡ 46 mod 100. Let me try to find u such that 14u ends with 46.

So, 14u mod 100 = 46.

Let me think about the last two digits.

Let me write u as a number, and 14 times that number ends with 46.

Let me think about the last digit first.

Let me consider the equation 14u ≡ 46 mod 100.

Looking at the last digit: 14u ends with 6.

Let me consider the last digit of 14u.

The last digit of 14u is determined by the last digit of u. Let me denote u as ...x, where x is the last digit.

Then, 14x mod 10.

Since 14 mod 10 is 4, so 4x ≡ 6 mod 10.

So, 4x ≡ 6 mod 10.

Divide both sides by 2: 2x ≡ 3 mod 5.

Multiply both sides by inverse of 2 mod 5. The inverse of 2 mod 5 is 3, since 2*3=6≡1 mod5.

Therefore, x ≡ 3*3 ≡ 9 ≡ 4 mod5.

Therefore, x ≡ 4 mod5. So, the last digit of u is 4 or 9.

Wait, because x ≡ 4 mod5, so possible last digits are 4 or 9.

So, possible last digits of u are 4 or 9.

Therefore, u can be written as 10k + 4 or 10k + 9 for some integer k.

Let me check both cases.

First case: u = 10k + 4.

Then, 14u = 14*(10k + 4) = 140k + 56.

We need 140k + 56 ≡ 46 mod 100.

So, 140k + 56 ≡ 46 mod 100.

Subtract 56: 140k ≡ -10 mod 100.

Which is 140k ≡ 90 mod 100.

Simplify 140k mod 100. Since 140 ≡ 40 mod 100, so 40k ≡ 90 mod 100.

Divide both sides by 10: 4k ≡ 9 mod 10.

So, 4k ≡ 9 mod 10.

Multiply both sides by inverse of 4 mod 10. The inverse of 4 mod 10 is... Well, 4 and 10 are not coprime, so inverse doesn't exist. Wait, but 4k ≡ 9 mod 10. Let me check if there are solutions.

Since gcd(4,10)=2, and 2 divides 9? No, 2 does not divide 9. Therefore, there are no solutions in this case.

Therefore, u cannot end with 4.

Therefore, the other case: u ends with 9. So, u = 10k + 9.

Then, 14u = 14*(10k + 9) = 140k + 126.

Compute 140k + 126 mod 100.

140k mod 100 is 40k, and 126 mod 100 is 26.

Therefore, 40k + 26 ≡ 46 mod 100.

Subtract 26: 40k ≡ 20 mod 100.

Divide both sides by 20: 2k ≡ 1 mod 5.

So, 2k ≡ 1 mod 5. Multiply both sides by inverse of 2 mod5, which is 3.

Therefore, k ≡ 3*1 ≡ 3 mod5.

Therefore, k = 5m + 3 for some integer m.

Therefore, u = 10k + 9 = 10*(5m + 3) + 9 = 50m + 30 + 9 = 50m + 39.

Therefore, u = 50m + 39.

Therefore, the solutions are u ≡ 39 mod50. Which matches our previous result.

Therefore, the solutions are 39, 89, 139,... So, the two smallest positive solutions are 39 and 89.

Therefore, average is (39 + 89)/2 = 64.

Therefore, the answer is 64.

But just to be thorough, let me check if there are any solutions between 1 and 39.

For example, let me check u= 39 - 50 = -11, which is negative. So, no.

But maybe there's another solution? Let me check u= 39 - 50*1 = -11, which is negative.

Alternatively, maybe there's another solution in the range 1-39?

But according to our previous analysis, the solutions are u=39 + 50m. So, for m=0, u=39. For m=-1, u=-11. So, no positive solutions less than 39.

Alternatively, maybe I can check some numbers. Let me check u= 9. 14*9=126. 126 mod100=26≠46.

u=19: 14*19=266. 266 mod100=66≠46.

u=29: 14*29=406. 406 mod100=6≠46.

u=39: 14*39=546. 546 mod100=46. Correct.

So, yes, 39 is the first solution.

Therefore, the two smallest positive solutions are 39 and 89.

Therefore, average is 64.

Alternatively, maybe I can think of the general solution. Since the solutions are u ≡ 39 mod50, then the solutions are 39, 89, 139,... So, the two smallest positive are 39 and 89.

Therefore, average is (39 + 89)/2 = 64.

Therefore, I think the answer is 64.

But just to make sure, let me check if there's another way someone might approach this problem and maybe get a different answer.

Alternatively, someone might try to solve 14u ≡ 46 mod100 by trial and error. But that would be tedious.

Alternatively, maybe using the Chinese Remainder Theorem? Let me see.

But since 100 = 4*25, and 4 and 25 are coprime.

So, solving 14u ≡ 46 mod4 and mod25, then combining solutions.

Let me try that.

First, solve 14u ≡ 46 mod4.

Compute 14 mod4: 14/4=3*4=12, remainder 2. So 14 ≡ 2 mod4.

Similarly, 46 mod4: 46/4=11*4=44, remainder 2. So 46 ≡ 2 mod4.

Therefore, the congruence becomes 2u ≡ 2 mod4.

Divide both sides by 2: u ≡ 1 mod2.

So, u must be odd.

Now, solve 14u ≡ 46 mod25.

Compute 14 mod25=14, 46 mod25=46-25=21. So, 14u ≡21 mod25.

We need to solve 14u ≡21 mod25.

First, check gcd(14,25)=1, since 14 and 25 are coprime.

Therefore, inverse of 14 mod25 exists. Let me find it.

Find x such that 14x ≡1 mod25.

Use extended Euclidean algorithm.

Compute gcd(14,25):

25 = 1*14 + 11

14 = 1*11 + 3

11 = 3*3 + 2

3 = 1*2 + 1

2 = 2*1 + 0

So, gcd is 1. Now backtracking:

1 = 3 - 1*2

But 2 = 11 - 3*3, so:

1 = 3 - 1*(11 - 3*3) = 4*3 - 1*11

But 3 = 14 - 1*11, so:

1 = 4*(14 - 1*11) - 1*11 = 4*14 - 5*11

But 11 = 25 - 1*14, so:

1 = 4*14 - 5*(25 - 1*14) = 4*14 -5*25 +5*14 = 9*14 -5*25

Therefore, 9*14 ≡1 mod25. Therefore, inverse of 14 mod25 is 9.

Therefore, multiplying both sides of 14u ≡21 mod25 by 9:

u ≡21*9 mod25.

Compute 21*9: 189. 189 divided by 25: 25*7=175, 189-175=14. Therefore, u ≡14 mod25.

Therefore, solutions to the original congruence must satisfy:

u ≡1 mod2 and u ≡14 mod25.

Now, solve this system.

Find u such that:

u ≡14 mod25

and u ≡1 mod2.

Let me write u =25k +14. Then, substitute into the second congruence:

25k +14 ≡1 mod2.

Compute 25 mod2=1, 14 mod2=0. Therefore:

1*k + 0 ≡1 mod2 => k ≡1 mod2.

Therefore, k is odd: k=2m +1 for some integer m.

Therefore, u=25*(2m +1) +14=50m +25 +14=50m +39.

Therefore, u≡39 mod50. Which is the same result as before.

Therefore, the solutions are u=39 +50m. Therefore, the two smallest positive solutions are 39 and 89.

Therefore, average is 64.

Therefore, regardless of the method, I get the same answer.

Therefore, I can be confident that the answer is 64.

**Final Answer**
The average of the two smallest positive integer solutions is \boxed{64}.

To solve the congruence:

$$
14u \equiv 46 \pmod{100}
$$

we begin by analyzing the structure of the equation. This is a linear congruence of the form $ ax \equiv b \pmod{m} $, where $ a = 14 $, $ b = 46 $, and $ m = 100 $.

---

### Step 1: Check for Solutions

We first check whether a solution exists. A solution exists if and only if $ \gcd(a, m) \mid b $. Here:

$$
\gcd(14, 100) = 2 \quad \text{and} \quad 2 \mid 46
$$

So, the congruence has solutions. The number of solutions modulo 100 is equal to $ \gcd(14, 100) = 2 $, meaning there are two distinct solutions modulo 100.

---

### Step 2: Simplify the Congruence

We divide the entire congruence by the gcd:

$$
\frac{14}{2}u \equiv \frac{46}{2} \pmod{\frac{100}{2}} \Rightarrow 7u \equiv 23 \pmod{50}
$$

Now, we solve the simplified congruence:

$$
7u \equiv 23 \pmod{50}
$$

---

### Step 3: Find the Modular Inverse of 7 Modulo 50

We need the inverse of 7 modulo 50, i.e., an integer $ x $ such that:

$$
7x \equiv 1 \pmod{50}
$$

Using the Extended Euclidean Algorithm:

$$
\begin{align*}
50 &= 7 \cdot 7 + 1 \\
7 &= 1 \cdot 7 + 0
\end{align*}
\Rightarrow 1 = 50 - 7 \cdot 7
$$

So, $ -7 \equiv 43 \pmod{50} $ is the inverse of 7 modulo 50.

---

### Step 4: Solve for $ u $

Multiply both sides of the congruence $ 7u \equiv 23 \pmod{50} $ by 43:

$$
u \equiv 23 \cdot 43 \pmod{50}
$$

Compute:

$$
23 \cdot 43 = 989 \Rightarrow 989 \mod 50 = 39
$$

So, the solution is:

$$
u \equiv 39 \pmod{50}
$$

---

### Step 5: Find the Two Smallest Positive Solutions

The general solution is:

$$
u = 39 + 50k \quad \text{for integer } k
$$

The two smallest positive solutions are:

- $ u = 39 $ (when $ k = 0 $)
- $ u = 89 $ (when $ k = 1 $)

---

### Step 6: Compute the Average

$$
\text{Average} = \frac{39 + 89}{2} = \frac{128}{2} = \boxed{64}
$$

---

### Final Answer

$$
\boxed{64}
$$

[Assistant]
Explanation:
Okay, so I need to find the average of the two smallest positive integer solutions to the congruence 14u ≡ 46 mod 100. Hmm, let me think about how to approach this.

First, I remember that solving a linear congruence like ax ≡ b mod m involves finding integers x that satisfy the equation. The general method is to find the modular inverse of a modulo m, provided that a and m are coprime. But wait, in this case, a is 14 and m is 100. Are they coprime? Let me check their greatest common divisor (gcd).

The gcd of 14 and 100. Let's see, factors of 14 are 1, 2, 7, 14. Factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, 100. The common factors are 1 and 2, so gcd(14, 100) is 2. Since the gcd is not 1, the congruence 14u ≡ 46 mod 100 might not have solutions unless 2 divides 46. Let me check that.

Yes, 46 divided by 2 is 23, which is an integer. So, since the gcd divides 46, there are solutions. The number of solutions modulo 100 should be equal to the gcd, which is 2. Therefore, there are two solutions modulo 100. But the question asks for the two smallest positive integer solutions. So, I need to find all solutions and pick the two smallest ones, then average them.

Alright, so how do I solve 14u ≡ 46 mod 100? Let me recall the steps. First, divide the entire congruence by the gcd, which is 2. That gives me:

(14/2)u ≡ 46/2 mod 100/2

Which simplifies to:

7u ≡ 23 mod 50

Now, since 7 and 50 are coprime (gcd(7,50)=1), there is a unique solution modulo 50. So, I need to find the inverse of 7 modulo 50.

To find the inverse of 7 mod 50, I need an integer x such that 7x ≡ 1 mod 50. Let me try to find that.

Let me use the extended Euclidean algorithm. Let me compute gcd(7,50) and express it as a linear combination of 7 and 50.

First, divide 50 by 7: 50 = 7*7 + 1. Then, divide 7 by 1: 7 = 1*7 + 0. So, gcd is 1. Now, backtracking:

1 = 50 - 7*7

Therefore, 1 = (-7)*7 + 1*50. So, the inverse of 7 mod 50 is -7. But -7 mod 50 is 43, since 50 - 7 = 43. Let me check: 7*43 = 301. Now, divide 301 by 50: 50*6 = 300, so 301 - 300 = 1. Yes, 7*43 ≡ 1 mod 50. Perfect.

So, the inverse of 7 mod 50 is 43. Therefore, multiplying both sides of the congruence 7u ≡ 23 mod 50 by 43 gives:

u ≡ 23*43 mod 50

Let me compute 23*43. Let me do 20*43 + 3*43. 20*43 is 860, 3*43 is 129. Adding them together: 860 + 129 = 989. Now, compute 989 mod 50.

Divide 989 by 50. 50*19 = 950. 989 - 950 = 39. So, 989 mod 50 is 39. Therefore, u ≡ 39 mod 50.

So, the solutions to the original congruence 14u ≡ 46 mod 100 are all integers congruent to 39 mod 50. But wait, since we divided the original congruence by 2, the solutions modulo 100 would be... Let me think.

When we have a congruence ax ≡ b mod m, and we divide by d = gcd(a, m), we get (a/d)x ≡ (b/d) mod (m/d). The solutions to the original congruence are the solutions to this new congruence plus multiples of m/d. Wait, so in this case, the solutions to 7u ≡ 23 mod 50 are u ≡ 39 mod 50. Therefore, the solutions to the original congruence 14u ≡ 46 mod 100 are u ≡ 39 mod 50 and u ≡ 39 + 50 mod 100? Wait, no. Wait, when you divide the modulus by d, the solutions are congruent modulo m/d. Wait, maybe I need to think again.

Let me recall that if we have ax ≡ b mod m, and d = gcd(a, m), then the solutions are given by x ≡ x0 + (m/d)*k, where k is an integer, and x0 is a particular solution.

In our case, after dividing by d=2, we had 7u ≡ 23 mod 50, which has a unique solution modulo 50, which is u ≡ 39 mod 50. Therefore, the solutions to the original congruence 14u ≡ 46 mod 100 are u ≡ 39 mod 50 and u ≡ 39 + 50 mod 100? Wait, no. Wait, the modulus for the original congruence is 100, and since we divided by d=2, the solutions are spaced by m/d = 100/2 = 50. Therefore, the solutions are u ≡ 39 mod 50. But since the original modulus is 100, the solutions are u ≡ 39 mod 50 and u ≡ 39 + 50 mod 100? Wait, no. Wait, if the solutions are u ≡ 39 mod 50, then modulo 100, the solutions would be 39 and 39 + 50 = 89. Because 50 is half of 100. Therefore, the solutions are 39 and 89 mod 100.

Wait, let me check that. Let me take u = 39. Then 14*39 = 546. Let me compute 546 mod 100. 546 divided by 100 is 5*100=500, so 546 - 500 = 46. So 546 ≡ 46 mod 100. Correct.

Now, take u = 39 + 50 = 89. 14*89. Let me compute 14*90 = 1260, so 14*89 = 1260 - 14 = 1246. 1246 mod 100 is 46. Correct. So, 89 is also a solution.

Therefore, the solutions are 39, 89, 139, 189, etc. So the two smallest positive integer solutions are 39 and 89. Therefore, the average is (39 + 89)/2 = 128/2 = 64.

Wait, but let me make sure there are no smaller solutions. For example, is there a solution less than 39? Let me check.

Suppose u is a positive integer less than 39. Let me check if there's another solution. Since the solutions are congruent to 39 mod 50, the next solution before 39 would be 39 - 50 = -11, which is negative. So, the smallest positive solution is 39, then 89. Therefore, the two smallest positive solutions are 39 and 89.

Therefore, their average is (39 + 89)/2 = 64.

But let me verify once again. Maybe I made a mistake in solving the congruence. Let me check my steps again.

Original congruence: 14u ≡ 46 mod 100.

Divide both sides by 2: 7u ≡ 23 mod 50.

Find inverse of 7 mod 50. As before, 7*43 = 301 ≡ 1 mod 50. So, multiplying both sides by 43: u ≡ 23*43 mod 50.

23*43: Let me compute 23*40 = 920, 23*3 = 69, so total is 920 + 69 = 989. 989 divided by 50: 50*19 = 950, 989 - 950 = 39. So, u ≡ 39 mod 50.

Therefore, solutions are u = 39 + 50k, where k is integer. Therefore, the smallest positive solutions are when k=0: 39, k=1: 89, k=2: 139, etc.

Therefore, the two smallest positive solutions are 39 and 89.

Average is (39 + 89)/2 = 128/2 = 64.

But let me check if there's another solution between 1 and 39. For example, maybe there's another solution? Let me check.

Suppose u is between 1 and 39. Let me check if 14u mod 100 equals 46.

Alternatively, since we know that the solutions are 39 mod 50, so the solutions are numbers congruent to 39 mod 50. So, numbers like 39, 89, 139,... So, there are no solutions between 1 and 39 except 39 itself. Therefore, 39 is the smallest positive solution, then 89.

Alternatively, maybe I can solve the original congruence another way. Let me try solving 14u ≡ 46 mod 100.

Let me write this as 14u - 46 = 100k for some integer k. Therefore, 14u - 100k = 46.

This is a linear Diophantine equation. Let me try solving it.

First, check if gcd(14, 100) divides 46. As before, gcd(14, 100) is 2, and 2 divides 46, so solutions exist.

Let me divide the equation by 2: 7u - 50k = 23.

So, 7u - 50k = 23.

We can write this as 7u = 50k + 23.

Looking for integer solutions u, k.

Let me solve for u: u = (50k + 23)/7.

We need (50k + 23) divisible by 7.

So, 50k + 23 ≡ 0 mod 7.

Compute 50 mod 7: 7*7=49, so 50 ≡ 1 mod 7.

Similarly, 23 mod 7: 7*3=21, so 23 ≡ 2 mod 7.

Therefore, the equation becomes 1*k + 2 ≡ 0 mod 7 => k + 2 ≡ 0 mod 7 => k ≡ -2 mod 7 => k ≡ 5 mod 7.

Therefore, k = 5 + 7m, where m is an integer.

Therefore, substituting back into u:

u = (50*(5 + 7m) + 23)/7 = (250 + 350m + 23)/7 = (273 + 350m)/7 = 273/7 + 350m/7 = 39 + 50m.

Therefore, u = 39 + 50m.

Therefore, the solutions are u = 39 + 50m, where m is integer.

Therefore, the smallest positive solutions are when m=0: 39, m=1: 89, m=2: 139, etc.

Therefore, the two smallest positive solutions are 39 and 89.

Therefore, their average is (39 + 89)/2 = 64.

So, that seems consistent with my previous result.

Alternatively, maybe I can check with another method. Let me try to solve 14u ≡ 46 mod 100.

Let me note that 14u ≡ 46 mod 100. Let me try to find u such that 14u ends with 46.

So, 14u mod 100 = 46.

Let me think about the last two digits.

Let me write u as a number, and 14 times that number ends with 46.

Let me think about the last digit first.

Let me consider the equation 14u ≡ 46 mod 100.

Looking at the last digit: 14u ends with 6.

Let me consider the last digit of 14u.

The last digit of 14u is determined by the last digit of u. Let me denote u as ...x, where x is the last digit.

Then, 14x mod 10.

Since 14 mod 10 is 4, so 4x ≡ 6 mod 10.

So, 4x ≡ 6 mod 10.

Divide both sides by 2: 2x ≡ 3 mod 5.

Multiply both sides by inverse of 2 mod 5. The inverse of 2 mod 5 is 3, since 2*3=6≡1 mod5.

Therefore, x ≡ 3*3 ≡ 9 ≡ 4 mod5.

Therefore, x ≡ 4 mod5. So, the last digit of u is 4 or 9.

Wait, because x ≡ 4 mod5, so possible last digits are 4 or 9.

So, possible last digits of u are 4 or 9.

Therefore, u can be written as 10k + 4 or 10k + 9 for some integer k.

Let me check both cases.

First case: u = 10k + 4.

Then, 14u = 14*(10k + 4) = 140k + 56.

We need 140k + 56 ≡ 46 mod 100.

So, 140k + 56 ≡ 46 mod 100.

Subtract 56: 140k ≡ -10 mod 100.

Which is 140k ≡ 90 mod 100.

Simplify 140k mod 100. Since 140 ≡ 40 mod 100, so 40k ≡ 90 mod 100.

Divide both sides by 10: 4k ≡ 9 mod 10.

So, 4k ≡ 9 mod 10.

Multiply both sides by inverse of 4 mod 10. The inverse of 4 mod 10 is... Well, 4 and 10 are not coprime, so inverse doesn't exist. Wait, but 4k ≡ 9 mod 10. Let me check if there are solutions.

Since gcd(4,10)=2, and 2 divides 9? No, 2 does not divide 9. Therefore, there are no solutions in this case.

Therefore, u cannot end with 4.

Therefore, the other case: u ends with 9. So, u = 10k + 9.

Then, 14u = 14*(10k + 9) = 140k + 126.

Compute 140k + 126 mod 100.

140k mod 100 is 40k, and 126 mod 100 is 26.

Therefore, 40k + 26 ≡ 46 mod 100.

Subtract 26: 40k ≡ 20 mod 100.

Divide both sides by 20: 2k ≡ 1 mod 5.

So, 2k ≡ 1 mod 5. Multiply both sides by inverse of 2 mod5, which is 3.

Therefore, k ≡ 3*1 ≡ 3 mod5.

Therefore, k = 5m + 3 for some integer m.

Therefore, u = 10k + 9 = 10*(5m + 3) + 9 = 50m + 30 + 9 = 50m + 39.

Therefore, u = 50m + 39.

Therefore, the solutions are u ≡ 39 mod50. Which matches our previous result.

Therefore, the solutions are 39, 89, 139,... So, the two smallest positive solutions are 39 and 89.

Therefore, average is (39 + 89)/2 = 64.

Therefore, the answer is 64.

But just to be thorough, let me check if there are any solutions between 1 and 39.

For example, let me check u= 39 - 50 = -11, which is negative. So, no.

But maybe there's another solution? Let me check u= 39 - 50*1 = -11, which is negative.

Alternatively, maybe there's another solution in the range 1-39?

But according to our previous analysis, the solutions are u=39 + 50m. So, for m=0, u=39. For m=-1, u=-11. So, no positive solutions less than 39.

Alternatively, maybe I can check some numbers. Let me check u= 9. 14*9=126. 126 mod100=26≠46.

u=19: 14*19=266. 266 mod100=66≠46.

u=29: 14*29=406. 406 mod100=6≠46.

u=39: 14*39=546. 546 mod100=46. Correct.

So, yes, 39 is the first solution.

Therefore, the two smallest positive solutions are 39 and 89.

Therefore, average is 64.

Alternatively, maybe I can think of the general solution. Since the solutions are u ≡ 39 mod50, then the solutions are 39, 89, 139,... So, the two smallest positive are 39 and 89.

Therefore, average is (39 + 89)/2 = 64.

Therefore, I think the answer is 64.

But just to make sure, let me check if there's another way someone might approach this problem and maybe get a different answer.

Alternatively, someone might try to solve 14u ≡ 46 mod100 by trial and error. But that would be tedious.

Alternatively, maybe using the Chinese Remainder Theorem? Let me see.

But since 100 = 4*25, and 4 and 25 are coprime.

So, solving 14u ≡ 46 mod4 and mod25, then combining solutions.

Let me try that.

First, solve 14u ≡ 46 mod4.

Compute 14 mod4: 14/4=3*4=12, remainder 2. So 14 ≡ 2 mod4.

Similarly, 46 mod4: 46/4=11*4=44, remainder 2. So 46 ≡ 2 mod4.

Therefore, the congruence becomes 2u ≡ 2 mod4.

Divide both sides by 2: u ≡ 1 mod2.

So, u must be odd.

Now, solve 14u ≡ 46 mod25.

Compute 14 mod25=14, 46 mod25=46-25=21. So, 14u ≡21 mod25.

We need to solve 14u ≡21 mod25.

First, check gcd(14,25)=1, since 14 and 25 are coprime.

Therefore, inverse of 14 mod25 exists. Let me find it.

Find x such that 14x ≡1 mod25.

Use extended Euclidean algorithm.

Compute gcd(14,25):

25 = 1*14 + 11

14 = 1*11 + 3

11 = 3*3 + 2

3 = 1*2 + 1

2 = 2*1 + 0

So, gcd is 1. Now backtracking:

1 = 3 - 1*2

But 2 = 11 - 3*3, so:

1 = 3 - 1*(11 - 3*3) = 4*3 - 1*11

But 3 = 14 - 1*11, so:

1 = 4*(14 - 1*11) - 1*11 = 4*14 - 5*11

But 11 = 25 - 1*14, so:

1 = 4*14 - 5*(25 - 1*14) = 4*14 -5*25 +5*14 = 9*14 -5*25

Therefore, 9*14 ≡1 mod25. Therefore, inverse of 14 mod25 is 9.

Therefore, multiplying both sides of 14u ≡21 mod25 by 9:

u ≡21*9 mod25.

Compute 21*9: 189. 189 divided by 25: 25*7=175, 189-175=14. Therefore, u ≡14 mod25.

Therefore, solutions to the original congruence must satisfy:

u ≡1 mod2 and u ≡14 mod25.

Now, solve this system.

Find u such that:

u ≡14 mod25

and u ≡1 mod2.

Let me write u =25k +14. Then, substitute into the second congruence:

25k +14 ≡1 mod2.

Compute 25 mod2=1, 14 mod2=0. Therefore:

1*k + 0 ≡1 mod2 => k ≡1 mod2.

Therefore, k is odd: k=2m +1 for some integer m.

Therefore, u=25*(2m +1) +14=50m +25 +14=50m +39.

Therefore, u≡39 mod50. Which is the same result as before.

Therefore, the solutions are u=39 +50m. Therefore, the two smallest positive solutions are 39 and 89.

Therefore, average is 64.

Therefore, regardless of the method, I get the same answer.

Therefore, I can be confident that the answer is 64.

**Final Answer**
The average of the two smallest positive integer solutions is \boxed{64}.

To solve the congruence:

$$
14u \equiv 46 \pmod{100}
$$

we begin by analyzing the structure of the equation. This is a linear congruence of the form $ ax \equiv b \pmod{m} $, where $ a = 14 $, $ b = 46 $, and $ m = 100 $.

---

### Step 1: Check for Solutions

We first check whether a solution exists. A solution exists if and only if $ \gcd(a, m) \mid b $. Here:

$$
\gcd(14, 100) = 2 \quad \text{and} \quad 2 \mid 46
$$

So, the congruence has solutions. The number of solutions modulo 100 is equal to $ \gcd(14, 100) = 2 $, meaning there are two distinct solutions modulo 100.

---

### Step 2: Simplify the Congruence

We divide the entire congruence by the gcd:

$$
\frac{14}{2}u \equiv \frac{46}{2} \pmod{\frac{100}{2}} \Rightarrow 7u \equiv 23 \pmod{50}
$$

Now, we solve the simplified congruence:

$$
7u \equiv 23 \pmod{50}
$$

---

### Step 3: Find the Modular Inverse of 7 Modulo 50

We need the inverse of 7 modulo 50, i.e., an integer $ x $ such that:

$$
7x \equiv 1 \pmod{50}
$$

Using the Extended Euclidean Algorithm:

$$
\begin{align*}
50 &= 7 \cdot 7 + 1 \\
7 &= 1 \cdot 7 + 0
\end{align*}
\Rightarrow 1 = 50 - 7 \cdot 7
$$

So, $ -7 \equiv 43 \pmod{50} $ is the inverse of 7 modulo 50.

---

### Step 4: Solve for $ u $

Multiply both sides of the congruence $ 7u \equiv 23 \pmod{50} $ by 43:

$$
u \equiv 23 \cdot 43 \pmod{50}
$$

Compute:

$$
23 \cdot 43 = 989 \Rightarrow 989 \mod 50 = 39
$$

So, the solution is:

$$
u \equiv 39 \pmod{50}
$$

---

### Step 5: Find the Two Smallest Positive Solutions

The general solution is:

$$
u = 39 + 50k \quad \text{for integer } k
$$

The two smallest positive solutions are:

- $ u = 39 $ (when $ k = 0 $)
- $ u = 89 $ (when $ k = 1 $)

---

### Step 6: Compute the Average

$$
\text{Average} = \frac{39 + 89}{2} = \frac{128}{2} = \boxed{64}
$$

---

### Final Answer

$$
\boxed{64}
$$

Solution:
Note that $14$, $46$, and $100$ all have a common factor of $2$, so we can divide it out: the solutions to $$14u \equiv 46 \pmod{100}$$ are identical to the solutions to $$7u \equiv 23 \pmod{50}.$$ Make sure you see why this is the case.

Now we can multiply both sides of the congruence by $7$ to obtain $$49u \equiv 161 \pmod{50},$$ which also has the same solutions as the previous congruence, since we could reverse the step above by multiplying both sides by $7^{-1}$. (We know that $7^{-1}$ exists modulo $50$ because $7$ and $50$ are relatively prime.)

Replacing each side of $49u\equiv 161$ by a $\pmod{50}$ equivalent, we have $$-u \equiv 11\pmod{50},$$ and thus $$u \equiv -11\pmod{50}.$$ This is the set of solutions to our original congruence. The two smallest positive solutions are $-11+50 = 39$ and $-11+2\cdot 50 = 89$. Their average is $\boxed{64}$.


Explanation:
Okay, so I need to find the image of the line y = 2x + 1 under the matrix M. But first, I don't know what the matrix M is. The problem gives me two vectors and their images under M. Let me write down what I know.

The matrix M transforms the vector [2; -1] to [9; 3], and the vector [1; -3] to [7; -1]. So, if I can figure out what matrix M is, then I can apply it to every point on the line y = 2x + 1 and find the equation of the transformed line.

Since M is a 2x2 matrix, let me denote it as:

M = [a  b]
      [c  d]

So, when we multiply M by a vector [x; y], we get [ax + by; cx + dy].

Given that M*[2; -1] = [9; 3], so:

2a - b = 9  ...(1)
2c - d = 3  ...(2)

Similarly, M*[1; -3] = [7; -1], so:

a - 3b = 7  ...(3)
c - 3d = -1 ...(4)

So now I have four equations with four unknowns: a, b, c, d.

Let me solve equations (1) and (3) first for a and b.

From equation (1): 2a - b = 9

From equation (3): a - 3b = 7

Let me solve equation (1) for b: b = 2a - 9

Substitute into equation (3):

a - 3*(2a - 9) = 7

Compute that:

a - 6a + 27 = 7

Combine like terms:

-5a + 27 = 7

Subtract 27:

-5a = 7 - 27 = -20

Divide by -5:

a = (-20)/(-5) = 4

So a = 4. Then from equation (1):

2*4 - b = 9 => 8 - b = 9 => -b = 1 => b = -1

So a = 4, b = -1.

Now let's solve equations (2) and (4) for c and d.

Equation (2): 2c - d = 3

Equation (4): c - 3d = -1

Let me solve equation (2) for d: d = 2c - 3

Substitute into equation (4):

c - 3*(2c - 3) = -1

Compute:

c - 6c + 9 = -1

Combine like terms:

-5c + 9 = -1

Subtract 9:

-5c = -1 - 9 = -10

Divide by -5:

c = (-10)/(-5) = 2

So c = 2. Then from equation (2):

2*2 - d = 3 => 4 - d = 3 => -d = -1 => d = 1

So c = 2, d = 1.

Therefore, the matrix M is:

[4  -1]
[2   1]

Let me check if this matrix works with the given vectors.

First vector [2; -1]:

M*[2; -1] = [4*2 + (-1)*(-1); 2*2 + 1*(-1)] = [8 + 1; 4 - 1] = [9; 3]. Correct.

Second vector [1; -3]:

M*[1; -3] = [4*1 + (-1)*(-3); 2*1 + 1*(-3)] = [4 + 3; 2 - 3] = [7; -1]. Correct. Great, so the matrix is correct.

Now, I need to find the image of the line y = 2x + 1 under M. So, essentially, every point (x, y) on the original line is transformed by M to a new point (x', y'), and I need to find the equation of the line that these transformed points lie on.

One approach is to parametrize the original line, apply the matrix to the parametrization, and then find the equation of the resulting line.

Alternatively, since linear transformations can transform lines into lines (unless the transformation is singular, but since we have two vectors that are transformed to different vectors, probably M is invertible), so the image of a line under a linear transformation is another line.

But since the original line is affine (not necessarily through the origin), we need to be careful. Wait, the original line is y = 2x + 1, which is an affine line. However, when we apply a linear transformation, it will transform the line into another affine line. However, since the matrix M is linear, it will map the origin to the origin, but our original line doesn't pass through the origin. So, the image will be an affine line.

But how do we find its equation?

One method is to take two points on the original line, apply the matrix to them, find the equation of the line through the transformed points.

Alternatively, we can parametrize the original line and then apply the matrix.

Let me try both approaches.

First approach: take two points on the original line, apply M, then find the equation of the line through the transformed points.

Let me pick two points on y = 2x + 1. For example, when x = 0, y = 1, so point (0, 1). When x = 1, y = 2*1 + 1 = 3, so point (1, 3). Let me check another point, maybe x = -1: y = 2*(-1) + 1 = -2 + 1 = -1, so (-1, -1). But maybe two points are enough.

So let's take (0, 1) and (1, 3). Let me apply M to these points.

First, (0, 1):

M*[0; 1] = [4*0 + (-1)*1; 2*0 + 1*1] = [ -1; 1 ]

Second, (1, 3):

M*[1; 3] = [4*1 + (-1)*3; 2*1 + 1*3] = [4 - 3; 2 + 3] = [1; 5]

So the transformed points are (-1, 1) and (1, 5). Now, I need to find the equation of the line passing through these two points.

First, find the slope m:

m = (5 - 1)/(1 - (-1)) = 4 / 2 = 2.

So the slope is 2. Then using point-slope form with point (-1, 1):

y - 1 = 2(x - (-1)) => y - 1 = 2(x + 1) => y = 2x + 2 + 1 => y = 2x + 3.

Wait, so the transformed line is y = 2x + 3? Let me check with another point to make sure.

Take the third point (-1, -1) on the original line. Let's apply M:

M*[-1; -1] = [4*(-1) + (-1)*(-1); 2*(-1) + 1*(-1)] = [-4 + 1; -2 -1] = [-3; -3]

Now, check if (-3, -3) lies on y = 2x + 3.

Left side y = -3. Right side: 2*(-3) + 3 = -6 + 3 = -3. Yes, it works. So that seems correct.

Wait, so the image of the line y = 2x + 1 under M is y = 2x + 3? But let me think again. Is there a way to verify this more formally?

Alternatively, maybe I can do this by parametrizing the original line and then applying the matrix.

Let me parametrize the original line y = 2x + 1. Let me let x = t, so y = 2t + 1. So any point on the line can be written as (t, 2t + 1), where t is a real number.

Now, apply matrix M to this vector:

M*[t; 2t + 1] = [4*t + (-1)*(2t + 1); 2*t + 1*(2t + 1)].

Compute each component:

First component: 4t - 2t - 1 = 2t - 1

Second component: 2t + 2t + 1 = 4t + 1

So the transformed point is (2t - 1, 4t + 1). Now, we need to find the equation of the line that these points satisfy.

Let me denote the transformed coordinates as (x', y'). So:

x' = 2t - 1

y' = 4t + 1

We can solve for t from the first equation:

From x' = 2t - 1 => 2t = x' + 1 => t = (x' + 1)/2

Substitute into the second equation:

y' = 4*( (x' + 1)/2 ) + 1 = 2*(x' + 1) + 1 = 2x' + 2 + 1 = 2x' + 3

Therefore, y' = 2x' + 3, which is the same as y = 2x + 3. So that confirms the earlier result.

Therefore, the image of the line y = 2x + 1 under M is y = 2x + 3.

Wait, but let me think again. The original line had slope 2, and the transformed line also has slope 2. Is that expected?

Well, linear transformations can preserve or change slopes depending on the transformation. However, in this case, since the matrix M has determinant ad - bc = (4)(1) - (-1)(2) = 4 + 2 = 6, which is non-zero, so M is invertible. However, the slope of the line is preserved? Wait, not necessarily. Wait, but in this case, the slope ended up being the same. Let me check with another example.

Suppose we have a line with slope m, and we apply a linear transformation. The image of the line will have a different slope unless the transformation preserves direction. However, in this case, the slope remained the same. Let me check why.

Alternatively, maybe the transformation is such that it scales the direction vector of the line. Let me think.

The original line has direction vector (1, 2). Let me apply M to this direction vector.

M*[1; 2] = [4*1 + (-1)*2; 2*1 + 1*2] = [4 - 2; 2 + 2] = [2; 4]. So the direction vector becomes [2; 4], which is equivalent to [1; 2] scaled by 2. Therefore, the direction vector is scaled by 2, but the direction is the same. Therefore, the slope remains 2. Therefore, the image line has the same slope as the original line, which is 2. Therefore, the transformed line has slope 2, and we found the intercept is 3. So that seems consistent.

But let me check if the intercept is correct. The original line had intercept 1, and the transformed line has intercept 3. Since the transformation is linear, but the original line is affine, not passing through the origin. So the transformation of the affine line would be another affine line. The intercept changes because the transformation moves the entire line.

Alternatively, another way to think about it is that the transformation M maps the original line y = 2x + 1 to y = 2x + 3. So, the answer is y = 2x + 3.

But let me check once more with another point. Let me take the point (2, 5) on the original line (since y = 2*2 + 1 = 5). Apply M:

M*[2; 5] = [4*2 + (-1)*5; 2*2 + 1*5] = [8 - 5; 4 + 5] = [3; 9]. Now check if (3, 9) lies on y = 2x + 3. 2*3 + 3 = 6 + 3 = 9. Yes, correct. So that works.

Therefore, I think the answer is y = 2x + 3.

But let me think again if there's another way to approach this problem, maybe using linear algebra concepts.

Suppose we consider the original line y = 2x + 1. Let me write this as a vector equation. Let me think of points on the line as vectors. Let me write the line as:

For any scalar t, the point is (t, 2t + 1). Which is the same as before.

Alternatively, we can think of the line as the set of points (x, y) such that y - 2x = 1. So, the equation is y - 2x - 1 = 0.

Now, applying the matrix M to the coordinates (x, y). Let me denote the transformed coordinates as (x', y'). Then:

x' = 4x - y

y' = 2x + y

Wait, because M is [4 -1; 2 1], so:

x' = 4x + (-1)y = 4x - y

y' = 2x + 1*y = 2x + y

So, given that the original point (x, y) satisfies y = 2x + 1, we can substitute y into the equations for x' and y'.

So, substitute y = 2x + 1 into x' and y':

x' = 4x - (2x + 1) = 4x - 2x - 1 = 2x - 1

y' = 2x + (2x + 1) = 4x + 1

So, we have x' = 2x - 1 and y' = 4x + 1. Now, we can express x in terms of x' and substitute into the equation for y'.

From x' = 2x - 1, solving for x:

x = (x' + 1)/2

Substitute into y':

y' = 4*( (x' + 1)/2 ) + 1 = 2(x' + 1) + 1 = 2x' + 2 + 1 = 2x' + 3

Therefore, y' = 2x' + 3, which is the same as y = 2x + 3. So this method also gives the same result. Therefore, this confirms the answer again.

Therefore, I think the answer is y = 2x + 3.

But let me check once more if there's any chance I made a mistake in the matrix multiplication or solving for the matrix M.

Earlier, I found that M is [4 -1; 2 1]. Let me check again with the given vectors.

First vector [2; -1]:

4*2 + (-1)*(-1) = 8 + 1 = 9. Correct.

2*2 + 1*(-1) = 4 - 1 = 3. Correct.

Second vector [1; -3]:

4*1 + (-1)*(-3) = 4 + 3 = 7. Correct.

2*1 + 1*(-3) = 2 - 3 = -1. Correct. So matrix is correct.

Therefore, the rest of the steps are correct. So the image of the line is y = 2x + 3.

Alternatively, maybe the problem expects a different answer? Let me think again.

Wait, another way to think about it: the transformation M can be considered as a linear transformation followed by a translation? Wait, no, because M is a linear transformation. However, the original line is affine, not linear. So, the image of an affine line under a linear transformation is another affine line. However, since the original line is not passing through the origin, the image will not pass through the origin unless the transformation maps the origin to the origin, which it does, but the line is shifted.

But in our case, the original line is y = 2x + 1. The transformation M maps the origin (0,0) to (0,0). However, the original line is not passing through the origin. So, the image of the line is another line, which we found as y = 2x + 3.

Alternatively, maybe we can think of the original line as a linear combination. Let me think.

Let me consider that any point on the original line can be written as a point on the line y = 2x + 1. Let me write this as a vector equation. Let me take a point (x, 2x + 1). This can be written as x*(1, 2) + (0, 1). So, the line is the set of points { t*(1, 2) + (0, 1) | t ∈ ℝ }.

Therefore, applying the matrix M to this, we get:

M*(t*(1, 2) + (0, 1)) = t*M*(1, 2) + M*(0, 1)

Earlier, we found that M*(1, 2) = (2, 4) and M*(0, 1) = (-1, 1). Therefore, the image of the line is { t*(2, 4) + (-1, 1) | t ∈ ℝ }.

So, parametric equations:

x' = 2t - 1

y' = 4t + 1

Which is the same as before. Then, eliminating t:

From x' = 2t - 1 => t = (x' + 1)/2

Substitute into y':

y' = 4*( (x' + 1)/2 ) + 1 = 2(x' + 1) + 1 = 2x' + 2 + 1 = 2x' + 3. So again, same result.

Therefore, I think this is solid. The answer is y = 2x + 3.

But just to make sure, let me check if there's a possibility that the question is expecting something else. For example, maybe I misapplied the matrix? Let me check once again.

Original line: y = 2x + 1. Take a general point (x, y) on the line. Apply M:

x' = 4x - y

y' = 2x + y

But since y = 2x + 1, substitute:

x' = 4x - (2x + 1) = 2x - 1

y' = 2x + (2x + 1) = 4x + 1

So, solving for x from x' = 2x - 1:

x = (x' + 1)/2

Then, substitute into y':

y' = 4*( (x' + 1)/2 ) + 1 = 2(x' + 1) + 1 = 2x' + 2 + 1 = 2x' + 3. So, y' = 2x' + 3. So, yes, same result.

Therefore, I think the answer is definitely y = 2x + 3.

But just to be thorough, let me check if there's a different interpretation of the problem. For example, maybe the question is about the image of the line under the transformation, which could be different if the line is considered as a set of points, but I think the approach is correct.

Alternatively, maybe the problem is expecting me to consider the line as a vector space, but since it's an affine line, not a vector space, the transformation is applied as above.

Alternatively, maybe I need to consider the transformation of the entire line, including the direction vector and the translation. Wait, but since the transformation is linear, it doesn't include translation. However, the original line is not a vector space, so the image is an affine line.

But in our case, since we have an affine line, the image is another affine line, which we found as y = 2x + 3.

Therefore, I think the answer is correct.

**Final Answer**
The image of the line under matrix $\mathbf{M}$ is \boxed{y = 2x + 3}.

To determine the image of the line $ y = 2x + 1 $ under the matrix transformation $ \mathbf{M} $, we proceed in a structured manner:

---

### **Step 1: Determine the Matrix $ \mathbf{M} $**

We are given that:

- $ \mathbf{M} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 9 \\ 3 \end{pmatrix} $
- $ \mathbf{M} \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 7 \\ -1 \end{pmatrix} $

Let $ \mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $. Then:

- From the first vector:
  $$
  \begin{cases}
  2a - b = 9 \\
  2c - d = 3
  \end{cases}
  $$

- From the second vector:
  $$
  \begin{cases}
  a - 3b = 7 \\
  c - 3d = -1
  \end{cases}
  $$

Solving these equations:

- From the first pair:
  $ b = 2a - 9 $, substitute into $ a - 3b = 7 $:
  $$
  a - 3(2a - 9) = 7 \Rightarrow a - 6a + 27 = 7 \Rightarrow -5a = -20 \Rightarrow a = 4
  $$
  Then $ b = 2(4) - 9 = -1 $

- From the second pair:
  $ d = 2c - 3 $, substitute into $ c - 3d = -1 $:
  $$
  c - 3(2c - 3) = -1 \Rightarrow c - 6c + 9 = -1 \Rightarrow -5c = -10 \Rightarrow c = 2
  $$
  Then $ d = 2(2) - 3 = 1 $

So, the matrix is:
$$
\mathbf{M} = \begin{pmatrix} 4 & -1 \\ 2 & 1 \end{pmatrix}
$$

---

### **Step 2: Apply the Matrix to the Line**

We are to find the image of the line $ y = 2x + 1 $ under $ \mathbf{M} $. This is an **affine line**, not a vector space, so the transformation will map it to another **affine line**.

Let’s parametrize the line as:
$$
\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} t \\ 2t + 1 \end{pmatrix}, \quad t \in \mathbb{R}
$$

Apply $ \mathbf{M} $ to this vector:
$$
\mathbf{M} \begin{pmatrix} t \\ 2t + 1 \end{pmatrix} = \begin{pmatrix} 4t - (2t + 1) \\ 2t + (2t + 1) \end{pmatrix} = \begin{pmatrix} 2t - 1 \\ 4t + 1 \end{pmatrix}
$$

Let the transformed coordinates be $ (x', y') $. Then:
$$
x' = 2t - 1, \quad y' = 4t + 1
$$

Solve for $ t $ from the first equation:
$$
t = \frac{x' + 1}{2}
$$

Substitute into the second equation:
$$
y' = 4\left(\frac{x' + 1}{2}\right) + 1 = 2(x' + 1) + 1 = 2x' + 3
$$

---

### **Step 3: Final Result**

Thus, the image of the line $ y = 2x + 1 $ under the transformation $ \mathbf{M} $ is the line:
$$
\boxed{y = 2x + 3}
$$

[Assistant]
Explanation:
Okay, so I need to find the image of the line y = 2x + 1 under the matrix M. But first, I don't know what the matrix M is. The problem gives me two vectors and their images under M. Let me write down what I know.

The matrix M transforms the vector [2; -1] to [9; 3], and the vector [1; -3] to [7; -1]. So, if I can figure out what matrix M is, then I can apply it to every point on the line y = 2x + 1 and find the equation of the transformed line.

Since M is a 2x2 matrix, let me denote it as:

M = [a  b]
      [c  d]

So, when we multiply M by a vector [x; y], we get [ax + by; cx + dy].

Given that M*[2; -1] = [9; 3], so:

2a - b = 9  ...(1)
2c - d = 3  ...(2)

Similarly, M*[1; -3] = [7; -1], so:

a - 3b = 7  ...(3)
c - 3d = -1 ...(4)

So now I have four equations with four unknowns: a, b, c, d.

Let me solve equations (1) and (3) first for a and b.

From equation (1): 2a - b = 9

From equation (3): a - 3b = 7

Let me solve equation (1) for b: b = 2a - 9

Substitute into equation (3):

a - 3*(2a - 9) = 7

Compute that:

a - 6a + 27 = 7

Combine like terms:

-5a + 27 = 7

Subtract 27:

-5a = 7 - 27 = -20

Divide by -5:

a = (-20)/(-5) = 4

So a = 4. Then from equation (1):

2*4 - b = 9 => 8 - b = 9 => -b = 1 => b = -1

So a = 4, b = -1.

Now let's solve equations (2) and (4) for c and d.

Equation (2): 2c - d = 3

Equation (4): c - 3d = -1

Let me solve equation (2) for d: d = 2c - 3

Substitute into equation (4):

c - 3*(2c - 3) = -1

Compute:

c - 6c + 9 = -1

Combine like terms:

-5c + 9 = -1

Subtract 9:

-5c = -1 - 9 = -10

Divide by -5:

c = (-10)/(-5) = 2

So c = 2. Then from equation (2):

2*2 - d = 3 => 4 - d = 3 => -d = -1 => d = 1

So c = 2, d = 1.

Therefore, the matrix M is:

[4  -1]
[2   1]

Let me check if this matrix works with the given vectors.

First vector [2; -1]:

M*[2; -1] = [4*2 + (-1)*(-1); 2*2 + 1*(-1)] = [8 + 1; 4 - 1] = [9; 3]. Correct.

Second vector [1; -3]:

M*[1; -3] = [4*1 + (-1)*(-3); 2*1 + 1*(-3)] = [4 + 3; 2 - 3] = [7; -1]. Correct. Great, so the matrix is correct.

Now, I need to find the image of the line y = 2x + 1 under M. So, essentially, every point (x, y) on the original line is transformed by M to a new point (x', y'), and I need to find the equation of the line that these transformed points lie on.

One approach is to parametrize the original line, apply the matrix to the parametrization, and then find the equation of the resulting line.

Alternatively, since linear transformations can transform lines into lines (unless the transformation is singular, but since we have two vectors that are transformed to different vectors, probably M is invertible), so the image of a line under a linear transformation is another line.

But since the original line is affine (not necessarily through the origin), we need to be careful. Wait, the original line is y = 2x + 1, which is an affine line. However, when we apply a linear transformation, it will transform the line into another affine line. However, since the matrix M is linear, it will map the origin to the origin, but our original line doesn't pass through the origin. So, the image will be an affine line.

But how do we find its equation?

One method is to take two points on the original line, apply the matrix to them, find the equation of the line through the transformed points.

Alternatively, we can parametrize the original line and then apply the matrix.

Let me try both approaches.

First approach: take two points on the original line, apply M, then find the equation of the line through the transformed points.

Let me pick two points on y = 2x + 1. For example, when x = 0, y = 1, so point (0, 1). When x = 1, y = 2*1 + 1 = 3, so point (1, 3). Let me check another point, maybe x = -1: y = 2*(-1) + 1 = -2 + 1 = -1, so (-1, -1). But maybe two points are enough.

So let's take (0, 1) and (1, 3). Let me apply M to these points.

First, (0, 1):

M*[0; 1] = [4*0 + (-1)*1; 2*0 + 1*1] = [ -1; 1 ]

Second, (1, 3):

M*[1; 3] = [4*1 + (-1)*3; 2*1 + 1*3] = [4 - 3; 2 + 3] = [1; 5]

So the transformed points are (-1, 1) and (1, 5). Now, I need to find the equation of the line passing through these two points.

First, find the slope m:

m = (5 - 1)/(1 - (-1)) = 4 / 2 = 2.

So the slope is 2. Then using point-slope form with point (-1, 1):

y - 1 = 2(x - (-1)) => y - 1 = 2(x + 1) => y = 2x + 2 + 1 => y = 2x + 3.

Wait, so the transformed line is y = 2x + 3? Let me check with another point to make sure.

Take the third point (-1, -1) on the original line. Let's apply M:

M*[-1; -1] = [4*(-1) + (-1)*(-1); 2*(-1) + 1*(-1)] = [-4 + 1; -2 -1] = [-3; -3]

Now, check if (-3, -3) lies on y = 2x + 3.

Left side y = -3. Right side: 2*(-3) + 3 = -6 + 3 = -3. Yes, it works. So that seems correct.

Wait, so the image of the line y = 2x + 1 under M is y = 2x + 3? But let me think again. Is there a way to verify this more formally?

Alternatively, maybe I can do this by parametrizing the original line and then applying the matrix.

Let me parametrize the original line y = 2x + 1. Let me let x = t, so y = 2t + 1. So any point on the line can be written as (t, 2t + 1), where t is a real number.

Now, apply matrix M to this vector:

M*[t; 2t + 1] = [4*t + (-1)*(2t + 1); 2*t + 1*(2t + 1)].

Compute each component:

First component: 4t - 2t - 1 = 2t - 1

Second component: 2t + 2t + 1 = 4t + 1

So the transformed point is (2t - 1, 4t + 1). Now, we need to find the equation of the line that these points satisfy.

Let me denote the transformed coordinates as (x', y'). So:

x' = 2t - 1

y' = 4t + 1

We can solve for t from the first equation:

From x' = 2t - 1 => 2t = x' + 1 => t = (x' + 1)/2

Substitute into the second equation:

y' = 4*( (x' + 1)/2 ) + 1 = 2*(x' + 1) + 1 = 2x' + 2 + 1 = 2x' + 3

Therefore, y' = 2x' + 3, which is the same as y = 2x + 3. So that confirms the earlier result.

Therefore, the image of the line y = 2x + 1 under M is y = 2x + 3.

Wait, but let me think again. The original line had slope 2, and the transformed line also has slope 2. Is that expected?

Well, linear transformations can preserve or change slopes depending on the transformation. However, in this case, since the matrix M has determinant ad - bc = (4)(1) - (-1)(2) = 4 + 2 = 6, which is non-zero, so M is invertible. However, the slope of the line is preserved? Wait, not necessarily. Wait, but in this case, the slope ended up being the same. Let me check with another example.

Suppose we have a line with slope m, and we apply a linear transformation. The image of the line will have a different slope unless the transformation preserves direction. However, in this case, the slope remained the same. Let me check why.

Alternatively, maybe the transformation is such that it scales the direction vector of the line. Let me think.

The original line has direction vector (1, 2). Let me apply M to this direction vector.

M*[1; 2] = [4*1 + (-1)*2; 2*1 + 1*2] = [4 - 2; 2 + 2] = [2; 4]. So the direction vector becomes [2; 4], which is equivalent to [1; 2] scaled by 2. Therefore, the direction vector is scaled by 2, but the direction is the same. Therefore, the slope remains 2. Therefore, the image line has the same slope as the original line, which is 2. Therefore, the transformed line has slope 2, and we found the intercept is 3. So that seems consistent.

But let me check if the intercept is correct. The original line had intercept 1, and the transformed line has intercept 3. Since the transformation is linear, but the original line is affine, not passing through the origin. So the transformation of the affine line would be another affine line. The intercept changes because the transformation moves the entire line.

Alternatively, another way to think about it is that the transformation M maps the original line y = 2x + 1 to y = 2x + 3. So, the answer is y = 2x + 3.

But let me check once more with another point. Let me take the point (2, 5) on the original line (since y = 2*2 + 1 = 5). Apply M:

M*[2; 5] = [4*2 + (-1)*5; 2*2 + 1*5] = [8 - 5; 4 + 5] = [3; 9]. Now check if (3, 9) lies on y = 2x + 3. 2*3 + 3 = 6 + 3 = 9. Yes, correct. So that works.

Therefore, I think the answer is y = 2x + 3.

But let me think again if there's another way to approach this problem, maybe using linear algebra concepts.

Suppose we consider the original line y = 2x + 1. Let me write this as a vector equation. Let me think of points on the line as vectors. Let me write the line as:

For any scalar t, the point is (t, 2t + 1). Which is the same as before.

Alternatively, we can think of the line as the set of points (x, y) such that y - 2x = 1. So, the equation is y - 2x - 1 = 0.

Now, applying the matrix M to the coordinates (x, y). Let me denote the transformed coordinates as (x', y'). Then:

x' = 4x - y

y' = 2x + y

Wait, because M is [4 -1; 2 1], so:

x' = 4x + (-1)y = 4x - y

y' = 2x + 1*y = 2x + y

So, given that the original point (x, y) satisfies y = 2x + 1, we can substitute y into the equations for x' and y'.

So, substitute y = 2x + 1 into x' and y':

x' = 4x - (2x + 1) = 4x - 2x - 1 = 2x - 1

y' = 2x + (2x + 1) = 4x + 1

So, we have x' = 2x - 1 and y' = 4x + 1. Now, we can express x in terms of x' and substitute into the equation for y'.

From x' = 2x - 1, solving for x:

x = (x' + 1)/2

Substitute into y':

y' = 4*( (x' + 1)/2 ) + 1 = 2(x' + 1) + 1 = 2x' + 2 + 1 = 2x' + 3

Therefore, y' = 2x' + 3, which is the same as y = 2x + 3. So this method also gives the same result. Therefore, this confirms the answer again.

Therefore, I think the answer is y = 2x + 3.

But let me check once more if there's any chance I made a mistake in the matrix multiplication or solving for the matrix M.

Earlier, I found that M is [4 -1; 2 1]. Let me check again with the given vectors.

First vector [2; -1]:

4*2 + (-1)*(-1) = 8 + 1 = 9. Correct.

2*2 + 1*(-1) = 4 - 1 = 3. Correct.

Second vector [1; -3]:

4*1 + (-1)*(-3) = 4 + 3 = 7. Correct.

2*1 + 1*(-3) = 2 - 3 = -1. Correct. So matrix is correct.

Therefore, the rest of the steps are correct. So the image of the line is y = 2x + 3.

Alternatively, maybe the problem expects a different answer? Let me think again.

Wait, another way to think about it: the transformation M can be considered as a linear transformation followed by a translation? Wait, no, because M is a linear transformation. However, the original line is affine, not linear. So, the image of an affine line under a linear transformation is another affine line. However, since the original line is not passing through the origin, the image will not pass through the origin unless the transformation maps the origin to the origin, which it does, but the line is shifted.

But in our case, the original line is y = 2x + 1. The transformation M maps the origin (0,0) to (0,0). However, the original line is not passing through the origin. So, the image of the line is another line, which we found as y = 2x + 3.

Alternatively, maybe we can think of the original line as a linear combination. Let me think.

Let me consider that any point on the original line can be written as a point on the line y = 2x + 1. Let me write this as a vector equation. Let me take a point (x, 2x + 1). This can be written as x*(1, 2) + (0, 1). So, the line is the set of points { t*(1, 2) + (0, 1) | t ∈ ℝ }.

Therefore, applying the matrix M to this, we get:

M*(t*(1, 2) + (0, 1)) = t*M*(1, 2) + M*(0, 1)

Earlier, we found that M*(1, 2) = (2, 4) and M*(0, 1) = (-1, 1). Therefore, the image of the line is { t*(2, 4) + (-1, 1) | t ∈ ℝ }.

So, parametric equations:

x' = 2t - 1

y' = 4t + 1

Which is the same as before. Then, eliminating t:

From x' = 2t - 1 => t = (x' + 1)/2

Substitute into y':

y' = 4*( (x' + 1)/2 ) + 1 = 2(x' + 1) + 1 = 2x' + 2 + 1 = 2x' + 3. So again, same result.

Therefore, I think this is solid. The answer is y = 2x + 3.

But just to make sure, let me check if there's a possibility that the question is expecting something else. For example, maybe I misapplied the matrix? Let me check once again.

Original line: y = 2x + 1. Take a general point (x, y) on the line. Apply M:

x' = 4x - y

y' = 2x + y

But since y = 2x + 1, substitute:

x' = 4x - (2x + 1) = 2x - 1

y' = 2x + (2x + 1) = 4x + 1

So, solving for x from x' = 2x - 1:

x = (x' + 1)/2

Then, substitute into y':

y' = 4*( (x' + 1)/2 ) + 1 = 2(x' + 1) + 1 = 2x' + 2 + 1 = 2x' + 3. So, y' = 2x' + 3. So, yes, same result.

Therefore, I think the answer is definitely y = 2x + 3.

But just to be thorough, let me check if there's a different interpretation of the problem. For example, maybe the question is about the image of the line under the transformation, which could be different if the line is considered as a set of points, but I think the approach is correct.

Alternatively, maybe the problem is expecting me to consider the line as a vector space, but since it's an affine line, not a vector space, the transformation is applied as above.

Alternatively, maybe I need to consider the transformation of the entire line, including the direction vector and the translation. Wait, but since the transformation is linear, it doesn't include translation. However, the original line is not a vector space, so the image is an affine line.

But in our case, since we have an affine line, the image is another affine line, which we found as y = 2x + 3.

Therefore, I think the answer is correct.

**Final Answer**
The image of the line under matrix $\mathbf{M}$ is \boxed{y = 2x + 3}.

To determine the image of the line $ y = 2x + 1 $ under the matrix transformation $ \mathbf{M} $, we proceed in a structured manner:

---

### **Step 1: Determine the Matrix $ \mathbf{M} $**

We are given that:

- $ \mathbf{M} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 9 \\ 3 \end{pmatrix} $
- $ \mathbf{M} \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 7 \\ -1 \end{pmatrix} $

Let $ \mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $. Then:

- From the first vector:
  $$
  \begin{cases}
  2a - b = 9 \\
  2c - d = 3
  \end{cases}
  $$

- From the second vector:
  $$
  \begin{cases}
  a - 3b = 7 \\
  c - 3d = -1
  \end{cases}
  $$

Solving these equations:

- From the first pair:
  $ b = 2a - 9 $, substitute into $ a - 3b = 7 $:
  $$
  a - 3(2a - 9) = 7 \Rightarrow a - 6a + 27 = 7 \Rightarrow -5a = -20 \Rightarrow a = 4
  $$
  Then $ b = 2(4) - 9 = -1 $

- From the second pair:
  $ d = 2c - 3 $, substitute into $ c - 3d = -1 $:
  $$
  c - 3(2c - 3) = -1 \Rightarrow c - 6c + 9 = -1 \Rightarrow -5c = -10 \Rightarrow c = 2
  $$
  Then $ d = 2(2) - 3 = 1 $

So, the matrix is:
$$
\mathbf{M} = \begin{pmatrix} 4 & -1 \\ 2 & 1 \end{pmatrix}
$$

---

### **Step 2: Apply the Matrix to the Line**

We are to find the image of the line $ y = 2x + 1 $ under $ \mathbf{M} $. This is an **affine line**, not a vector space, so the transformation will map it to another **affine line**.

Let’s parametrize the line as:
$$
\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} t \\ 2t + 1 \end{pmatrix}, \quad t \in \mathbb{R}
$$

Apply $ \mathbf{M} $ to this vector:
$$
\mathbf{M} \begin{pmatrix} t \\ 2t + 1 \end{pmatrix} = \begin{pmatrix} 4t - (2t + 1) \\ 2t + (2t + 1) \end{pmatrix} = \begin{pmatrix} 2t - 1 \\ 4t + 1 \end{pmatrix}
$$

Let the transformed coordinates be $ (x', y') $. Then:
$$
x' = 2t - 1, \quad y' = 4t + 1
$$

Solve for $ t $ from the first equation:
$$
t = \frac{x' + 1}{2}
$$

Substitute into the second equation:
$$
y' = 4\left(\frac{x' + 1}{2}\right) + 1 = 2(x' + 1) + 1 = 2x' + 3
$$

---

### **Step 3: Final Result**

Thus, the image of the line $ y = 2x + 1 $ under the transformation $ \mathbf{M} $ is the line:
$$
\boxed{y = 2x + 3}
$$

Solution:
We have that $\mathbf{M} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 9 \\ 3 \end{pmatrix}$ and $\mathbf{M} \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 7 \\ -1 \end{pmatrix}.$  Then $\mathbf{M} \begin{pmatrix} 6 \\ -3 \end{pmatrix} = \begin{pmatrix} 27 \\ 9 \end{pmatrix},$ so
\[\mathbf{M} \begin{pmatrix} 6 \\ -3 \end{pmatrix} - \mathbf{M} \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 27 \\ 9 \end{pmatrix} - \begin{pmatrix} 7 \\ -1 \end{pmatrix}.\]This gives us $\mathbf{M} \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 20 \\ 10 \end{pmatrix},$ so
\[\mathbf{M} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}.\]Then
\[\mathbf{M} \begin{pmatrix} 1 \\ 0 \end{pmatrix} - \mathbf{M} \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix} - \begin{pmatrix} 7 \\ -1 \end{pmatrix}.\]This gives us $\mathbf{M} \begin{pmatrix} 0 \\ 3 \end{pmatrix} = \begin{pmatrix} -3 \\ 3 \end{pmatrix},$ so
\[\mathbf{M} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}.\]Finally,
\begin{align*}
\mathbf{M} \begin{pmatrix} 1 \\ 3 \end{pmatrix} &= \mathbf{M} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + 3 \mathbf{M} \begin{pmatrix} 0 \\ 1 \end{pmatrix} \\
&= \begin{pmatrix} 4 \\ 2 \end{pmatrix} + 3 \begin{pmatrix} -1 \\ 1 \end{pmatrix} \\
&= \begin{pmatrix} 1 \\ 5 \end{pmatrix}.
\end{align*}Since $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 3 \end{pmatrix}$ lie on the line $y = 2x + 1,$ we want to compute the equation of the line through $\begin{pmatrix} -1 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 5 \end{pmatrix}.$  The equation of this line is $\boxed{y = 2x + 3}.$