Explanation:
Okay, so I need to find the constant term in the expansion of $\left(10x^3 - \frac{1}{2x^2}\right)^5$. Hmm, let me think about how to approach this.

First, I remember that when you have a binomial raised to a power, like $(a + b)^n$, you can use the binomial theorem to expand it. The binomial theorem says that each term in the expansion is of the form $\binom{n}{k} a^{n - k} b^k$, where $k$ ranges from 0 to $n$.

In this case, my $a$ is $10x^3$ and my $b$ is $-\frac{1}{2x^2}$, and the exponent $n$ is 5. So, the general term in the expansion should be $\binom{5}{k} (10x^3)^{5 - k} \left(-\frac{1}{2x^2}\right)^k$.

Now, I need to find the term that is a constant, meaning the term where the exponent of $x$ is zero. So, my goal is to find the value of $k$ such that the exponent of $x$ in the term is zero.

Let me figure out the exponent of $x$ in the general term.

First, let's break down the exponents from each part.

The term $(10x^3)^{5 - k}$ will contribute an exponent of $3(5 - k)$ for $x$.

Then, the term $\left(-\frac{1}{2x^2}\right)^k$ can be rewritten as $(-1)^k \cdot \frac{1}{2^k} \cdot x^{-2k}$. So, the exponent of $x$ here is $-2k$.

Therefore, combining both parts, the total exponent of $x$ in the general term is $3(5 - k) - 2k$.

Let me compute that:

$3(5 - k) - 2k = 15 - 3k - 2k = 15 - 5k$.

We want this exponent to be zero for the constant term. So, set up the equation:

$15 - 5k = 0$

Solving for $k$:

$15 = 5k$

$k = 3$

Okay, so when $k = 3$, the term will be the constant term. Now, I need to compute that term.

Let me write out the term when $k = 3$:

$\binom{5}{3} (10x^3)^{5 - 3} \left(-\frac{1}{2x^2}\right)^3$

First, compute the binomial coefficient $\binom{5}{3}$.

$\binom{5}{3} = \frac{5!}{3!2!} = \frac{120}{6 \cdot 2} = \frac{120}{12} = 10$.

Okay, so that's 10.

Next, compute $(10x^3)^{5 - 3} = (10x^3)^2$. Let me calculate that:

$(10x^3)^2 = 10^2 \cdot x^{6} = 100x^6$.

Then, compute $\left(-\frac{1}{2x^2}\right)^3$. Let me break that down:

First, the coefficient part: $-\frac{1}{2}$ raised to the 3rd power is $(-1)^3 \cdot \frac{1}{2^3} = -1 \cdot \frac{1}{8} = -\frac{1}{8}$.

Then, the $x$ part: $x^{-2}$ raised to the 3rd power is $x^{-6}$.

So, altogether, $\left(-\frac{1}{2x^2}\right)^3 = -\frac{1}{8}x^{-6}$.

Now, multiply all the parts together:

The binomial coefficient is 10, multiplied by $100x^6$, multiplied by $-\frac{1}{8}x^{-6}$.

Let me compute the coefficients first:

$10 \times 100 = 1000$, and then $1000 \times (-\frac{1}{8}) = -\frac{1000}{8} = -125$.

Now, the $x$ terms: $x^6 \times x^{-6} = x^{0} = 1$.

So, the entire term simplifies to $-125 \times 1 = -125$.

Therefore, the constant term is $-125$.

Wait, let me check my steps again to make sure I didn't make a mistake.

First, binomial coefficient: $\binom{5}{3} = 10$ – that's correct.

Then, $(10x^3)^2 = 100x^6$ – correct.

Then, $\left(-\frac{1}{2x^2}\right)^3$: the coefficient is $(-1/2)^3 = -1/8$, and the exponent is $-2*3 = -6$, so that's correct.

Multiplying coefficients: 10 * 100 = 1000, then 1000 * (-1/8) = -125.

Yes, that seems right.

But let me check if I did the exponent calculation correctly.

Original exponent was 15 - 5k. For k=3, 15 - 15 = 0. Correct.

So, the exponent is zero, so that term is indeed constant.

Therefore, the constant term is -125.

Wait, but let me think again. Maybe I made a mistake in the sign? Let me check the sign again.

The term is $\left(-\frac{1}{2x^2}\right)^3$. Since the original binomial is $10x^3 - \frac{1}{2x^2}$, so the second term is negative. So when we raise it to the power of k, which is 3, we have (-1)^3, which is -1. So that's correct.

So, the coefficient is 10 * 100 * (-1/8) = -125.

Yes, that seems correct.

Alternatively, maybe I can compute the entire expansion step by step for verification? Although that might take longer, but let me try for k=3.

Alternatively, maybe I can check with another approach.

Alternatively, think about the general term again.

Alternatively, maybe I can compute the expansion for smaller exponents? Wait, but maybe not necessary.

Alternatively, think about the entire expansion.

But given that the exponent calculation is correct, and the coefficients are calculated correctly, I think -125 is the right answer.

But let me check once more.

Let me recompute the coefficient:

$\binom{5}{3} = 10$.

$(10x^3)^{2} = 10^2 x^{6} = 100x^6$.

$\left(-\frac{1}{2x^2}\right)^3 = (-1)^3 \times \frac{1}{2^3} \times x^{-6} = -\frac{1}{8}x^{-6}$.

Multiply all together: 10 * 100 * (-1/8) * x^{6 - 6} = 10 * 100 * (-1/8) * x^0.

Calculating 10 * 100 = 1000.

1000 * (-1/8) = -125.

Yes, that's correct.

Therefore, the constant term is indeed -125.

Wait, but let me check if there's another term that could be constant? For example, maybe another value of k?

Wait, the exponent was 15 - 5k = 0, so k=3. Since 15 -5k=0 only when k=3. So there is only one term with exponent zero. Therefore, that is the only constant term.

Therefore, the answer is -125.

But let me check with another method. Maybe expanding the expression step by step? Although with exponent 5, it might be tedious, but let me try.

Alternatively, think of the expansion as sum_{k=0}^5 [C(5,k)*(10x^3)^{5 -k}*(-1/(2x^2))^k].

So, for each term, the exponent of x is 3*(5 -k) - 2k = 15 - 5k.

Set 15 -5k = 0 => k=3.

So, only k=3 gives the constant term.

Therefore, the coefficient is C(5,3)*(10)^{2}*(-1/2)^3.

Wait, that's another way to write it.

C(5,3) is 10, 10^2 is 100, (-1/2)^3 is -1/8.

So, 10 * 100 * (-1/8) = -1000/8 = -125.

Same result.

Therefore, I think that's correct.

Alternatively, maybe I can compute the entire expansion? Let me try for k=0 to 5.

But that would take time, but let me try for k=0,1,2,3,4,5.

For k=0: term is C(5,0)*(10x^3)^5*(-1/(2x^2))^0 = 1 * 10^5 x^{15} * 1 = 100000x^{15}.

Exponent 15.

k=1: C(5,1)*(10x^3)^4*(-1/(2x^2))^1 = 5 * 10^4 x^{12} * (-1/(2x^2)) = 5*10000*(-1/2)x^{10} = 5*10000*(-1/2)x^{10} = -25000x^{10}.

Exponent 10.

k=2: C(5,2)*(10x^3)^3*(-1/(2x^2))^2 = 10 * 10^3 x^9 * (1/(4x^4)) = 10*1000*(1/4)x^{5} = 10*1000/4 x^5 = 2500x^5.

Exponent 5.

k=3: C(5,3)*(10x^3)^2*(-1/(2x^2))^3 = 10 * 100x^6 * (-1/(8x^6)) = 10*100*(-1/8)x^0 = -125.

Yes, that's the constant term.

k=4: C(5,4)*(10x^3)^1*(-1/(2x^2))^4 = 5 * 10x^3 * (1/(16x^8)) = 5*10/(16) x^{-5} = 50/16 x^{-5} = 25/8 x^{-5}.

Exponent -5.

k=5: C(5,5)*(10x^3)^0*(-1/(2x^2))^5 = 1 * 1 * (-1)^5/(32x^{10}) = -1/(32x^{10}).

Exponent -10.

So, indeed, only when k=3, the exponent is zero, and the term is -125.

Therefore, the constant term is -125.

So, I think that's correct.

But just to make sure, let me check if I made any arithmetic errors.

For k=3:

C(5,3) is 10.

(10x^3)^2 is 100x^6.

(-1/(2x^2))^3 is (-1)^3/(2^3 x^6) = -1/(8x^6).

Multiply them all together: 10 * 100x^6 * (-1/(8x^6)) = 10 * 100 * (-1/8) * x^{6 - 6} = 10 * (-100/8) = 10 * (-12.5) = -125.

Yes, that's correct.

Alternatively, 10 * 100 is 1000, 1000 divided by 8 is 125, and with the negative sign, it's -125.

Yes.

Therefore, I think the answer is definitely -125.

I don't see any mistakes in my reasoning. So, I think that's the correct constant term.

**Final Answer**
The constant term is \boxed{-125}.

To find the **constant term** in the expansion of
$$
\left(10x^3 - \frac{1}{2x^2}\right)^5,
$$
we apply the **Binomial Theorem**, which states that for any binomial expression $(a + b)^n$, the general term is given by:

$$
T_k = \binom{n}{k} a^{n-k} b^k.
$$

---

### Step 1: Identify the components

Let:
- $ a = 10x^3 $
- $ b = -\frac{1}{2x^2} $
- $ n = 5 $

Then, the general term in the expansion is:

$$
T_k = \binom{5}{k} (10x^3)^{5-k} \left(-\frac{1}{2x^2}\right)^k.
$$

---

### Step 2: Determine the exponent of $ x $

We are interested in the **constant term**, which means the exponent of $ x $ in the term must be zero.

Let’s compute the exponent of $ x $ in the general term:

- From $ (10x^3)^{5-k} $, the exponent of $ x $ is $ 3(5 - k) $.
- From $ \left(-\frac{1}{2x^2}\right)^k $, the exponent of $ x $ is $ -2k $.

So the total exponent of $ x $ is:

$$
3(5 - k) - 2k = 15 - 5k.
$$

Set this equal to zero to find the value of $ k $ that gives the constant term:

$$
15 - 5k = 0 \Rightarrow k = 3.
$$

---

### Step 3: Compute the constant term

Now substitute $ k = 3 $ into the general term:

$$
T_3 = \binom{5}{3} (10x^3)^{2} \left(-\frac{1}{2x^2}\right)^3.
$$

Compute each part:

- $ \binom{5}{3} = 10 $
- $ (10x^3)^2 = 100x^6 $
- $ \left(-\frac{1}{2x^2}\right)^3 = -\frac{1}{8x^6} $

Now multiply:

$$
T_3 = 10 \cdot 100x^6 \cdot \left(-\frac{1}{8x^6}\right) = 10 \cdot 100 \cdot \left(-\frac{1}{8}\right) \cdot x^{6 - 6} = -\frac{1000}{8} = -125.
$$

---

### Final Answer

$$
\boxed{-125}
$$

[Assistant]
Explanation:
Okay, so I need to find the constant term in the expansion of $\left(10x^3 - \frac{1}{2x^2}\right)^5$. Hmm, let me think about how to approach this.

First, I remember that when you have a binomial raised to a power, like $(a + b)^n$, you can use the binomial theorem to expand it. The binomial theorem says that each term in the expansion is of the form $\binom{n}{k} a^{n - k} b^k$, where $k$ ranges from 0 to $n$.

In this case, my $a$ is $10x^3$ and my $b$ is $-\frac{1}{2x^2}$, and the exponent $n$ is 5. So, the general term in the expansion should be $\binom{5}{k} (10x^3)^{5 - k} \left(-\frac{1}{2x^2}\right)^k$.

Now, I need to find the term that is a constant, meaning the term where the exponent of $x$ is zero. So, my goal is to find the value of $k$ such that the exponent of $x$ in the term is zero.

Let me figure out the exponent of $x$ in the general term.

First, let's break down the exponents from each part.

The term $(10x^3)^{5 - k}$ will contribute an exponent of $3(5 - k)$ for $x$.

Then, the term $\left(-\frac{1}{2x^2}\right)^k$ can be rewritten as $(-1)^k \cdot \frac{1}{2^k} \cdot x^{-2k}$. So, the exponent of $x$ here is $-2k$.

Therefore, combining both parts, the total exponent of $x$ in the general term is $3(5 - k) - 2k$.

Let me compute that:

$3(5 - k) - 2k = 15 - 3k - 2k = 15 - 5k$.

We want this exponent to be zero for the constant term. So, set up the equation:

$15 - 5k = 0$

Solving for $k$:

$15 = 5k$

$k = 3$

Okay, so when $k = 3$, the term will be the constant term. Now, I need to compute that term.

Let me write out the term when $k = 3$:

$\binom{5}{3} (10x^3)^{5 - 3} \left(-\frac{1}{2x^2}\right)^3$

First, compute the binomial coefficient $\binom{5}{3}$.

$\binom{5}{3} = \frac{5!}{3!2!} = \frac{120}{6 \cdot 2} = \frac{120}{12} = 10$.

Okay, so that's 10.

Next, compute $(10x^3)^{5 - 3} = (10x^3)^2$. Let me calculate that:

$(10x^3)^2 = 10^2 \cdot x^{6} = 100x^6$.

Then, compute $\left(-\frac{1}{2x^2}\right)^3$. Let me break that down:

First, the coefficient part: $-\frac{1}{2}$ raised to the 3rd power is $(-1)^3 \cdot \frac{1}{2^3} = -1 \cdot \frac{1}{8} = -\frac{1}{8}$.

Then, the $x$ part: $x^{-2}$ raised to the 3rd power is $x^{-6}$.

So, altogether, $\left(-\frac{1}{2x^2}\right)^3 = -\frac{1}{8}x^{-6}$.

Now, multiply all the parts together:

The binomial coefficient is 10, multiplied by $100x^6$, multiplied by $-\frac{1}{8}x^{-6}$.

Let me compute the coefficients first:

$10 \times 100 = 1000$, and then $1000 \times (-\frac{1}{8}) = -\frac{1000}{8} = -125$.

Now, the $x$ terms: $x^6 \times x^{-6} = x^{0} = 1$.

So, the entire term simplifies to $-125 \times 1 = -125$.

Therefore, the constant term is $-125$.

Wait, let me check my steps again to make sure I didn't make a mistake.

First, binomial coefficient: $\binom{5}{3} = 10$ – that's correct.

Then, $(10x^3)^2 = 100x^6$ – correct.

Then, $\left(-\frac{1}{2x^2}\right)^3$: the coefficient is $(-1/2)^3 = -1/8$, and the exponent is $-2*3 = -6$, so that's correct.

Multiplying coefficients: 10 * 100 = 1000, then 1000 * (-1/8) = -125.

Yes, that seems right.

But let me check if I did the exponent calculation correctly.

Original exponent was 15 - 5k. For k=3, 15 - 15 = 0. Correct.

So, the exponent is zero, so that term is indeed constant.

Therefore, the constant term is -125.

Wait, but let me think again. Maybe I made a mistake in the sign? Let me check the sign again.

The term is $\left(-\frac{1}{2x^2}\right)^3$. Since the original binomial is $10x^3 - \frac{1}{2x^2}$, so the second term is negative. So when we raise it to the power of k, which is 3, we have (-1)^3, which is -1. So that's correct.

So, the coefficient is 10 * 100 * (-1/8) = -125.

Yes, that seems correct.

Alternatively, maybe I can compute the entire expansion step by step for verification? Although that might take longer, but let me try for k=3.

Alternatively, maybe I can check with another approach.

Alternatively, think about the general term again.

Alternatively, maybe I can compute the expansion for smaller exponents? Wait, but maybe not necessary.

Alternatively, think about the entire expansion.

But given that the exponent calculation is correct, and the coefficients are calculated correctly, I think -125 is the right answer.

But let me check once more.

Let me recompute the coefficient:

$\binom{5}{3} = 10$.

$(10x^3)^{2} = 10^2 x^{6} = 100x^6$.

$\left(-\frac{1}{2x^2}\right)^3 = (-1)^3 \times \frac{1}{2^3} \times x^{-6} = -\frac{1}{8}x^{-6}$.

Multiply all together: 10 * 100 * (-1/8) * x^{6 - 6} = 10 * 100 * (-1/8) * x^0.

Calculating 10 * 100 = 1000.

1000 * (-1/8) = -125.

Yes, that's correct.

Therefore, the constant term is indeed -125.

Wait, but let me check if there's another term that could be constant? For example, maybe another value of k?

Wait, the exponent was 15 - 5k = 0, so k=3. Since 15 -5k=0 only when k=3. So there is only one term with exponent zero. Therefore, that is the only constant term.

Therefore, the answer is -125.

But let me check with another method. Maybe expanding the expression step by step? Although with exponent 5, it might be tedious, but let me try.

Alternatively, think of the expansion as sum_{k=0}^5 [C(5,k)*(10x^3)^{5 -k}*(-1/(2x^2))^k].

So, for each term, the exponent of x is 3*(5 -k) - 2k = 15 - 5k.

Set 15 -5k = 0 => k=3.

So, only k=3 gives the constant term.

Therefore, the coefficient is C(5,3)*(10)^{2}*(-1/2)^3.

Wait, that's another way to write it.

C(5,3) is 10, 10^2 is 100, (-1/2)^3 is -1/8.

So, 10 * 100 * (-1/8) = -1000/8 = -125.

Same result.

Therefore, I think that's correct.

Alternatively, maybe I can compute the entire expansion? Let me try for k=0 to 5.

But that would take time, but let me try for k=0,1,2,3,4,5.

For k=0: term is C(5,0)*(10x^3)^5*(-1/(2x^2))^0 = 1 * 10^5 x^{15} * 1 = 100000x^{15}.

Exponent 15.

k=1: C(5,1)*(10x^3)^4*(-1/(2x^2))^1 = 5 * 10^4 x^{12} * (-1/(2x^2)) = 5*10000*(-1/2)x^{10} = 5*10000*(-1/2)x^{10} = -25000x^{10}.

Exponent 10.

k=2: C(5,2)*(10x^3)^3*(-1/(2x^2))^2 = 10 * 10^3 x^9 * (1/(4x^4)) = 10*1000*(1/4)x^{5} = 10*1000/4 x^5 = 2500x^5.

Exponent 5.

k=3: C(5,3)*(10x^3)^2*(-1/(2x^2))^3 = 10 * 100x^6 * (-1/(8x^6)) = 10*100*(-1/8)x^0 = -125.

Yes, that's the constant term.

k=4: C(5,4)*(10x^3)^1*(-1/(2x^2))^4 = 5 * 10x^3 * (1/(16x^8)) = 5*10/(16) x^{-5} = 50/16 x^{-5} = 25/8 x^{-5}.

Exponent -5.

k=5: C(5,5)*(10x^3)^0*(-1/(2x^2))^5 = 1 * 1 * (-1)^5/(32x^{10}) = -1/(32x^{10}).

Exponent -10.

So, indeed, only when k=3, the exponent is zero, and the term is -125.

Therefore, the constant term is -125.

So, I think that's correct.

But just to make sure, let me check if I made any arithmetic errors.

For k=3:

C(5,3) is 10.

(10x^3)^2 is 100x^6.

(-1/(2x^2))^3 is (-1)^3/(2^3 x^6) = -1/(8x^6).

Multiply them all together: 10 * 100x^6 * (-1/(8x^6)) = 10 * 100 * (-1/8) * x^{6 - 6} = 10 * (-100/8) = 10 * (-12.5) = -125.

Yes, that's correct.

Alternatively, 10 * 100 is 1000, 1000 divided by 8 is 125, and with the negative sign, it's -125.

Yes.

Therefore, I think the answer is definitely -125.

I don't see any mistakes in my reasoning. So, I think that's the correct constant term.

**Final Answer**
The constant term is \boxed{-125}.

To find the **constant term** in the expansion of
$$
\left(10x^3 - \frac{1}{2x^2}\right)^5,
$$
we apply the **Binomial Theorem**, which states that for any binomial expression $(a + b)^n$, the general term is given by:

$$
T_k = \binom{n}{k} a^{n-k} b^k.
$$

---

### Step 1: Identify the components

Let:
- $ a = 10x^3 $
- $ b = -\frac{1}{2x^2} $
- $ n = 5 $

Then, the general term in the expansion is:

$$
T_k = \binom{5}{k} (10x^3)^{5-k} \left(-\frac{1}{2x^2}\right)^k.
$$

---

### Step 2: Determine the exponent of $ x $

We are interested in the **constant term**, which means the exponent of $ x $ in the term must be zero.

Let’s compute the exponent of $ x $ in the general term:

- From $ (10x^3)^{5-k} $, the exponent of $ x $ is $ 3(5 - k) $.
- From $ \left(-\frac{1}{2x^2}\right)^k $, the exponent of $ x $ is $ -2k $.

So the total exponent of $ x $ is:

$$
3(5 - k) - 2k = 15 - 5k.
$$

Set this equal to zero to find the value of $ k $ that gives the constant term:

$$
15 - 5k = 0 \Rightarrow k = 3.
$$

---

### Step 3: Compute the constant term

Now substitute $ k = 3 $ into the general term:

$$
T_3 = \binom{5}{3} (10x^3)^{2} \left(-\frac{1}{2x^2}\right)^3.
$$

Compute each part:

- $ \binom{5}{3} = 10 $
- $ (10x^3)^2 = 100x^6 $
- $ \left(-\frac{1}{2x^2}\right)^3 = -\frac{1}{8x^6} $

Now multiply:

$$
T_3 = 10 \cdot 100x^6 \cdot \left(-\frac{1}{8x^6}\right) = 10 \cdot 100 \cdot \left(-\frac{1}{8}\right) \cdot x^{6 - 6} = -\frac{1000}{8} = -125.
$$

---

### Final Answer

$$
\boxed{-125}
$$

Solution: To get a constant term, the exponents of $x$ must cancel.  If we take the term with 2 $x^3$'s and 3 $\frac{1}{x^2}$'s, then they will cancel.  By the Binomial Theorem, this term is $$\binom52 (10x^3)^2\left(-\frac{1}{2x^2}\right)^3=10\cdot100\cdot-\frac{1}{8}\cdot x^6\cdot\frac{1}{x^6}$$$$\Rightarrow \frac{1000}{-8}=\boxed{-125}$$


Explanation:
Okay, so I need to find the coordinates (x, y) of the fourth vertex of a parallelogram, given three of its vertices: (5, 3), (6, 8), and (7, 4). Also, it's mentioned that x > 7. Then, I have to find x + y. Alright, let me think about how to approach this.

First, I remember that in a parallelogram, opposite sides are equal and parallel. Also, the diagonals of a parallelogram bisect each other. Maybe I can use these properties to find the missing coordinate.

But wait, I need to figure out which points are connected. Since there are three points given, I need to determine how they are connected to form the parallelogram. There are different possibilities for which points are adjacent or opposite. Let me try to visualize or sketch the points.

Let me label the given points as A(5, 3), B(6, 8), and C(7, 4). The fourth point is D(x, y). Now, in a parallelogram, the order of the points matters. Depending on the order, the fourth point can be different. So, I need to figure out the correct order.

One way to approach this is to consider the vectors between the points. In a parallelogram, the vector from A to B should be the same as the vector from D to C, and the vector from B to C should be the same as the vector from A to D. Alternatively, maybe the other way around. Let me try to think.

Alternatively, since the diagonals bisect each other, the midpoint of diagonal AC should be the same as the midpoint of diagonal BD. Wait, but I don't know which points are connected by diagonals. Maybe I need to check different possibilities.

Let me try to consider different cases. Let me assume that the three points are connected in order, so that AB and BC are sides, and then find the fourth point. But since there are three points, there are different ways they can be arranged.

Alternatively, maybe I can use the property that in a parallelogram, the coordinates of the fourth vertex can be found by vector addition. For example, if we have points A, B, C, D, then vector AB equals vector DC, and vector AD equals vector BC.

But since I don't know the order, maybe I need to check different possibilities.

Let me try to list all possible cases for the order of the points. Let me assume that the three given points are A, B, C, and D is the fourth. Depending on which sides are connected, the fourth point will be different.

Case 1: Suppose the points are in order A, B, C, D. Then, AB and BC are adjacent sides, and then CD and DA are the other sides. But since we have three points, maybe the fourth point can be found by vector addition.

Alternatively, maybe using coordinates. Let me think.

Another approach: In a parallelogram, the coordinates of the fourth vertex can be found by adding the vectors. For example, if we have three points, say A, B, C, then the fourth point D can be found by D = A + C - B, or something like that. Wait, let me recall.

If we have points A, B, C, D in order, then the midpoint of AC should be the same as the midpoint of BD. So, if I can figure out which points are diagonally opposite, then I can use that.

But since I don't know which points are diagonally opposite, maybe I need to check different possibilities.

Let me try to compute midpoints for different pairs.

First, let me consider that maybe A and C are diagonally opposite, and B and D are diagonally opposite. Then, the midpoint of AC should equal the midpoint of BD.

Midpoint of AC: ((5 + 7)/2, (3 + 4)/2) = (12/2, 7/2) = (6, 3.5)

Midpoint of BD: ((6 + x)/2, (8 + y)/2). If this is equal to (6, 3.5), then:

(6 + x)/2 = 6 => 6 + x = 12 => x = 6

(8 + y)/2 = 3.5 => 8 + y = 7 => y = -1

But then x = 6, which is not greater than 7. The problem says x > 7, so this case is invalid. Therefore, this case is not possible.

Alternatively, maybe A and B are diagonally opposite, and C and D are diagonally opposite. Then midpoint of AB should equal midpoint of CD.

Midpoint of AB: ((5 + 6)/2, (3 + 8)/2) = (11/2, 11/2) = (5.5, 5.5)

Midpoint of CD: ((7 + x)/2, (4 + y)/2). Setting equal:

(7 + x)/2 = 5.5 => 7 + x = 11 => x = 4

(4 + y)/2 = 5.5 => 4 + y = 11 => y = 7

So x = 4, which is less than 7. Again, not satisfying x > 7. So this case is invalid.

Third possibility: A and D are diagonally opposite, and B and C are diagonally opposite. Then midpoint of AD equals midpoint of BC.

Midpoint of BC: ((6 + 7)/2, (8 + 4)/2) = (13/2, 12/2) = (6.5, 6)

Midpoint of AD: ((5 + x)/2, (3 + y)/2). Setting equal:

(5 + x)/2 = 6.5 => 5 + x = 13 => x = 8

(3 + y)/2 = 6 => 3 + y = 12 => y = 9

So x = 8, y = 9. Then x = 8 > 7, which satisfies the condition. So this seems like a valid case. Therefore, the coordinates of D would be (8, 9), so x + y = 17. Is this the answer? Wait, but let me check if this is correct.

But before that, let me check if there are other possibilities. Because maybe there are other configurations where the points are connected differently. For example, maybe the three given points are not in order, so maybe different pairings.

Wait, but in a parallelogram, there are only two diagonals. So if we have three points, there are three possibilities for which two are diagonally opposite. But we tried all three possibilities, and only one of them gave x > 7. So that must be the answer. But let me verify.

Alternatively, maybe I can use vectors. Let me try that approach.

Suppose we have points A(5, 3), B(6, 8), C(7, 4), and D(x, y). Let me assume that AB and DC are sides, and AD and BC are sides. Then vector AB = vector DC.

Vector AB is B - A = (6 - 5, 8 - 3) = (1, 5)

Vector DC is C - D = (7 - x, 4 - y). So if AB = DC, then:

1 = 7 - x => x = 7 - 1 = 6

5 = 4 - y => y = 4 - 5 = -1

So D would be (6, -1). But again, x = 6 < 7, which is invalid. So this case is invalid.

Alternatively, if vector BC = vector AD.

Vector BC = C - B = (7 - 6, 4 - 8) = (1, -4)

Vector AD = D - A = (x - 5, y - 3). So setting equal:

x - 5 = 1 => x = 6

y - 3 = -4 => y = -1

Again, same result (6, -1). Not valid.

Alternatively, maybe AB is equal to CD. Wait, vector AB = (1, 5). Vector CD = D - C = (x - 7, y - 4). So if AB = CD:

x - 7 = 1 => x = 8

y - 4 = 5 => y = 9

So D is (8, 9). Which is the same as before. So this gives x = 8 > 7, which is valid. So this seems correct.

Alternatively, if we consider BC = AD. Wait, we did that already. So that gives (6, -1). So that's invalid.

Alternatively, maybe AC is a side? Wait, but if we consider different sides. Let me think.

Alternatively, maybe the three points are not in order. For example, suppose that the parallelogram is A, B, D, C. Then, depending on the order, the vectors would be different.

Alternatively, maybe I can use the property that in a parallelogram, the coordinates can be found by adding the coordinates of two adjacent vertices and subtracting the third. Wait, like if you have three points, then the fourth is A + C - B or something like that.

Wait, earlier when I considered midpoints, when I assumed that A and D are diagonally opposite, and B and C are diagonally opposite, then midpoint of AD equals midpoint of BC. Which gave me D = (8, 9). So that seems to be the case.

Alternatively, if I use vector addition. Let me think. If AB and AC are adjacent sides, then the fourth vertex would be B + AC. Wait, but I need to be careful.

Alternatively, if I consider that in a parallelogram, the fourth vertex can be found by adding the vector from A to B to point C, or adding the vector from A to C to point B. Wait, let me think.

Suppose we have points A, B, and C. If AB and AC are adjacent sides, then the fourth vertex would be B + AC or C + AB. Wait, but in reality, if AB and AC are adjacent sides, then the fourth vertex would be B + AC. Let me compute that.

Vector AB is (1, 5). Vector AC is (7 - 5, 4 - 3) = (2, 1). So adding vector AC to point B: (6 + 2, 8 + 1) = (8, 9). So that gives D as (8, 9). Which is the same as before. So that seems to be the case.

Alternatively, adding vector AB to point C: (7 + 1, 4 + 5) = (8, 9). Same result. So that gives D as (8, 9). So that seems to be the correct answer.

But let me check if this actually forms a parallelogram. Let me verify the coordinates.

So points are A(5, 3), B(6, 8), C(7, 4), D(8, 9). Let me check if opposite sides are equal and parallel.

First, AB vector is (1, 5). DC vector is C to D: (8 - 7, 9 - 4) = (1, 5). So AB equals DC. Good.

AD vector is D - A = (8 - 5, 9 - 3) = (3, 6). BC vector is C - B = (7 - 6, 4 - 8) = (1, -4). Wait, that's not equal. Wait, hold on. Wait, if AB and DC are equal, then AD and BC should be equal?

Wait, no. Wait, in a parallelogram, opposite sides are equal. So AB should equal DC, and AD should equal BC. Wait, but here AD is (3, 6) and BC is (1, -4). These are not equal. So something is wrong here. Wait, that can't be. So maybe my assumption is wrong?

Wait, so if AB and DC are equal, then AD and BC should also be equal. But according to this, they are not. So that suggests that my previous conclusion is wrong? Wait, but how?

Wait, maybe I made a mistake in the order of the points. If the points are A, B, D, C, then the sides would be AB, BD, DC, and CA? Wait, no. Wait, maybe I need to check the order of the points.

Wait, if the points are A(5,3), B(6,8), D(8,9), C(7,4). Let me check the sides. If it's a parallelogram, then the order should be such that AB is adjacent to BC, etc. Wait, maybe the order is A, B, C, D? But then AB, BC, CD, DA. Wait, but if AB is (1,5), BC is (1, -4), CD is (-1, -5), DA is (-3, 6). Which are not equal. So that's not a parallelogram.

Wait, so maybe my previous approach is wrong? How come?

Wait, but earlier when I used the midpoint method, assuming that A and D are diagonally opposite, and B and C are diagonally opposite, then the midpoint of AD equals midpoint of BC. Which gave me D(8,9). But when I check the sides, it's not a parallelogram. So there must be a mistake here.

Wait, let me check the midpoints again. If A(5,3) and D(8,9) are diagonally opposite, then midpoint is ((5 + 8)/2, (3 + 9)/2) = (13/2, 12/2) = (6.5, 6). Midpoint of BC: B(6,8) and C(7,4). Midpoint is ((6 + 7)/2, (8 + 4)/2) = (13/2, 12/2) = (6.5, 6). So midpoints are equal. So that's correct. So diagonals bisect each other, so it is a parallelogram. But when I checked the sides, they didn't seem to match. Wait, maybe I messed up the sides.

Wait, if diagonals are AC and BD? Wait, no. If A and D are diagonally opposite, then the other diagonal is B and C. So diagonals are AD and BC. Wait, but in a parallelogram, diagonals bisect each other, but they are not necessarily equal unless it's a rectangle. So if the midpoints are equal, then it's a parallelogram. So maybe my mistake was in checking the sides.

Wait, let me check again. If the points are A(5,3), B(6,8), C(7,4), D(8,9). Let me check the sides. If the order is A, B, C, D, then sides AB, BC, CD, DA. But if it's a parallelogram, then AB should be equal and parallel to CD, and BC equal and parallel to DA.

AB vector: (6 - 5, 8 - 3) = (1, 5)

CD vector: (8 - 7, 9 - 4) = (1, 5). So AB equals CD. Good.

BC vector: (7 - 6, 4 - 8) = (1, -4)

DA vector: (5 - 8, 3 - 9) = (-3, -6). Wait, that's not equal to BC. Wait, so BC is (1, -4), DA is (-3, -6). These are not equal. So something is wrong here. But according to the midpoint, it should be a parallelogram. What's going on?

Wait, maybe the order of the points is different. Maybe the parallelogram is A, B, D, C? Let me check.

So sides AB, BD, DC, CA? Wait, no. If the order is A, B, D, C, then sides AB, BD, DC, and CA? Wait, that doesn't make sense. Wait, maybe the order is A, B, D, C? Let me check the sides AB, BD, DC, and CA? Wait, no, that's not a parallelogram. Wait, maybe the order is A, D, C, B? Let me try.

If the order is A, D, C, B. Then sides AD, DC, CB, BA.

AD vector: D - A = (8 - 5, 9 - 3) = (3, 6)

DC vector: C - D = (7 - 8, 4 - 9) = (-1, -5)

CB vector: B - C = (6 - 7, 8 - 4) = (-1, 4)

BA vector: A - B = (5 - 6, 3 - 8) = (-1, -5)

Wait, so AD is (3,6), DC is (-1, -5), CB is (-1, 4), BA is (-1, -5). Not equal. Not a parallelogram.

Wait, maybe the order is A, C, B, D? Let me check.

Sides AC, CB, BD, DA.

AC vector: C - A = (2, 1)

CB vector: B - C = (-1, 4)

BD vector: D - B = (8 - 6, 9 - 8) = (2, 1)

DA vector: A - D = (-3, -6)

So AC and BD are both (2,1) and (2,1). So those are equal. Then CB and DA? CB is (-1, 4), DA is (-3, -6). Not equal. So again, not a parallelogram.

Wait, so why does the midpoint method say it's a parallelogram? Because if diagonals bisect each other, then it's a parallelogram. So maybe my mistake is in the order of the points? Maybe the sides are AB and AD, and then BC and DC?

Wait, if the diagonals are AD and BC, then the sides would be AB, BD, DC, and CA? Wait, no. Wait, if diagonals are AD and BC, then the sides are AB, BD, DC, and CA? That seems not. Wait, maybe I need to think differently.

Wait, in a parallelogram, the diagonals are the connections between opposite vertices. So if diagonals are AD and BC, then the vertices are A, B, C, D connected as A-B-C-D-A? Wait, no. Wait, if diagonals are AD and BC, then the vertices are A, B, C, D connected as A-B-C-D-A? Wait, but then diagonals would be AC and BD. Wait, confusion arises here.

Wait, maybe I need to draw it mentally. Let me think of the coordinates:

A(5,3), B(6,8), C(7,4), D(8,9). Let me plot these points approximately.

Point A is at (5,3), which is somewhere in the lower left. Point B is at (6,8), upper right of A. Point C is at (7,4), which is to the right of B but lower. Point D is at (8,9), which is to the right of C and higher.

If I connect A to B to C to D to A, then AB is up and right, BC is down and right, CD is up and right, DA is down and left. Wait, not sure. Alternatively, maybe connecting A to B to D to C to A? Let me check.

AB is from A(5,3) to B(6,8). Then BD is from B(6,8) to D(8,9). Then DC is from D(8,9) to C(7,4). Then CA is from C(7,4) back to A(5,3). Hmm, not sure if that's a parallelogram.

Alternatively, maybe the sides are AB and DC, and AD and BC. Wait, AB is (1,5), DC is (1,5). So AB and DC are equal and parallel. Then AD is (3,6), BC is (1, -4). Not equal. So that's not a parallelogram. Wait, but according to the midpoint, diagonals bisect each other. So why isn't it a parallelogram?

Wait, maybe I made a mistake in assuming which points are diagonally opposite. If diagonals are AC and BD, then midpoints should be equal. Let me check.

Midpoint of AC: ((5 + 7)/2, (3 + 4)/2) = (6, 3.5)

Midpoint of BD: ((6 + 8)/2, (8 + 9)/2) = (14/2, 17/2) = (7, 8.5). These are not equal. So that's not the case. Wait, but earlier when I considered diagonals AD and BC, their midpoints were equal. So if diagonals are AD and BC, then midpoints are equal. So that would mean that the figure is a parallelogram. But when I check the sides, they don't seem to match. Wait, maybe I need to check the sides again.

If diagonals are AD and BC, then the sides would be AB, BD, DC, and CA? Wait, no. Wait, in a parallelogram, the sides are AB, BC, CD, DA. Wait, but if diagonals are AD and BC, then the sides are AB, BC, CD, DA. Wait, but then diagonals are AC and BD. Wait, confusion again.

Wait, maybe I need to recall that in a parallelogram, the diagonals are the lines connecting opposite vertices. So if the vertices are A, B, C, D in order, then diagonals are AC and BD. So if I have diagonals AD and BC, that would mean that the vertices are not in order A, B, C, D. Maybe the order is A, B, D, C? Then diagonals would be AD and BC. So in that case, the sides would be AB, BD, DC, and CA. Wait, but BD and CA are diagonals? No, BD is a side? Wait, no. If the order is A, B, D, C, then sides are AB, BD, DC, and CA. But BD and CA are not necessarily sides. Wait, this is getting too confusing.

Alternatively, maybe I need to use vectors again. If AB and DC are equal and parallel, and AD and BC are equal and parallel, then it's a parallelogram. So if AB = DC and AD = BC, then it's a parallelogram. So let me check with D(8,9):

AB vector is (1,5). DC vector is (8 - 7, 9 - 4) = (1,5). So AB = DC. Good.

AD vector is (8 - 5, 9 - 3) = (3,6). BC vector is (7 - 6, 4 - 8) = (1, -4). These are not equal. So AD ≠ BC. Therefore, it's not a parallelogram? But according to the midpoint, diagonals bisect each other. Wait, but if diagonals bisect each other, then it's a parallelogram. So there's a contradiction here. What is wrong?

Wait, maybe my calculation of the vectors is wrong. Wait, if diagonals are AD and BC, then the midpoints are equal. So if diagonals are AD and BC, then the figure is a parallelogram. But in that case, the sides would be AB, BD, DC, and CA? Wait, but BD and CA are diagonals? No, if diagonals are AD and BC, then the sides are AB, BC, CD, DA? Wait, no. Wait, if diagonals are AD and BC, then the vertices are connected as A to B to C to D to A? Wait, but then diagonals would be AC and BD. Wait, I think I need to clarify.

Let me recall that in a parallelogram, the diagonals are the lines connecting opposite vertices. So if the vertices are A, B, C, D in order, then diagonals are AC and BD. If diagonals are AD and BC, then the order of the vertices is different. For example, if the vertices are A, B, D, C, then diagonals would be AD and BC. So in this case, the sides would be AB, BD, DC, and CA. Wait, but BD is a diagonal? No, if the order is A, B, D, C, then sides are AB, BD, DC, and CA. Wait, but BD is a side? Wait, no. If you have a quadrilateral A, B, D, C, then sides are AB, BD, DC, and CA. But BD is a side? Wait, that would make the quadrilateral A-B-D-C-A. So sides AB, BD, DC, and CA. But BD is a side? That seems odd. Wait, but if you connect A to B to D to C to A, then BD is a side. But in that case, the sides AB and DC are not necessarily parallel. Wait, but according to the midpoint, diagonals AD and BC bisect each other, so it's a parallelogram. But how?

Wait, maybe I need to think of the sides as AB and DC, and AD and BC. Wait, if the order is A, B, C, D, then sides AB, BC, CD, DA. If diagonals are AC and BD. But if the order is A, B, D, C, then sides AB, BD, DC, CA. Then diagonals would be AD and BC. So in this case, sides AB and DC are not necessarily parallel. Wait, but if diagonals bisect each other, then it's a parallelogram. So maybe in this case, even though the sides AB and DC are not parallel, but the other sides BD and CA are?

Wait, this is getting too confusing. Maybe I need to use another approach. Let me check if the figure with points A(5,3), B(6,8), C(7,4), D(8,9) is a parallelogram by checking slopes.

Slope of AB: (8 - 3)/(6 - 5) = 5/1 = 5

Slope of DC: (9 - 4)/(8 - 7) = 5/1 = 5. So AB || DC.

Slope of AD: (9 - 3)/(8 - 5) = 6/3 = 2

Slope of BC: (4 - 8)/(7 - 6) = (-4)/1 = -4. So AD is not parallel to BC. Wait, so only one pair of sides is parallel. So it's not a parallelogram? But according to the midpoint, diagonals bisect each other. Wait, but if diagonals bisect each other, then it must be a parallelogram. So there's a contradiction here. What is wrong?

Wait, maybe I made a mistake in calculating the midpoint. Let me check again.

If diagonals are AD and BC, then midpoint of AD is ((5 + 8)/2, (3 + 9)/2) = (13/2, 12/2) = (6.5, 6). Midpoint of BC is ((6 + 7)/2, (8 + 4)/2) = (13/2, 12/2) = (6.5, 6). So midpoints are equal. Therefore, diagonals bisect each other, so it must be a parallelogram. But according to slopes, only one pair of sides is parallel. How is that possible?

Wait, maybe I miscalculated the slopes. Let me check again.

Slope of AB: (8 - 3)/(6 - 5) = 5/1 = 5. Correct.

Slope of DC: (9 - 4)/(8 - 7) = 5/1 = 5. Correct. So AB || DC.

Slope of AD: (9 - 3)/(8 - 5) = 6/3 = 2. Correct.

Slope of BC: (4 - 8)/(7 - 6) = (-4)/1 = -4. Correct. So AD is not parallel to BC. Wait, but in a parallelogram, both pairs of opposite sides should be parallel. So why is only one pair?

Wait, maybe the other pair of sides is BD and AC? Wait, BD is from B(6,8) to D(8,9). Slope is (9 - 8)/(8 - 6) = 1/2. AC is from A(5,3) to C(7,4). Slope is (4 - 3)/(7 - 5) = 1/2. So BD and AC are parallel. So BD || AC. So the other pair of sides are BD and AC. Wait, but BD and AC are diagonals, not sides. Wait, but in the figure, if the sides are AB, DC, BD, and AC? No, that doesn't make sense. Wait, maybe the sides are AB, DC, and the other sides are AD and BC? But AD and BC are not parallel. Wait, but BD and AC are diagonals. So if diagonals are parallel, then it's a parallelogram? No, in a parallelogram, diagonals are not necessarily parallel. Wait, but in this case, they are parallel? Wait, BD has slope 1/2, AC has slope 1/2. So they are parallel. But in a parallelogram, diagonals are not parallel. Wait, unless it's a degenerate parallelogram? Wait, no. Wait, in a general parallelogram, diagonals are not parallel. So if diagonals are parallel, then it's a special case? Wait, but if diagonals are parallel, then the figure is a parallelogram? Wait, no. Wait, if diagonals are parallel, then the figure is a parallelogram? Or is that a different condition?

Wait, let me think. In a parallelogram, diagonals bisect each other, but they are not necessarily parallel. If diagonals are parallel, then the figure would have to be a degenerate parallelogram? Wait, no. Wait, suppose diagonals are parallel. Let me take an example. Suppose we have a parallelogram with diagonals AC and BD. If AC || BD, then what does that imply?

In a parallelogram, diagonals bisect each other. If they are also parallel, then the only way this can happen is if the parallelogram is a rectangle? No, in a rectangle, diagonals are equal and bisect each other but are not parallel. Wait, in a rectangle, diagonals are congruent and bisect each other but are not parallel. In a rhombus, diagonals are not parallel. In a square, same as rectangle. Wait, so if diagonals are parallel, then the figure is not a parallelogram? Wait, but according to the midpoint condition, if diagonals bisect each other, then it's a parallelogram. So if diagonals bisect each other and are parallel, then it's a parallelogram. But in that case, if diagonals are parallel and bisect each other, then the figure is a parallelogram. But in that case, the sides would be... Wait, maybe in this case, the figure is a parallelogram with diagonals parallel? But how?

Wait, maybe my confusion comes from the fact that I'm considering different pairs of sides. If diagonals are AD and BC, which are parallel and bisect each other, then the sides would be AB, BD, DC, and CA. But BD and CA are diagonals? Wait, no. If the diagonals are AD and BC, then the sides are AB, BC, CD, DA. Wait, but then diagonals would be AC and BD. Wait, I think I need to stop here and accept that there is a contradiction in my calculations, which suggests that I made a mistake somewhere.

Wait, let me check once again. If the diagonals are AD and BC, which bisect each other, then the figure is a parallelogram. Therefore, the sides must be AB, BC, CD, DA. Wait, but then diagonals are AC and BD. Wait, but earlier I thought diagonals were AD and BC. So there is confusion here. Maybe I need to clarify.

Let me recall that in a parallelogram, the diagonals are the lines connecting opposite vertices. So if the vertices are A, B, C, D in order, then diagonals are AC and BD. If the diagonals are AD and BC, then the order of the vertices is different. For example, if the vertices are A, B, D, C, then diagonals are AD and BC. In this case, the sides are AB, BD, DC, and CA. But BD and CA are diagonals? No, BD is a side? Wait, if the order is A, B, D, C, then sides are AB, BD, DC, and CA. So BD is a side, and CA is a side. But in this case, BD and CA are not diagonals. Wait, but if diagonals are AD and BC, then they are connecting A to D and B to C. So in this case, the sides are AB, BD, DC, and CA. So BD is a side, and CA is a side. But then, in this case, the sides AB and DC are parallel (slope 5), and sides BD and CA are parallel (slope 1/2). Therefore, both pairs of opposite sides are parallel. Therefore, it is a parallelogram. Wait, but earlier I thought that sides AD and BC were supposed to be parallel, but in this case, sides BD and CA are parallel. So maybe I was wrong in thinking that AD and BC are sides. So in this case, the sides are AB, BD, DC, and CA. So AB || DC and BD || CA. Therefore, it is a parallelogram. So my mistake was in thinking that AD and BC were sides, but they are actually diagonals. So the sides are AB, BD, DC, and CA. Therefore, it is a parallelogram. Therefore, the coordinates D(8,9) are correct.

Therefore, x + y = 8 + 9 = 17. So the answer is 17.

But just to make sure, let me check the lengths of the sides. AB length: sqrt(1² + 5²) = sqrt(26). DC length: sqrt(1² + 5²) = sqrt(26). BD length: sqrt((8 - 6)² + (9 - 8)²) = sqrt(4 + 1) = sqrt(5). CA length: sqrt((7 - 5)² + (4 - 3)²) = sqrt(4 + 1) = sqrt(5). So opposite sides are equal. Therefore, it is a parallelogram. So my initial confusion was due to misidentifying which sides are which. So the answer is indeed (8,9), so x + y = 17.

Therefore, the value of x + y is 17.

**Final Answer**
The value of $x + y$ is \boxed{17}.

To determine the value of $ x + y $ for the fourth vertex of a parallelogram given three of its vertices: $ A(5, 3) $, $ B(6, 8) $, and $ C(7, 4) $, we can use the geometric property that **in a parallelogram, the diagonals bisect each other**. This means that the **midpoint of one diagonal is the same as the midpoint of the other diagonal**.

---

### Step 1: Assume the fourth vertex is $ D(x, y) $

We are told that $ x > 7 $, so we will look for a configuration where this condition is satisfied.

Let’s consider the possibility that the diagonals are $ AC $ and $ BD $. That is, the diagonals connect $ A $ to $ C $ and $ B $ to $ D $.

---

### Step 2: Compute the midpoint of diagonal $ AC $

$$
\text{Midpoint of } AC = \left( \frac{5 + 7}{2}, \frac{3 + 4}{2} \right) = (6, 3.5)
$$

---

### Step 3: Set the midpoint of diagonal $ BD $ equal to that of $ AC $

$$
\text{Midpoint of } BD = \left( \frac{6 + x}{2}, \frac{8 + y}{2} \right)
$$

Set this equal to $ (6, 3.5) $:

$$
\frac{6 + x}{2} = 6 \Rightarrow 6 + x = 12 \Rightarrow x = 6
$$
$$
\frac{8 + y}{2} = 3.5 \Rightarrow 8 + y = 7 \Rightarrow y = -1
$$

But this gives $ x = 6 $, which **does not satisfy** the condition $ x > 7 $. So this configuration is invalid.

---

### Step 4: Try another configuration — diagonals $ AD $ and $ BC $

Now, assume the diagonals are $ AD $ and $ BC $, and compute their midpoints.

$$
\text{Midpoint of } BC = \left( \frac{6 + 7}{2}, \frac{8 + 4}{2} \right) = (6.5, 6)
$$

$$
\text{Midpoint of } AD = \left( \frac{5 + x}{2}, \frac{3 + y}{2} \right)
$$

Set this equal to $ (6.5, 6) $:

$$
\frac{5 + x}{2} = 6.5 \Rightarrow 5 + x = 13 \Rightarrow x = 8
$$
$$
\frac{3 + y}{2} = 6 \Rightarrow 3 + y = 12 \Rightarrow y = 9
$$

Now, $ x = 8 $, which **does satisfy** the condition $ x > 7 $.

---

### Step 5: Verify that the figure is a parallelogram

Let’s check if the sides are equal and parallel.

- Vector $ \vec{AB} = (6 - 5, 8 - 3) = (1, 5) $
- Vector $ \vec{DC} = (8 - 7, 9 - 4) = (1, 5) $ → $ \vec{AB} = \vec{DC} $

- Vector $ \vec{AD} = (8 - 5, 9 - 3) = (3, 6) $
- Vector $ \vec{BC} = (7 - 6, 4 - 8) = (1, -4) $ → Not equal, but we are checking **opposite sides**.

Wait — we should check the **other pair of opposite sides**:

- Vector $ \vec{BD} = (8 - 6, 9 - 8) = (2, 1) $
- Vector $ \vec{AC} = (7 - 5, 4 - 3) = (2, 1) $ → $ \vec{BD} = \vec{AC} $

So both pairs of opposite sides are equal and parallel, confirming that the figure is a **parallelogram**.

---

### Final Step: Compute $ x + y $

$$
x = 8, \quad y = 9 \Rightarrow x + y = 8 + 9 = \boxed{17}
$$

---

### ✅ Final Answer:

$$
\boxed{17}
$$

[Assistant]
Explanation:
Okay, so I need to find the coordinates (x, y) of the fourth vertex of a parallelogram, given three of its vertices: (5, 3), (6, 8), and (7, 4). Also, it's mentioned that x > 7. Then, I have to find x + y. Alright, let me think about how to approach this.

First, I remember that in a parallelogram, opposite sides are equal and parallel. Also, the diagonals of a parallelogram bisect each other. Maybe I can use these properties to find the missing coordinate.

But wait, I need to figure out which points are connected. Since there are three points given, I need to determine how they are connected to form the parallelogram. There are different possibilities for which points are adjacent or opposite. Let me try to visualize or sketch the points.

Let me label the given points as A(5, 3), B(6, 8), and C(7, 4). The fourth point is D(x, y). Now, in a parallelogram, the order of the points matters. Depending on the order, the fourth point can be different. So, I need to figure out the correct order.

One way to approach this is to consider the vectors between the points. In a parallelogram, the vector from A to B should be the same as the vector from D to C, and the vector from B to C should be the same as the vector from A to D. Alternatively, maybe the other way around. Let me try to think.

Alternatively, since the diagonals bisect each other, the midpoint of diagonal AC should be the same as the midpoint of diagonal BD. Wait, but I don't know which points are connected by diagonals. Maybe I need to check different possibilities.

Let me try to consider different cases. Let me assume that the three points are connected in order, so that AB and BC are sides, and then find the fourth point. But since there are three points, there are different ways they can be arranged.

Alternatively, maybe I can use the property that in a parallelogram, the coordinates of the fourth vertex can be found by vector addition. For example, if we have points A, B, C, D, then vector AB equals vector DC, and vector AD equals vector BC.

But since I don't know the order, maybe I need to check different possibilities.

Let me try to list all possible cases for the order of the points. Let me assume that the three given points are A, B, C, and D is the fourth. Depending on which sides are connected, the fourth point will be different.

Case 1: Suppose the points are in order A, B, C, D. Then, AB and BC are adjacent sides, and then CD and DA are the other sides. But since we have three points, maybe the fourth point can be found by vector addition.

Alternatively, maybe using coordinates. Let me think.

Another approach: In a parallelogram, the coordinates of the fourth vertex can be found by adding the vectors. For example, if we have three points, say A, B, C, then the fourth point D can be found by D = A + C - B, or something like that. Wait, let me recall.

If we have points A, B, C, D in order, then the midpoint of AC should be the same as the midpoint of BD. So, if I can figure out which points are diagonally opposite, then I can use that.

But since I don't know which points are diagonally opposite, maybe I need to check different possibilities.

Let me try to compute midpoints for different pairs.

First, let me consider that maybe A and C are diagonally opposite, and B and D are diagonally opposite. Then, the midpoint of AC should equal the midpoint of BD.

Midpoint of AC: ((5 + 7)/2, (3 + 4)/2) = (12/2, 7/2) = (6, 3.5)

Midpoint of BD: ((6 + x)/2, (8 + y)/2). If this is equal to (6, 3.5), then:

(6 + x)/2 = 6 => 6 + x = 12 => x = 6

(8 + y)/2 = 3.5 => 8 + y = 7 => y = -1

But then x = 6, which is not greater than 7. The problem says x > 7, so this case is invalid. Therefore, this case is not possible.

Alternatively, maybe A and B are diagonally opposite, and C and D are diagonally opposite. Then midpoint of AB should equal midpoint of CD.

Midpoint of AB: ((5 + 6)/2, (3 + 8)/2) = (11/2, 11/2) = (5.5, 5.5)

Midpoint of CD: ((7 + x)/2, (4 + y)/2). Setting equal:

(7 + x)/2 = 5.5 => 7 + x = 11 => x = 4

(4 + y)/2 = 5.5 => 4 + y = 11 => y = 7

So x = 4, which is less than 7. Again, not satisfying x > 7. So this case is invalid.

Third possibility: A and D are diagonally opposite, and B and C are diagonally opposite. Then midpoint of AD equals midpoint of BC.

Midpoint of BC: ((6 + 7)/2, (8 + 4)/2) = (13/2, 12/2) = (6.5, 6)

Midpoint of AD: ((5 + x)/2, (3 + y)/2). Setting equal:

(5 + x)/2 = 6.5 => 5 + x = 13 => x = 8

(3 + y)/2 = 6 => 3 + y = 12 => y = 9

So x = 8, y = 9. Then x = 8 > 7, which satisfies the condition. So this seems like a valid case. Therefore, the coordinates of D would be (8, 9), so x + y = 17. Is this the answer? Wait, but let me check if this is correct.

But before that, let me check if there are other possibilities. Because maybe there are other configurations where the points are connected differently. For example, maybe the three given points are not in order, so maybe different pairings.

Wait, but in a parallelogram, there are only two diagonals. So if we have three points, there are three possibilities for which two are diagonally opposite. But we tried all three possibilities, and only one of them gave x > 7. So that must be the answer. But let me verify.

Alternatively, maybe I can use vectors. Let me try that approach.

Suppose we have points A(5, 3), B(6, 8), C(7, 4), and D(x, y). Let me assume that AB and DC are sides, and AD and BC are sides. Then vector AB = vector DC.

Vector AB is B - A = (6 - 5, 8 - 3) = (1, 5)

Vector DC is C - D = (7 - x, 4 - y). So if AB = DC, then:

1 = 7 - x => x = 7 - 1 = 6

5 = 4 - y => y = 4 - 5 = -1

So D would be (6, -1). But again, x = 6 < 7, which is invalid. So this case is invalid.

Alternatively, if vector BC = vector AD.

Vector BC = C - B = (7 - 6, 4 - 8) = (1, -4)

Vector AD = D - A = (x - 5, y - 3). So setting equal:

x - 5 = 1 => x = 6

y - 3 = -4 => y = -1

Again, same result (6, -1). Not valid.

Alternatively, maybe AB is equal to CD. Wait, vector AB = (1, 5). Vector CD = D - C = (x - 7, y - 4). So if AB = CD:

x - 7 = 1 => x = 8

y - 4 = 5 => y = 9

So D is (8, 9). Which is the same as before. So this gives x = 8 > 7, which is valid. So this seems correct.

Alternatively, if we consider BC = AD. Wait, we did that already. So that gives (6, -1). So that's invalid.

Alternatively, maybe AC is a side? Wait, but if we consider different sides. Let me think.

Alternatively, maybe the three points are not in order. For example, suppose that the parallelogram is A, B, D, C. Then, depending on the order, the vectors would be different.

Alternatively, maybe I can use the property that in a parallelogram, the coordinates can be found by adding the coordinates of two adjacent vertices and subtracting the third. Wait, like if you have three points, then the fourth is A + C - B or something like that.

Wait, earlier when I considered midpoints, when I assumed that A and D are diagonally opposite, and B and C are diagonally opposite, then midpoint of AD equals midpoint of BC. Which gave me D = (8, 9). So that seems to be the case.

Alternatively, if I use vector addition. Let me think. If AB and AC are adjacent sides, then the fourth vertex would be B + AC. Wait, but I need to be careful.

Alternatively, if I consider that in a parallelogram, the fourth vertex can be found by adding the vector from A to B to point C, or adding the vector from A to C to point B. Wait, let me think.

Suppose we have points A, B, and C. If AB and AC are adjacent sides, then the fourth vertex would be B + AC or C + AB. Wait, but in reality, if AB and AC are adjacent sides, then the fourth vertex would be B + AC. Let me compute that.

Vector AB is (1, 5). Vector AC is (7 - 5, 4 - 3) = (2, 1). So adding vector AC to point B: (6 + 2, 8 + 1) = (8, 9). So that gives D as (8, 9). Which is the same as before. So that seems to be the case.

Alternatively, adding vector AB to point C: (7 + 1, 4 + 5) = (8, 9). Same result. So that gives D as (8, 9). So that seems to be the correct answer.

But let me check if this actually forms a parallelogram. Let me verify the coordinates.

So points are A(5, 3), B(6, 8), C(7, 4), D(8, 9). Let me check if opposite sides are equal and parallel.

First, AB vector is (1, 5). DC vector is C to D: (8 - 7, 9 - 4) = (1, 5). So AB equals DC. Good.

AD vector is D - A = (8 - 5, 9 - 3) = (3, 6). BC vector is C - B = (7 - 6, 4 - 8) = (1, -4). Wait, that's not equal. Wait, hold on. Wait, if AB and DC are equal, then AD and BC should be equal?

Wait, no. Wait, in a parallelogram, opposite sides are equal. So AB should equal DC, and AD should equal BC. Wait, but here AD is (3, 6) and BC is (1, -4). These are not equal. So something is wrong here. Wait, that can't be. So maybe my assumption is wrong?

Wait, so if AB and DC are equal, then AD and BC should also be equal. But according to this, they are not. So that suggests that my previous conclusion is wrong? Wait, but how?

Wait, maybe I made a mistake in the order of the points. If the points are A, B, D, C, then the sides would be AB, BD, DC, and CA? Wait, no. Wait, maybe I need to check the order of the points.

Wait, if the points are A(5,3), B(6,8), D(8,9), C(7,4). Let me check the sides. If it's a parallelogram, then the order should be such that AB is adjacent to BC, etc. Wait, maybe the order is A, B, C, D? But then AB, BC, CD, DA. Wait, but if AB is (1,5), BC is (1, -4), CD is (-1, -5), DA is (-3, 6). Which are not equal. So that's not a parallelogram.

Wait, so maybe my previous approach is wrong? How come?

Wait, but earlier when I used the midpoint method, assuming that A and D are diagonally opposite, and B and C are diagonally opposite, then the midpoint of AD equals midpoint of BC. Which gave me D(8,9). But when I check the sides, it's not a parallelogram. So there must be a mistake here.

Wait, let me check the midpoints again. If A(5,3) and D(8,9) are diagonally opposite, then midpoint is ((5 + 8)/2, (3 + 9)/2) = (13/2, 12/2) = (6.5, 6). Midpoint of BC: B(6,8) and C(7,4). Midpoint is ((6 + 7)/2, (8 + 4)/2) = (13/2, 12/2) = (6.5, 6). So midpoints are equal. So that's correct. So diagonals bisect each other, so it is a parallelogram. But when I checked the sides, they didn't seem to match. Wait, maybe I messed up the sides.

Wait, if diagonals are AC and BD? Wait, no. If A and D are diagonally opposite, then the other diagonal is B and C. So diagonals are AD and BC. Wait, but in a parallelogram, diagonals bisect each other, but they are not necessarily equal unless it's a rectangle. So if the midpoints are equal, then it's a parallelogram. So maybe my mistake was in checking the sides.

Wait, let me check again. If the points are A(5,3), B(6,8), C(7,4), D(8,9). Let me check the sides. If the order is A, B, C, D, then sides AB, BC, CD, DA. But if it's a parallelogram, then AB should be equal and parallel to CD, and BC equal and parallel to DA.

AB vector: (6 - 5, 8 - 3) = (1, 5)

CD vector: (8 - 7, 9 - 4) = (1, 5). So AB equals CD. Good.

BC vector: (7 - 6, 4 - 8) = (1, -4)

DA vector: (5 - 8, 3 - 9) = (-3, -6). Wait, that's not equal to BC. Wait, so BC is (1, -4), DA is (-3, -6). These are not equal. So something is wrong here. But according to the midpoint, it should be a parallelogram. What's going on?

Wait, maybe the order of the points is different. Maybe the parallelogram is A, B, D, C? Let me check.

So sides AB, BD, DC, CA? Wait, no. If the order is A, B, D, C, then sides AB, BD, DC, and CA? Wait, that doesn't make sense. Wait, maybe the order is A, B, D, C? Let me check the sides AB, BD, DC, and CA? Wait, no, that's not a parallelogram. Wait, maybe the order is A, D, C, B? Let me try.

If the order is A, D, C, B. Then sides AD, DC, CB, BA.

AD vector: D - A = (8 - 5, 9 - 3) = (3, 6)

DC vector: C - D = (7 - 8, 4 - 9) = (-1, -5)

CB vector: B - C = (6 - 7, 8 - 4) = (-1, 4)

BA vector: A - B = (5 - 6, 3 - 8) = (-1, -5)

Wait, so AD is (3,6), DC is (-1, -5), CB is (-1, 4), BA is (-1, -5). Not equal. Not a parallelogram.

Wait, maybe the order is A, C, B, D? Let me check.

Sides AC, CB, BD, DA.

AC vector: C - A = (2, 1)

CB vector: B - C = (-1, 4)

BD vector: D - B = (8 - 6, 9 - 8) = (2, 1)

DA vector: A - D = (-3, -6)

So AC and BD are both (2,1) and (2,1). So those are equal. Then CB and DA? CB is (-1, 4), DA is (-3, -6). Not equal. So again, not a parallelogram.

Wait, so why does the midpoint method say it's a parallelogram? Because if diagonals bisect each other, then it's a parallelogram. So maybe my mistake is in the order of the points? Maybe the sides are AB and AD, and then BC and DC?

Wait, if the diagonals are AD and BC, then the sides would be AB, BD, DC, and CA? Wait, no. Wait, if diagonals are AD and BC, then the sides are AB, BD, DC, and CA? That seems not. Wait, maybe I need to think differently.

Wait, in a parallelogram, the diagonals are the connections between opposite vertices. So if diagonals are AD and BC, then the vertices are A, B, C, D connected as A-B-C-D-A? Wait, no. Wait, if diagonals are AD and BC, then the vertices are A, B, C, D connected as A-B-C-D-A? Wait, but then diagonals would be AC and BD. Wait, confusion arises here.

Wait, maybe I need to draw it mentally. Let me think of the coordinates:

A(5,3), B(6,8), C(7,4), D(8,9). Let me plot these points approximately.

Point A is at (5,3), which is somewhere in the lower left. Point B is at (6,8), upper right of A. Point C is at (7,4), which is to the right of B but lower. Point D is at (8,9), which is to the right of C and higher.

If I connect A to B to C to D to A, then AB is up and right, BC is down and right, CD is up and right, DA is down and left. Wait, not sure. Alternatively, maybe connecting A to B to D to C to A? Let me check.

AB is from A(5,3) to B(6,8). Then BD is from B(6,8) to D(8,9). Then DC is from D(8,9) to C(7,4). Then CA is from C(7,4) back to A(5,3). Hmm, not sure if that's a parallelogram.

Alternatively, maybe the sides are AB and DC, and AD and BC. Wait, AB is (1,5), DC is (1,5). So AB and DC are equal and parallel. Then AD is (3,6), BC is (1, -4). Not equal. So that's not a parallelogram. Wait, but according to the midpoint, diagonals bisect each other. So why isn't it a parallelogram?

Wait, maybe I made a mistake in assuming which points are diagonally opposite. If diagonals are AC and BD, then midpoints should be equal. Let me check.

Midpoint of AC: ((5 + 7)/2, (3 + 4)/2) = (6, 3.5)

Midpoint of BD: ((6 + 8)/2, (8 + 9)/2) = (14/2, 17/2) = (7, 8.5). These are not equal. So that's not the case. Wait, but earlier when I considered diagonals AD and BC, their midpoints were equal. So if diagonals are AD and BC, then midpoints are equal. So that would mean that the figure is a parallelogram. But when I check the sides, they don't seem to match. Wait, maybe I need to check the sides again.

If diagonals are AD and BC, then the sides would be AB, BD, DC, and CA? Wait, no. Wait, in a parallelogram, the sides are AB, BC, CD, DA. Wait, but if diagonals are AD and BC, then the sides are AB, BC, CD, DA. Wait, but then diagonals are AC and BD. Wait, confusion again.

Wait, maybe I need to recall that in a parallelogram, the diagonals are the lines connecting opposite vertices. So if the vertices are A, B, C, D in order, then diagonals are AC and BD. So if I have diagonals AD and BC, that would mean that the vertices are not in order A, B, C, D. Maybe the order is A, B, D, C? Then diagonals would be AD and BC. So in that case, the sides would be AB, BD, DC, and CA. Wait, but BD and CA are diagonals? No, BD is a side? Wait, no. If the order is A, B, D, C, then sides are AB, BD, DC, and CA. But BD and CA are not necessarily sides. Wait, this is getting too confusing.

Alternatively, maybe I need to use vectors again. If AB and DC are equal and parallel, and AD and BC are equal and parallel, then it's a parallelogram. So if AB = DC and AD = BC, then it's a parallelogram. So let me check with D(8,9):

AB vector is (1,5). DC vector is (8 - 7, 9 - 4) = (1,5). So AB = DC. Good.

AD vector is (8 - 5, 9 - 3) = (3,6). BC vector is (7 - 6, 4 - 8) = (1, -4). These are not equal. So AD ≠ BC. Therefore, it's not a parallelogram? But according to the midpoint, diagonals bisect each other. Wait, but if diagonals bisect each other, then it's a parallelogram. So there's a contradiction here. What is wrong?

Wait, maybe my calculation of the vectors is wrong. Wait, if diagonals are AD and BC, then the midpoints are equal. So if diagonals are AD and BC, then the figure is a parallelogram. But in that case, the sides would be AB, BD, DC, and CA? Wait, but BD and CA are diagonals? No, if diagonals are AD and BC, then the sides are AB, BC, CD, DA? Wait, no. Wait, if diagonals are AD and BC, then the vertices are connected as A to B to C to D to A? Wait, but then diagonals would be AC and BD. Wait, I think I need to clarify.

Let me recall that in a parallelogram, the diagonals are the lines connecting opposite vertices. So if the vertices are A, B, C, D in order, then diagonals are AC and BD. If diagonals are AD and BC, then the order of the vertices is different. For example, if the vertices are A, B, D, C, then diagonals would be AD and BC. So in this case, the sides would be AB, BD, DC, and CA. Wait, but BD is a diagonal? No, if the order is A, B, D, C, then sides are AB, BD, DC, and CA. Wait, but BD is a side? Wait, no. If you have a quadrilateral A, B, D, C, then sides are AB, BD, DC, and CA. But BD is a side? Wait, that would make the quadrilateral A-B-D-C-A. So sides AB, BD, DC, and CA. But BD is a side? That seems odd. Wait, but if you connect A to B to D to C to A, then BD is a side. But in that case, the sides AB and DC are not necessarily parallel. Wait, but according to the midpoint, diagonals AD and BC bisect each other, so it's a parallelogram. But how?

Wait, maybe I need to think of the sides as AB and DC, and AD and BC. Wait, if the order is A, B, C, D, then sides AB, BC, CD, DA. If diagonals are AC and BD. But if the order is A, B, D, C, then sides AB, BD, DC, CA. Then diagonals would be AD and BC. So in this case, sides AB and DC are not necessarily parallel. Wait, but if diagonals bisect each other, then it's a parallelogram. So maybe in this case, even though the sides AB and DC are not parallel, but the other sides BD and CA are?

Wait, this is getting too confusing. Maybe I need to use another approach. Let me check if the figure with points A(5,3), B(6,8), C(7,4), D(8,9) is a parallelogram by checking slopes.

Slope of AB: (8 - 3)/(6 - 5) = 5/1 = 5

Slope of DC: (9 - 4)/(8 - 7) = 5/1 = 5. So AB || DC.

Slope of AD: (9 - 3)/(8 - 5) = 6/3 = 2

Slope of BC: (4 - 8)/(7 - 6) = (-4)/1 = -4. So AD is not parallel to BC. Wait, so only one pair of sides is parallel. So it's not a parallelogram? But according to the midpoint, diagonals bisect each other. Wait, but if diagonals bisect each other, then it must be a parallelogram. So there's a contradiction here. What is wrong?

Wait, maybe I made a mistake in calculating the midpoint. Let me check again.

If diagonals are AD and BC, then midpoint of AD is ((5 + 8)/2, (3 + 9)/2) = (13/2, 12/2) = (6.5, 6). Midpoint of BC is ((6 + 7)/2, (8 + 4)/2) = (13/2, 12/2) = (6.5, 6). So midpoints are equal. Therefore, diagonals bisect each other, so it must be a parallelogram. But according to slopes, only one pair of sides is parallel. How is that possible?

Wait, maybe I miscalculated the slopes. Let me check again.

Slope of AB: (8 - 3)/(6 - 5) = 5/1 = 5. Correct.

Slope of DC: (9 - 4)/(8 - 7) = 5/1 = 5. Correct. So AB || DC.

Slope of AD: (9 - 3)/(8 - 5) = 6/3 = 2. Correct.

Slope of BC: (4 - 8)/(7 - 6) = (-4)/1 = -4. Correct. So AD is not parallel to BC. Wait, but in a parallelogram, both pairs of opposite sides should be parallel. So why is only one pair?

Wait, maybe the other pair of sides is BD and AC? Wait, BD is from B(6,8) to D(8,9). Slope is (9 - 8)/(8 - 6) = 1/2. AC is from A(5,3) to C(7,4). Slope is (4 - 3)/(7 - 5) = 1/2. So BD and AC are parallel. So BD || AC. So the other pair of sides are BD and AC. Wait, but BD and AC are diagonals, not sides. Wait, but in the figure, if the sides are AB, DC, BD, and AC? No, that doesn't make sense. Wait, maybe the sides are AB, DC, and the other sides are AD and BC? But AD and BC are not parallel. Wait, but BD and AC are diagonals. So if diagonals are parallel, then it's a parallelogram? No, in a parallelogram, diagonals are not necessarily parallel. Wait, but in this case, they are parallel? Wait, BD has slope 1/2, AC has slope 1/2. So they are parallel. But in a parallelogram, diagonals are not parallel. Wait, unless it's a degenerate parallelogram? Wait, no. Wait, in a general parallelogram, diagonals are not parallel. So if diagonals are parallel, then it's a special case? Wait, but if diagonals are parallel, then the figure is a parallelogram? Wait, no. Wait, if diagonals are parallel, then the figure is a parallelogram? Or is that a different condition?

Wait, let me think. In a parallelogram, diagonals bisect each other, but they are not necessarily parallel. If diagonals are parallel, then the figure would have to be a degenerate parallelogram? Wait, no. Wait, suppose diagonals are parallel. Let me take an example. Suppose we have a parallelogram with diagonals AC and BD. If AC || BD, then what does that imply?

In a parallelogram, diagonals bisect each other. If they are also parallel, then the only way this can happen is if the parallelogram is a rectangle? No, in a rectangle, diagonals are equal and bisect each other but are not parallel. Wait, in a rectangle, diagonals are congruent and bisect each other but are not parallel. In a rhombus, diagonals are not parallel. In a square, same as rectangle. Wait, so if diagonals are parallel, then the figure is not a parallelogram? Wait, but according to the midpoint condition, if diagonals bisect each other, then it's a parallelogram. So if diagonals bisect each other and are parallel, then it's a parallelogram. But in that case, if diagonals are parallel and bisect each other, then the figure is a parallelogram. But in that case, the sides would be... Wait, maybe in this case, the figure is a parallelogram with diagonals parallel? But how?

Wait, maybe my confusion comes from the fact that I'm considering different pairs of sides. If diagonals are AD and BC, which are parallel and bisect each other, then the sides would be AB, BD, DC, and CA. But BD and CA are diagonals? Wait, no. If the diagonals are AD and BC, then the sides are AB, BC, CD, DA. Wait, but then diagonals would be AC and BD. Wait, I think I need to stop here and accept that there is a contradiction in my calculations, which suggests that I made a mistake somewhere.

Wait, let me check once again. If the diagonals are AD and BC, which bisect each other, then the figure is a parallelogram. Therefore, the sides must be AB, BC, CD, DA. Wait, but then diagonals are AC and BD. Wait, but earlier I thought diagonals were AD and BC. So there is confusion here. Maybe I need to clarify.

Let me recall that in a parallelogram, the diagonals are the lines connecting opposite vertices. So if the vertices are A, B, C, D in order, then diagonals are AC and BD. If the diagonals are AD and BC, then the order of the vertices is different. For example, if the vertices are A, B, D, C, then diagonals are AD and BC. In this case, the sides are AB, BD, DC, and CA. But BD and CA are diagonals? No, BD is a side? Wait, if the order is A, B, D, C, then sides are AB, BD, DC, and CA. So BD is a side, and CA is a side. But in this case, BD and CA are not diagonals. Wait, but if diagonals are AD and BC, then they are connecting A to D and B to C. So in this case, the sides are AB, BD, DC, and CA. So BD is a side, and CA is a side. But then, in this case, the sides AB and DC are parallel (slope 5), and sides BD and CA are parallel (slope 1/2). Therefore, both pairs of opposite sides are parallel. Therefore, it is a parallelogram. Wait, but earlier I thought that sides AD and BC were supposed to be parallel, but in this case, sides BD and CA are parallel. So maybe I was wrong in thinking that AD and BC are sides. So in this case, the sides are AB, BD, DC, and CA. So AB || DC and BD || CA. Therefore, it is a parallelogram. So my mistake was in thinking that AD and BC were sides, but they are actually diagonals. So the sides are AB, BD, DC, and CA. Therefore, it is a parallelogram. Therefore, the coordinates D(8,9) are correct.

Therefore, x + y = 8 + 9 = 17. So the answer is 17.

But just to make sure, let me check the lengths of the sides. AB length: sqrt(1² + 5²) = sqrt(26). DC length: sqrt(1² + 5²) = sqrt(26). BD length: sqrt((8 - 6)² + (9 - 8)²) = sqrt(4 + 1) = sqrt(5). CA length: sqrt((7 - 5)² + (4 - 3)²) = sqrt(4 + 1) = sqrt(5). So opposite sides are equal. Therefore, it is a parallelogram. So my initial confusion was due to misidentifying which sides are which. So the answer is indeed (8,9), so x + y = 17.

Therefore, the value of x + y is 17.

**Final Answer**
The value of $x + y$ is \boxed{17}.

To determine the value of $ x + y $ for the fourth vertex of a parallelogram given three of its vertices: $ A(5, 3) $, $ B(6, 8) $, and $ C(7, 4) $, we can use the geometric property that **in a parallelogram, the diagonals bisect each other**. This means that the **midpoint of one diagonal is the same as the midpoint of the other diagonal**.

---

### Step 1: Assume the fourth vertex is $ D(x, y) $

We are told that $ x > 7 $, so we will look for a configuration where this condition is satisfied.

Let’s consider the possibility that the diagonals are $ AC $ and $ BD $. That is, the diagonals connect $ A $ to $ C $ and $ B $ to $ D $.

---

### Step 2: Compute the midpoint of diagonal $ AC $

$$
\text{Midpoint of } AC = \left( \frac{5 + 7}{2}, \frac{3 + 4}{2} \right) = (6, 3.5)
$$

---

### Step 3: Set the midpoint of diagonal $ BD $ equal to that of $ AC $

$$
\text{Midpoint of } BD = \left( \frac{6 + x}{2}, \frac{8 + y}{2} \right)
$$

Set this equal to $ (6, 3.5) $:

$$
\frac{6 + x}{2} = 6 \Rightarrow 6 + x = 12 \Rightarrow x = 6
$$
$$
\frac{8 + y}{2} = 3.5 \Rightarrow 8 + y = 7 \Rightarrow y = -1
$$

But this gives $ x = 6 $, which **does not satisfy** the condition $ x > 7 $. So this configuration is invalid.

---

### Step 4: Try another configuration — diagonals $ AD $ and $ BC $

Now, assume the diagonals are $ AD $ and $ BC $, and compute their midpoints.

$$
\text{Midpoint of } BC = \left( \frac{6 + 7}{2}, \frac{8 + 4}{2} \right) = (6.5, 6)
$$

$$
\text{Midpoint of } AD = \left( \frac{5 + x}{2}, \frac{3 + y}{2} \right)
$$

Set this equal to $ (6.5, 6) $:

$$
\frac{5 + x}{2} = 6.5 \Rightarrow 5 + x = 13 \Rightarrow x = 8
$$
$$
\frac{3 + y}{2} = 6 \Rightarrow 3 + y = 12 \Rightarrow y = 9
$$

Now, $ x = 8 $, which **does satisfy** the condition $ x > 7 $.

---

### Step 5: Verify that the figure is a parallelogram

Let’s check if the sides are equal and parallel.

- Vector $ \vec{AB} = (6 - 5, 8 - 3) = (1, 5) $
- Vector $ \vec{DC} = (8 - 7, 9 - 4) = (1, 5) $ → $ \vec{AB} = \vec{DC} $

- Vector $ \vec{AD} = (8 - 5, 9 - 3) = (3, 6) $
- Vector $ \vec{BC} = (7 - 6, 4 - 8) = (1, -4) $ → Not equal, but we are checking **opposite sides**.

Wait — we should check the **other pair of opposite sides**:

- Vector $ \vec{BD} = (8 - 6, 9 - 8) = (2, 1) $
- Vector $ \vec{AC} = (7 - 5, 4 - 3) = (2, 1) $ → $ \vec{BD} = \vec{AC} $

So both pairs of opposite sides are equal and parallel, confirming that the figure is a **parallelogram**.

---

### Final Step: Compute $ x + y $

$$
x = 8, \quad y = 9 \Rightarrow x + y = 8 + 9 = \boxed{17}
$$

---

### ✅ Final Answer:

$$
\boxed{17}
$$

Solution:
Name the points $A(5,3)$, $B(6,8)$, $C(7,4)$, and $D(x,y)$ and sketch the first three.  We find that there are three possible locations for $D$ (see figure).  Only the one to the right has an $x$-coordinate greater than 7.  Since $AC$ is parallel to $BD$ and equal in length to it, $D$ is two units to the right and one unit up from $B$, just as $C$ is two units to the right and one unit up from $A$.  Therefore, the coordinates of $D$ are $(8,9)$, and $x+y=8+9=\boxed{17}$.

[asy]
size(5cm);
import graph;
defaultpen(linewidth(0.7)+fontsize(10));
dotfactor=5;
real x = 7;
pair A=(5,3), B=(6,8), C=(7,4), D1=(8,9), D2=(4,7),D3=(6,-1);
pair[] dots = {A,B,C};
dot(dots);
xaxis(-2,10,Ticks(" ",1.0,begin=false,end=false,NoZero),Arrows(4));
yaxis(-2,10,Ticks(" ",1.0,begin=false,end=false,NoZero),Arrows(4));
draw(A--C--D1--B--cycle);//linetype("8 8"));
draw(A--D3--C);
draw(A--C--B--D2--cycle);//,linetype("1 2 3 1"));
label("$A(5,3)$",A,SW);
label("$B(6,8)$",B,NW);
label("$C(7,4)$",C,E);
dot(D1,UnFill);
dot(D2,UnFill);
dot(D3,UnFill);[/asy]


Explanation:
Okay, so I need to solve this equation: Half the value of 3x minus 9 is equal to x plus 37. Let me try to parse that. The problem says "Half the value of 3x - 9 is x + 37." Hmm. So, translating that into an equation, half of (3x - 9) equals x + 37. Let me write that down.

So, half of (3x - 9) would be (1/2)(3x - 9). And that's equal to x + 37. So the equation is:

(1/2)(3x - 9) = x + 37

Alright, now I need to solve for x. Let me think about how to approach this. Maybe I can start by eliminating the fraction to make it easier. If I multiply both sides of the equation by 2, that should get rid of the denominator. Let me try that.

Multiplying both sides by 2:

2 * [ (1/2)(3x - 9) ] = 2 * (x + 37)

Simplifying the left side: The 2 and the 1/2 cancel each other out, so it becomes 3x - 9. On the right side, distributing the 2 gives 2x + 74. So now the equation is:

3x - 9 = 2x + 74

Okay, now I need to solve for x. Let me subtract 2x from both sides to get the x terms on one side. So:

3x - 2x - 9 = 74

Which simplifies to:

x - 9 = 74

Then, adding 9 to both sides:

x = 74 + 9

Which is:

x = 83

Wait, let me check if that makes sense. Let me plug x = 83 back into the original equation to verify.

Original equation: Half of (3x - 9) equals x + 37.

Left side: (1/2)(3*83 - 9) = (1/2)(249 - 9) = (1/2)(240) = 120

Right side: 83 + 37 = 120

Yes, both sides equal 120. So x = 83 is correct. Hmm, seems straightforward. Let me just go through the steps again to make sure I didn't skip anything or make a mistake in arithmetic.

Starting with (1/2)(3x - 9) = x + 37. Multiply both sides by 2: 3x - 9 = 2x + 74. Subtract 2x: x - 9 = 74. Add 9: x = 83. Yep, that's correct. And plugging back in works. So I think that's the right answer.

But just to make sure, let me try solving it another way. Maybe expanding or something else. Let's see. Alternatively, I could have distributed the 1/2 first. Let me try that approach.

Original equation: (1/2)(3x - 9) = x + 37

Distribute the 1/2: (1/2)*3x - (1/2)*9 = x + 37

Which is (3/2)x - 9/2 = x + 37

Now, maybe subtract x from both sides to get the x terms on one side. Let's do that:

(3/2)x - x - 9/2 = 37

Calculating (3/2)x - x: that's (3/2 - 2/2)x = (1/2)x

So now we have (1/2)x - 9/2 = 37

Then, add 9/2 to both sides:

(1/2)x = 37 + 9/2

Convert 37 to halves: 37 is 74/2, so 74/2 + 9/2 = 83/2

Therefore, (1/2)x = 83/2

Multiply both sides by 2:

x = 83

Same result. So that confirms it again. So x is definitely 83. I think that's solid. Maybe I can check with another method? Let's see. Maybe using substitution or something else. But I think these two methods are sufficient. Both lead to x = 83, and plugging back in works. So I think that's the answer.

Alternatively, maybe I can think about the problem in words. If half of (3x - 9) is equal to x + 37, then maybe I can think of it as:

If I take 3x - 9 and divide by 2, I get x + 37. So, 3x - 9 divided by 2 equals x + 37. Which is the same equation. So, same steps. So yeah, same result. So I think that's correct.

Another way to check is to think about the difference between the two sides. Let me see. Suppose I didn't know the answer, but I can think about how much more one side is than the other. But maybe that's complicating. Alternatively, maybe I can think of the equation as:

(3x - 9)/2 - (x + 37) = 0

But that's just rearranging the equation. Let me compute that:

(3x - 9)/2 - x - 37 = 0

Multiply through by 2 to eliminate denominator:

3x - 9 - 2x - 74 = 0

Which simplifies to x - 83 = 0, so x = 83. Yep, same answer. So that's another way to look at it. So, regardless of the method, I get x = 83. Therefore, I can be confident that the answer is 83.

I think that's thorough enough. I considered different methods, checked my work by plugging in, and all roads lead to x = 83. So I think that's the correct answer.

**Final Answer**
The value of $ x $ is \boxed{83}.

To solve the equation:

$$
\text{Half the value of } 3x - 9 \text{ is } x + 37
$$

we first translate the verbal statement into a mathematical equation. The phrase "half the value of $ 3x - 9 $" means:

$$
\frac{1}{2}(3x - 9)
$$

So the equation becomes:

$$
\frac{1}{2}(3x - 9) = x + 37
$$

---

### Step 1: Eliminate the Fraction

To simplify, multiply both sides of the equation by 2 to eliminate the denominator:

$$
2 \cdot \frac{1}{2}(3x - 9) = 2 \cdot (x + 37)
$$

This simplifies to:

$$
3x - 9 = 2x + 74
$$

---

### Step 2: Solve for $ x $

Subtract $ 2x $ from both sides:

$$
3x - 2x - 9 = 74
$$

$$
x - 9 = 74
$$

Add 9 to both sides:

$$
x = 74 + 9 = 83
$$

---

### Step 3: Verify the Solution

Plug $ x = 83 $ back into the original equation to confirm:

**Left side:**

$$
\frac{1}{2}(3x - 9) = \frac{1}{2}(3 \cdot 83 - 9) = \frac{1}{2}(249 - 9) = \frac{1}{2}(240) = 120
$$

**Right side:**

$$
x + 37 = 83 + 37 = 120
$$

Both sides are equal, confirming the solution is correct.

---

### Final Answer

$$
\boxed{83}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation: Half the value of 3x minus 9 is equal to x plus 37. Let me try to parse that. The problem says "Half the value of 3x - 9 is x + 37." Hmm. So, translating that into an equation, half of (3x - 9) equals x + 37. Let me write that down.

So, half of (3x - 9) would be (1/2)(3x - 9). And that's equal to x + 37. So the equation is:

(1/2)(3x - 9) = x + 37

Alright, now I need to solve for x. Let me think about how to approach this. Maybe I can start by eliminating the fraction to make it easier. If I multiply both sides of the equation by 2, that should get rid of the denominator. Let me try that.

Multiplying both sides by 2:

2 * [ (1/2)(3x - 9) ] = 2 * (x + 37)

Simplifying the left side: The 2 and the 1/2 cancel each other out, so it becomes 3x - 9. On the right side, distributing the 2 gives 2x + 74. So now the equation is:

3x - 9 = 2x + 74

Okay, now I need to solve for x. Let me subtract 2x from both sides to get the x terms on one side. So:

3x - 2x - 9 = 74

Which simplifies to:

x - 9 = 74

Then, adding 9 to both sides:

x = 74 + 9

Which is:

x = 83

Wait, let me check if that makes sense. Let me plug x = 83 back into the original equation to verify.

Original equation: Half of (3x - 9) equals x + 37.

Left side: (1/2)(3*83 - 9) = (1/2)(249 - 9) = (1/2)(240) = 120

Right side: 83 + 37 = 120

Yes, both sides equal 120. So x = 83 is correct. Hmm, seems straightforward. Let me just go through the steps again to make sure I didn't skip anything or make a mistake in arithmetic.

Starting with (1/2)(3x - 9) = x + 37. Multiply both sides by 2: 3x - 9 = 2x + 74. Subtract 2x: x - 9 = 74. Add 9: x = 83. Yep, that's correct. And plugging back in works. So I think that's the right answer.

But just to make sure, let me try solving it another way. Maybe expanding or something else. Let's see. Alternatively, I could have distributed the 1/2 first. Let me try that approach.

Original equation: (1/2)(3x - 9) = x + 37

Distribute the 1/2: (1/2)*3x - (1/2)*9 = x + 37

Which is (3/2)x - 9/2 = x + 37

Now, maybe subtract x from both sides to get the x terms on one side. Let's do that:

(3/2)x - x - 9/2 = 37

Calculating (3/2)x - x: that's (3/2 - 2/2)x = (1/2)x

So now we have (1/2)x - 9/2 = 37

Then, add 9/2 to both sides:

(1/2)x = 37 + 9/2

Convert 37 to halves: 37 is 74/2, so 74/2 + 9/2 = 83/2

Therefore, (1/2)x = 83/2

Multiply both sides by 2:

x = 83

Same result. So that confirms it again. So x is definitely 83. I think that's solid. Maybe I can check with another method? Let's see. Maybe using substitution or something else. But I think these two methods are sufficient. Both lead to x = 83, and plugging back in works. So I think that's the answer.

Alternatively, maybe I can think about the problem in words. If half of (3x - 9) is equal to x + 37, then maybe I can think of it as:

If I take 3x - 9 and divide by 2, I get x + 37. So, 3x - 9 divided by 2 equals x + 37. Which is the same equation. So, same steps. So yeah, same result. So I think that's correct.

Another way to check is to think about the difference between the two sides. Let me see. Suppose I didn't know the answer, but I can think about how much more one side is than the other. But maybe that's complicating. Alternatively, maybe I can think of the equation as:

(3x - 9)/2 - (x + 37) = 0

But that's just rearranging the equation. Let me compute that:

(3x - 9)/2 - x - 37 = 0

Multiply through by 2 to eliminate denominator:

3x - 9 - 2x - 74 = 0

Which simplifies to x - 83 = 0, so x = 83. Yep, same answer. So that's another way to look at it. So, regardless of the method, I get x = 83. Therefore, I can be confident that the answer is 83.

I think that's thorough enough. I considered different methods, checked my work by plugging in, and all roads lead to x = 83. So I think that's the correct answer.

**Final Answer**
The value of $ x $ is \boxed{83}.

To solve the equation:

$$
\text{Half the value of } 3x - 9 \text{ is } x + 37
$$

we first translate the verbal statement into a mathematical equation. The phrase "half the value of $ 3x - 9 $" means:

$$
\frac{1}{2}(3x - 9)
$$

So the equation becomes:

$$
\frac{1}{2}(3x - 9) = x + 37
$$

---

### Step 1: Eliminate the Fraction

To simplify, multiply both sides of the equation by 2 to eliminate the denominator:

$$
2 \cdot \frac{1}{2}(3x - 9) = 2 \cdot (x + 37)
$$

This simplifies to:

$$
3x - 9 = 2x + 74
$$

---

### Step 2: Solve for $ x $

Subtract $ 2x $ from both sides:

$$
3x - 2x - 9 = 74
$$

$$
x - 9 = 74
$$

Add 9 to both sides:

$$
x = 74 + 9 = 83
$$

---

### Step 3: Verify the Solution

Plug $ x = 83 $ back into the original equation to confirm:

**Left side:**

$$
\frac{1}{2}(3x - 9) = \frac{1}{2}(3 \cdot 83 - 9) = \frac{1}{2}(249 - 9) = \frac{1}{2}(240) = 120
$$

**Right side:**

$$
x + 37 = 83 + 37 = 120
$$

Both sides are equal, confirming the solution is correct.

---

### Final Answer

$$
\boxed{83}
$$

Solution: We translate the problem to the equation $\frac{1}{2}(3x-9) = x+37$. Multiplying both sides by 2 gives $3x-9 = 2x+74$. Subtracting $2x$ from both sides yield $x-9 = 74$. Adding $9$ to both sides yields $ x = \boxed{83}$.


Explanation:
Okay, so I need to figure out the degree of the polynomial f. The problem says that (x - 1) multiplied by f(x) equals this polynomial: 3x⁴ + x³ - 25x² + 38x - 17. Hmm, let me think about how to approach this.

First, I remember that when you multiply two polynomials, the degree of the product is the sum of the degrees of the two polynomials. So, if I have (x - 1) times f(x), then the degree of the product should be equal to the degree of (x - 1) plus the degree of f(x).

Let me write that down. Let’s denote the degree of f(x) as n. Then, the degree of (x - 1) is 1, since it's a linear polynomial. Therefore, the degree of the product (x - 1) * f(x) should be 1 + n.

But the problem gives me the product as 3x⁴ + x³ - 25x² + 38x - 17. Let me check the degree of that polynomial. The highest power of x here is 4, so the degree is 4.

So, according to the earlier reasoning, 1 + n = 4. Solving for n, that would be n = 4 - 1 = 3. Therefore, the degree of f(x) should be 3? Wait, is that all? Let me make sure there isn't a trick here.

But hold on, maybe there's something else. For example, if (x - 1) is a factor of the polynomial on the right-hand side, then f(x) would be that polynomial divided by (x - 1). But if (x - 1) is a factor, then when we divide, there shouldn't be any remainder. However, the problem doesn't state that (x - 1) is a factor, but actually, it's given that (x - 1) multiplied by f(x) equals that polynomial. Therefore, (x - 1) must be a factor, so the division should result in a polynomial f(x) with no remainder.

But maybe there's a case where even if (x - 1) is not a factor, but the product is given? Wait, no. The equation is (x - 1) * f(x) equals that polynomial. Therefore, (x - 1) must divide that polynomial. Therefore, the division should be exact.

But maybe I need to check if (x - 1) is actually a factor of the given polynomial. Let me try plugging in x = 1 into the polynomial to see if it equals zero. If it does, then (x - 1) is a factor.

Let me compute the value of the polynomial at x = 1:

3(1)^4 + (1)^3 - 25(1)^2 + 38(1) - 17.

Calculating each term:

3(1) = 3

1^3 = 1

-25(1)^2 = -25

38(1) = 38

-17 remains.

Adding them up: 3 + 1 - 25 + 38 - 17.

Let me compute step by step:

Start with 3 + 1 = 4

4 - 25 = -21

-21 + 38 = 17

17 - 17 = 0.

Oh! So the polynomial evaluates to zero when x = 1. Therefore, (x - 1) is indeed a factor. Therefore, f(x) is the quotient when we divide the given polynomial by (x - 1).

So, if I divide 3x⁴ + x³ - 25x² + 38x - 17 by (x - 1), the quotient will be f(x). Since the degree of the original polynomial is 4 and we are dividing by a degree 1 polynomial, the quotient should have degree 4 - 1 = 3. Therefore, f(x) is a cubic polynomial, so its degree is 3.

But let me confirm this by actually performing the division, just to be thorough. Maybe there's a mistake in my reasoning?

Let me try polynomial long division. Dividing 3x⁴ + x³ - 25x² + 38x - 17 by (x - 1).

First, divide 3x⁴ by x, which gives 3x³. Multiply (x - 1) by 3x³: 3x⁴ - 3x³. Subtract this from the original polynomial:

(3x⁴ + x³ - 25x² + 38x - 17) - (3x⁴ - 3x³) =

3x⁴ - 3x⁴ + x³ + 3x³ - 25x² + 38x - 17 = 4x³ - 25x² + 38x - 17.

Next term: divide 4x³ by x, which is 4x². Multiply (x - 1) by 4x²: 4x³ - 4x². Subtract:

(4x³ - 25x² + 38x - 17) - (4x³ - 4x²) =

4x³ - 4x³ -25x² + 4x² + 38x -17 = -21x² + 38x -17.

Next term: divide -21x² by x, which is -21x. Multiply (x - 1) by -21x: -21x² + 21x. Subtract:

(-21x² + 38x -17) - (-21x² + 21x) =

-21x² + 21x² + 38x - 21x -17 = 17x -17.

Next term: divide 17x by x, which is 17. Multiply (x - 1) by 17: 17x -17. Subtract:

(17x -17) - (17x -17) = 0.

So the division comes out exact, and the quotient is 3x³ + 4x² -21x +17. Therefore, f(x) is 3x³ + 4x² -21x +17, which is indeed a cubic polynomial. Therefore, the degree of f is 3.

Wait, but let me check if there's another way to think about this. For example, if I didn't do the division, but just used the degrees. Since (x - 1) * f(x) is degree 4, then f(x) must be degree 3. So that's straightforward. But maybe there's a case where the leading terms cancel? For example, if the leading term of f(x) is such that when multiplied by x, it cancels with something? But no, because when you multiply (x - 1) by f(x), the leading term is x times the leading term of f(x). So if the product has leading term 3x⁴, then the leading term of f(x) must be 3x³. Therefore, the degree of f(x) is 3.

Therefore, regardless of the method, the answer is 3.

But just to make sure, let me check if maybe f(x) could have a lower degree? Suppose f(x) was degree 2. Then (x - 1)*f(x) would be degree 3, but the given polynomial is degree 4. Therefore, that's impossible. Similarly, if f(x) was degree 4, then the product would be degree 5, which is higher than 4. Therefore, the only possible degree is 3.

Therefore, the answer is 3.

But let me check once again with the division result. The quotient was 3x³ + 4x² -21x +17. Let me multiply this by (x - 1) to check if I get the original polynomial.

Multiply 3x³ + 4x² -21x +17 by (x - 1):

First, multiply 3x³ by (x - 1): 3x⁴ - 3x³.

Then, multiply 4x² by (x - 1): 4x³ - 4x².

Then, multiply -21x by (x - 1): -21x² + 21x.

Then, multiply 17 by (x - 1): 17x -17.

Now, add all these together:

3x⁴ - 3x³ + 4x³ - 4x² -21x² + 21x +17x -17.

Combine like terms:

For x⁴: 3x⁴.

For x³: (-3x³ + 4x³) = 1x³.

For x²: (-4x² -21x²) = -25x².

For x terms: (21x +17x) = 38x.

Constants: -17.

So altogether, 3x⁴ + x³ -25x² +38x -17, which matches the given polynomial. Perfect! So the division was correct, and f(x) is indeed a cubic polynomial. Therefore, the degree is 3.

Therefore, the answer is 3.

**Final Answer**
The degree of $ f $ is \boxed{3}.

To determine the degree of the polynomial $ f(x) $, we are given the equation:

$$
(x - 1) \cdot f(x) = 3x^4 + x^3 - 25x^2 + 38x - 17
$$

---

### Step 1: Understand the Degree Relationship

When two polynomials are multiplied, the degree of the product is the **sum** of the degrees of the two polynomials. That is:

$$
\deg((x - 1) \cdot f(x)) = \deg(x - 1) + \deg(f(x))
$$

- The degree of $ x - 1 $ is 1 (since it's a linear polynomial).
- The degree of the right-hand side polynomial is 4 (the highest power of $ x $ is 4).

So, we have:

$$
\deg((x - 1) \cdot f(x)) = 1 + \deg(f(x)) = 4
$$

Solving for $ \deg(f(x)) $:

$$
\deg(f(x)) = 4 - 1 = 3
$$

---

### Step 2: Confirm the Polynomial Division

To ensure that this result is valid, we can perform polynomial division: divide the given polynomial by $ x - 1 $ to find $ f(x) $.

Let’s divide:

$$
\frac{3x^4 + x^3 - 25x^2 + 38x - 17}{x - 1}
$$

Using polynomial long division or synthetic division, we find that the quotient is:

$$
f(x) = 3x^3 + 4x^2 - 21x + 17
$$

This is a cubic polynomial, confirming that the degree of $ f(x) $ is indeed 3.

---

### Step 3: Final Verification

To double-check, we can multiply $ f(x) $ by $ x - 1 $ and verify that the result is the original polynomial:

$$
(3x^3 + 4x^2 - 21x + 17)(x - 1) = 3x^4 + x^3 - 25x^2 + 38x - 17
$$

This matches the given polynomial, confirming that the division was correct and that $ f(x) $ is a cubic polynomial.

---

### Final Answer

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the degree of the polynomial f. The problem says that (x - 1) multiplied by f(x) equals this polynomial: 3x⁴ + x³ - 25x² + 38x - 17. Hmm, let me think about how to approach this.

First, I remember that when you multiply two polynomials, the degree of the product is the sum of the degrees of the two polynomials. So, if I have (x - 1) times f(x), then the degree of the product should be equal to the degree of (x - 1) plus the degree of f(x).

Let me write that down. Let’s denote the degree of f(x) as n. Then, the degree of (x - 1) is 1, since it's a linear polynomial. Therefore, the degree of the product (x - 1) * f(x) should be 1 + n.

But the problem gives me the product as 3x⁴ + x³ - 25x² + 38x - 17. Let me check the degree of that polynomial. The highest power of x here is 4, so the degree is 4.

So, according to the earlier reasoning, 1 + n = 4. Solving for n, that would be n = 4 - 1 = 3. Therefore, the degree of f(x) should be 3? Wait, is that all? Let me make sure there isn't a trick here.

But hold on, maybe there's something else. For example, if (x - 1) is a factor of the polynomial on the right-hand side, then f(x) would be that polynomial divided by (x - 1). But if (x - 1) is a factor, then when we divide, there shouldn't be any remainder. However, the problem doesn't state that (x - 1) is a factor, but actually, it's given that (x - 1) multiplied by f(x) equals that polynomial. Therefore, (x - 1) must be a factor, so the division should result in a polynomial f(x) with no remainder.

But maybe there's a case where even if (x - 1) is not a factor, but the product is given? Wait, no. The equation is (x - 1) * f(x) equals that polynomial. Therefore, (x - 1) must divide that polynomial. Therefore, the division should be exact.

But maybe I need to check if (x - 1) is actually a factor of the given polynomial. Let me try plugging in x = 1 into the polynomial to see if it equals zero. If it does, then (x - 1) is a factor.

Let me compute the value of the polynomial at x = 1:

3(1)^4 + (1)^3 - 25(1)^2 + 38(1) - 17.

Calculating each term:

3(1) = 3

1^3 = 1

-25(1)^2 = -25

38(1) = 38

-17 remains.

Adding them up: 3 + 1 - 25 + 38 - 17.

Let me compute step by step:

Start with 3 + 1 = 4

4 - 25 = -21

-21 + 38 = 17

17 - 17 = 0.

Oh! So the polynomial evaluates to zero when x = 1. Therefore, (x - 1) is indeed a factor. Therefore, f(x) is the quotient when we divide the given polynomial by (x - 1).

So, if I divide 3x⁴ + x³ - 25x² + 38x - 17 by (x - 1), the quotient will be f(x). Since the degree of the original polynomial is 4 and we are dividing by a degree 1 polynomial, the quotient should have degree 4 - 1 = 3. Therefore, f(x) is a cubic polynomial, so its degree is 3.

But let me confirm this by actually performing the division, just to be thorough. Maybe there's a mistake in my reasoning?

Let me try polynomial long division. Dividing 3x⁴ + x³ - 25x² + 38x - 17 by (x - 1).

First, divide 3x⁴ by x, which gives 3x³. Multiply (x - 1) by 3x³: 3x⁴ - 3x³. Subtract this from the original polynomial:

(3x⁴ + x³ - 25x² + 38x - 17) - (3x⁴ - 3x³) =

3x⁴ - 3x⁴ + x³ + 3x³ - 25x² + 38x - 17 = 4x³ - 25x² + 38x - 17.

Next term: divide 4x³ by x, which is 4x². Multiply (x - 1) by 4x²: 4x³ - 4x². Subtract:

(4x³ - 25x² + 38x - 17) - (4x³ - 4x²) =

4x³ - 4x³ -25x² + 4x² + 38x -17 = -21x² + 38x -17.

Next term: divide -21x² by x, which is -21x. Multiply (x - 1) by -21x: -21x² + 21x. Subtract:

(-21x² + 38x -17) - (-21x² + 21x) =

-21x² + 21x² + 38x - 21x -17 = 17x -17.

Next term: divide 17x by x, which is 17. Multiply (x - 1) by 17: 17x -17. Subtract:

(17x -17) - (17x -17) = 0.

So the division comes out exact, and the quotient is 3x³ + 4x² -21x +17. Therefore, f(x) is 3x³ + 4x² -21x +17, which is indeed a cubic polynomial. Therefore, the degree of f is 3.

Wait, but let me check if there's another way to think about this. For example, if I didn't do the division, but just used the degrees. Since (x - 1) * f(x) is degree 4, then f(x) must be degree 3. So that's straightforward. But maybe there's a case where the leading terms cancel? For example, if the leading term of f(x) is such that when multiplied by x, it cancels with something? But no, because when you multiply (x - 1) by f(x), the leading term is x times the leading term of f(x). So if the product has leading term 3x⁴, then the leading term of f(x) must be 3x³. Therefore, the degree of f(x) is 3.

Therefore, regardless of the method, the answer is 3.

But just to make sure, let me check if maybe f(x) could have a lower degree? Suppose f(x) was degree 2. Then (x - 1)*f(x) would be degree 3, but the given polynomial is degree 4. Therefore, that's impossible. Similarly, if f(x) was degree 4, then the product would be degree 5, which is higher than 4. Therefore, the only possible degree is 3.

Therefore, the answer is 3.

But let me check once again with the division result. The quotient was 3x³ + 4x² -21x +17. Let me multiply this by (x - 1) to check if I get the original polynomial.

Multiply 3x³ + 4x² -21x +17 by (x - 1):

First, multiply 3x³ by (x - 1): 3x⁴ - 3x³.

Then, multiply 4x² by (x - 1): 4x³ - 4x².

Then, multiply -21x by (x - 1): -21x² + 21x.

Then, multiply 17 by (x - 1): 17x -17.

Now, add all these together:

3x⁴ - 3x³ + 4x³ - 4x² -21x² + 21x +17x -17.

Combine like terms:

For x⁴: 3x⁴.

For x³: (-3x³ + 4x³) = 1x³.

For x²: (-4x² -21x²) = -25x².

For x terms: (21x +17x) = 38x.

Constants: -17.

So altogether, 3x⁴ + x³ -25x² +38x -17, which matches the given polynomial. Perfect! So the division was correct, and f(x) is indeed a cubic polynomial. Therefore, the degree is 3.

Therefore, the answer is 3.

**Final Answer**
The degree of $ f $ is \boxed{3}.

To determine the degree of the polynomial $ f(x) $, we are given the equation:

$$
(x - 1) \cdot f(x) = 3x^4 + x^3 - 25x^2 + 38x - 17
$$

---

### Step 1: Understand the Degree Relationship

When two polynomials are multiplied, the degree of the product is the **sum** of the degrees of the two polynomials. That is:

$$
\deg((x - 1) \cdot f(x)) = \deg(x - 1) + \deg(f(x))
$$

- The degree of $ x - 1 $ is 1 (since it's a linear polynomial).
- The degree of the right-hand side polynomial is 4 (the highest power of $ x $ is 4).

So, we have:

$$
\deg((x - 1) \cdot f(x)) = 1 + \deg(f(x)) = 4
$$

Solving for $ \deg(f(x)) $:

$$
\deg(f(x)) = 4 - 1 = 3
$$

---

### Step 2: Confirm the Polynomial Division

To ensure that this result is valid, we can perform polynomial division: divide the given polynomial by $ x - 1 $ to find $ f(x) $.

Let’s divide:

$$
\frac{3x^4 + x^3 - 25x^2 + 38x - 17}{x - 1}
$$

Using polynomial long division or synthetic division, we find that the quotient is:

$$
f(x) = 3x^3 + 4x^2 - 21x + 17
$$

This is a cubic polynomial, confirming that the degree of $ f(x) $ is indeed 3.

---

### Step 3: Final Verification

To double-check, we can multiply $ f(x) $ by $ x - 1 $ and verify that the result is the original polynomial:

$$
(3x^3 + 4x^2 - 21x + 17)(x - 1) = 3x^4 + x^3 - 25x^2 + 38x - 17
$$

This matches the given polynomial, confirming that the division was correct and that $ f(x) $ is a cubic polynomial.

---

### Final Answer

$$
\boxed{3}
$$

Solution: Since the product of $f$ and a polynomial with degree 1 equals a polynomial with degree 4, we know that $f$ is a polynomial with degree $4-1=\boxed{3}$.