Explanation:
Okay, so I need to find the eighth term of this geometric sequence: 125/9, 25/3, 5, 3, ... Hmm, let me recall what a geometric sequence is. It's a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio, right? So, first, I need to figure out what the common ratio (r) is here.

Let me start by looking at the first few terms. The first term is 125/9, the second term is 25/3, the third term is 5, and the fourth term is 3. To find the common ratio, I can divide the second term by the first term. Let me do that:

r = (25/3) / (125/9). Dividing by a fraction is the same as multiplying by its reciprocal, so that would be (25/3) * (9/125). Let me compute that.

First, simplify the numbers. 25 and 125: 25 divides into 125 five times. So, 25/125 is 1/5. Then, 9/3 is 3. So, multiplying those together: (1/5) * 3 = 3/5. Wait, so the common ratio is 3/5? Let me check that with the next terms to make sure I didn't make a mistake.

So, if the second term is 25/3, multiplying by 3/5 should give the third term. Let me compute that: (25/3) * (3/5). The 3s cancel out, and 25 divided by 5 is 5. So that gives 5, which matches the third term. Good. Now, let's check from the third term to the fourth term. Third term is 5, multiplying by 3/5: 5 * (3/5) = 3, which is the fourth term. Perfect, so the common ratio is indeed 3/5.

Okay, so now that I know the first term (a₁) is 125/9 and the common ratio (r) is 3/5, I need to find the eighth term. The formula for the nth term of a geometric sequence is a_n = a₁ * r^(n-1). So, for the eighth term, n is 8. Let me write that down:

a₈ = a₁ * r^(8-1) = (125/9) * (3/5)^7.

Now, I need to compute this. Let me first compute (3/5)^7. But before I do that, maybe I can simplify the expression before calculating. Let me see:

(125/9) * (3/5)^7. Let me note that 125 is 5³, and 9 is 3². Let me write that:

a₈ = (5³ / 3²) * (3^7 / 5^7). Because (3/5)^7 is 3^7 / 5^7.

So, combining these fractions:

(5³ / 3²) * (3^7 / 5^7) = (5³ * 3^7) / (3² * 5^7).

Now, let me simplify the exponents. For the 5s: 5³ / 5^7 = 5^(3-7) = 5^(-4) = 1 / 5^4. For the 3s: 3^7 / 3² = 3^(7-2) = 3^5.

Therefore, the entire expression becomes (3^5) / (5^4).

Now, let me compute 3^5 and 5^4.

3^5 is 3*3*3*3*3. Let me compute step by step:

3*3 = 9, 9*3 = 27, 27*3 = 81, 81*3 = 243. So, 3^5 = 243.

5^4 is 5*5*5*5. Let me compute that:

5*5 = 25, 25*5 = 125, 125*5 = 625. So, 5^4 = 625.

Therefore, a₈ = 243 / 625.

Wait, let me check if that's correct. Let me verify my steps again to make sure I didn't make a mistake.

Original expression: (125/9)*(3/5)^7.

Expressed as (5³ / 3²) * (3^7 / 5^7) = 5³ * 3^7 / (3² * 5^7) = 3^(7-2) * 5^(3-7) = 3^5 / 5^4. Yes, that's correct. Then 3^5 is 243, 5^4 is 625. So 243/625.

But let me check if that's the eighth term. Let me try another approach. Maybe compute each term step by step up to the eighth term? Although that might take longer, but maybe I can verify.

First term: 125/9 ≈ 13.888...

Second term: 125/9 * 3/5 = (125*3)/(9*5) = 375/45 = 25/3 ≈ 8.333... Correct.

Third term: 25/3 * 3/5 = (25*3)/(3*5) = 75/15 = 5. Correct.

Fourth term: 5 * 3/5 = 3. Correct.

Fifth term: 3 * 3/5 = 9/5 = 1.8

Sixth term: 9/5 * 3/5 = 27/25 = 1.08

Seventh term: 27/25 * 3/5 = 81/125 = 0.648

Eighth term: 81/125 * 3/5 = 243/625 ≈ 0.3888...

So, according to this step-by-step calculation, the eighth term is 243/625, which matches my previous result. Therefore, that seems correct.

Alternatively, maybe I can check using exponents. Let me think. The first term is a₁ = 125/9. Then, each term is multiplied by 3/5. So, term 2 is a₁*(3/5), term 3 is a₁*(3/5)^2, term 4 is a₁*(3/5)^3, ..., term n is a₁*(3/5)^{n-1}. Therefore, term 8 is a₁*(3/5)^7. Which is exactly what I did earlier. So, that's correct.

Therefore, the answer is 243/625. Let me check if this fraction can be simplified. 243 and 625. Let me see. 243 is 3^5, and 625 is 5^4. They have no common factors, since 3 and 5 are primes. Therefore, 243/625 is already in simplest form.

So, I think that's the answer. Let me just recap all steps once again to make sure I didn't skip anything or make a calculation error.

1. Found common ratio by dividing second term by first term: (25/3)/(125/9) = 3/5. Checked with subsequent terms, confirmed ratio is 3/5.

2. Applied formula for nth term: a_n = a₁ * r^{n-1}.

3. Plugged in n=8: a₈ = (125/9)*(3/5)^7.

4. Expressed 125 as 5³ and 9 as 3², then expanded (3/5)^7 as 3^7/5^7.

5. Combined terms: (5³ * 3^7) / (3² * 5^7) = 3^5 / 5^4 = 243/625.

6. Verified by calculating each term step by step up to the eighth term, which gave the same result.

7. Checked that 243 and 625 have no common factors, so fraction is simplified.

Therefore, I'm confident that the eighth term is 243/625.

**Final Answer**
The eighth term of the sequence is \boxed{\dfrac{243}{625}}.

To find the eighth term of the geometric sequence:

$$
\frac{125}{9}, \frac{25}{3}, 5, 3, \ldots
$$

we begin by identifying the **common ratio** $ r $, which is the constant factor between consecutive terms.

---

### Step 1: Find the Common Ratio

We compute the ratio between the second and first terms:

$$
r = \frac{\frac{25}{3}}{\frac{125}{9}} = \frac{25}{3} \cdot \frac{9}{125} = \frac{225}{375} = \frac{3}{5}
$$

To confirm, we check the ratio between the third and second terms:

$$
\frac{5}{\frac{25}{3}} = 5 \cdot \frac{3}{25} = \frac{15}{25} = \frac{3}{5}
$$

And between the fourth and third terms:

$$
\frac{3}{5} = \frac{3}{5}
$$

So, the common ratio is indeed $ r = \frac{3}{5} $.

---

### Step 2: Use the Formula for the $ n $-th Term

The general formula for the $ n $-th term of a geometric sequence is:

$$
a_n = a_1 \cdot r^{n-1}
$$

Here, the first term is:

$$
a_1 = \frac{125}{9}
$$

We are looking for the 8th term, so $ n = 8 $:

$$
a_8 = \frac{125}{9} \cdot \left( \frac{3}{5} \right)^7
$$

---

### Step 3: Simplify the Expression

We simplify the expression step by step:

$$
a_8 = \frac{125}{9} \cdot \frac{3^7}{5^7}
$$

Note that:

- $ 125 = 5^3 $
- $ 9 = 3^2 $

So:

$$
a_8 = \frac{5^3}{3^2} \cdot \frac{3^7}{5^7} = \frac{5^3 \cdot 3^7}{3^2 \cdot 5^7}
$$

Now simplify the exponents:

- $ \frac{5^3}{5^7} = 5^{3 - 7} = 5^{-4} = \frac{1}{5^4} $
- $ \frac{3^7}{3^2} = 3^{7 - 2} = 3^5 $

So:

$$
a_8 = \frac{3^5}{5^4}
$$

Now compute the powers:

- $ 3^5 = 243 $
- $ 5^4 = 625 $

Thus:

$$
a_8 = \frac{243}{625}
$$

---

### Step 4: Final Answer

The fraction $ \frac{243}{625} $ is already in its simplest form, as 243 and 625 have no common factors (243 = $ 3^5 $, 625 = $ 5^4 $).

---

### Final Answer

$$
\boxed{\dfrac{243}{625}}
$$

[Assistant]
Explanation:
Okay, so I need to find the eighth term of this geometric sequence: 125/9, 25/3, 5, 3, ... Hmm, let me recall what a geometric sequence is. It's a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio, right? So, first, I need to figure out what the common ratio (r) is here.

Let me start by looking at the first few terms. The first term is 125/9, the second term is 25/3, the third term is 5, and the fourth term is 3. To find the common ratio, I can divide the second term by the first term. Let me do that:

r = (25/3) / (125/9). Dividing by a fraction is the same as multiplying by its reciprocal, so that would be (25/3) * (9/125). Let me compute that.

First, simplify the numbers. 25 and 125: 25 divides into 125 five times. So, 25/125 is 1/5. Then, 9/3 is 3. So, multiplying those together: (1/5) * 3 = 3/5. Wait, so the common ratio is 3/5? Let me check that with the next terms to make sure I didn't make a mistake.

So, if the second term is 25/3, multiplying by 3/5 should give the third term. Let me compute that: (25/3) * (3/5). The 3s cancel out, and 25 divided by 5 is 5. So that gives 5, which matches the third term. Good. Now, let's check from the third term to the fourth term. Third term is 5, multiplying by 3/5: 5 * (3/5) = 3, which is the fourth term. Perfect, so the common ratio is indeed 3/5.

Okay, so now that I know the first term (a₁) is 125/9 and the common ratio (r) is 3/5, I need to find the eighth term. The formula for the nth term of a geometric sequence is a_n = a₁ * r^(n-1). So, for the eighth term, n is 8. Let me write that down:

a₈ = a₁ * r^(8-1) = (125/9) * (3/5)^7.

Now, I need to compute this. Let me first compute (3/5)^7. But before I do that, maybe I can simplify the expression before calculating. Let me see:

(125/9) * (3/5)^7. Let me note that 125 is 5³, and 9 is 3². Let me write that:

a₈ = (5³ / 3²) * (3^7 / 5^7). Because (3/5)^7 is 3^7 / 5^7.

So, combining these fractions:

(5³ / 3²) * (3^7 / 5^7) = (5³ * 3^7) / (3² * 5^7).

Now, let me simplify the exponents. For the 5s: 5³ / 5^7 = 5^(3-7) = 5^(-4) = 1 / 5^4. For the 3s: 3^7 / 3² = 3^(7-2) = 3^5.

Therefore, the entire expression becomes (3^5) / (5^4).

Now, let me compute 3^5 and 5^4.

3^5 is 3*3*3*3*3. Let me compute step by step:

3*3 = 9, 9*3 = 27, 27*3 = 81, 81*3 = 243. So, 3^5 = 243.

5^4 is 5*5*5*5. Let me compute that:

5*5 = 25, 25*5 = 125, 125*5 = 625. So, 5^4 = 625.

Therefore, a₈ = 243 / 625.

Wait, let me check if that's correct. Let me verify my steps again to make sure I didn't make a mistake.

Original expression: (125/9)*(3/5)^7.

Expressed as (5³ / 3²) * (3^7 / 5^7) = 5³ * 3^7 / (3² * 5^7) = 3^(7-2) * 5^(3-7) = 3^5 / 5^4. Yes, that's correct. Then 3^5 is 243, 5^4 is 625. So 243/625.

But let me check if that's the eighth term. Let me try another approach. Maybe compute each term step by step up to the eighth term? Although that might take longer, but maybe I can verify.

First term: 125/9 ≈ 13.888...

Second term: 125/9 * 3/5 = (125*3)/(9*5) = 375/45 = 25/3 ≈ 8.333... Correct.

Third term: 25/3 * 3/5 = (25*3)/(3*5) = 75/15 = 5. Correct.

Fourth term: 5 * 3/5 = 3. Correct.

Fifth term: 3 * 3/5 = 9/5 = 1.8

Sixth term: 9/5 * 3/5 = 27/25 = 1.08

Seventh term: 27/25 * 3/5 = 81/125 = 0.648

Eighth term: 81/125 * 3/5 = 243/625 ≈ 0.3888...

So, according to this step-by-step calculation, the eighth term is 243/625, which matches my previous result. Therefore, that seems correct.

Alternatively, maybe I can check using exponents. Let me think. The first term is a₁ = 125/9. Then, each term is multiplied by 3/5. So, term 2 is a₁*(3/5), term 3 is a₁*(3/5)^2, term 4 is a₁*(3/5)^3, ..., term n is a₁*(3/5)^{n-1}. Therefore, term 8 is a₁*(3/5)^7. Which is exactly what I did earlier. So, that's correct.

Therefore, the answer is 243/625. Let me check if this fraction can be simplified. 243 and 625. Let me see. 243 is 3^5, and 625 is 5^4. They have no common factors, since 3 and 5 are primes. Therefore, 243/625 is already in simplest form.

So, I think that's the answer. Let me just recap all steps once again to make sure I didn't skip anything or make a calculation error.

1. Found common ratio by dividing second term by first term: (25/3)/(125/9) = 3/5. Checked with subsequent terms, confirmed ratio is 3/5.

2. Applied formula for nth term: a_n = a₁ * r^{n-1}.

3. Plugged in n=8: a₈ = (125/9)*(3/5)^7.

4. Expressed 125 as 5³ and 9 as 3², then expanded (3/5)^7 as 3^7/5^7.

5. Combined terms: (5³ * 3^7) / (3² * 5^7) = 3^5 / 5^4 = 243/625.

6. Verified by calculating each term step by step up to the eighth term, which gave the same result.

7. Checked that 243 and 625 have no common factors, so fraction is simplified.

Therefore, I'm confident that the eighth term is 243/625.

**Final Answer**
The eighth term of the sequence is \boxed{\dfrac{243}{625}}.

To find the eighth term of the geometric sequence:

$$
\frac{125}{9}, \frac{25}{3}, 5, 3, \ldots
$$

we begin by identifying the **common ratio** $ r $, which is the constant factor between consecutive terms.

---

### Step 1: Find the Common Ratio

We compute the ratio between the second and first terms:

$$
r = \frac{\frac{25}{3}}{\frac{125}{9}} = \frac{25}{3} \cdot \frac{9}{125} = \frac{225}{375} = \frac{3}{5}
$$

To confirm, we check the ratio between the third and second terms:

$$
\frac{5}{\frac{25}{3}} = 5 \cdot \frac{3}{25} = \frac{15}{25} = \frac{3}{5}
$$

And between the fourth and third terms:

$$
\frac{3}{5} = \frac{3}{5}
$$

So, the common ratio is indeed $ r = \frac{3}{5} $.

---

### Step 2: Use the Formula for the $ n $-th Term

The general formula for the $ n $-th term of a geometric sequence is:

$$
a_n = a_1 \cdot r^{n-1}
$$

Here, the first term is:

$$
a_1 = \frac{125}{9}
$$

We are looking for the 8th term, so $ n = 8 $:

$$
a_8 = \frac{125}{9} \cdot \left( \frac{3}{5} \right)^7
$$

---

### Step 3: Simplify the Expression

We simplify the expression step by step:

$$
a_8 = \frac{125}{9} \cdot \frac{3^7}{5^7}
$$

Note that:

- $ 125 = 5^3 $
- $ 9 = 3^2 $

So:

$$
a_8 = \frac{5^3}{3^2} \cdot \frac{3^7}{5^7} = \frac{5^3 \cdot 3^7}{3^2 \cdot 5^7}
$$

Now simplify the exponents:

- $ \frac{5^3}{5^7} = 5^{3 - 7} = 5^{-4} = \frac{1}{5^4} $
- $ \frac{3^7}{3^2} = 3^{7 - 2} = 3^5 $

So:

$$
a_8 = \frac{3^5}{5^4}
$$

Now compute the powers:

- $ 3^5 = 243 $
- $ 5^4 = 625 $

Thus:

$$
a_8 = \frac{243}{625}
$$

---

### Step 4: Final Answer

The fraction $ \frac{243}{625} $ is already in its simplest form, as 243 and 625 have no common factors (243 = $ 3^5 $, 625 = $ 5^4 $).

---

### Final Answer

$$
\boxed{\dfrac{243}{625}}
$$

Solution:
The common ratio between consecutive terms is $\frac{3}{5}$ (you can choose any two consecutive terms and divide the second one by the first to find the common ratio). So the $n^\text{th}$ term of the sequence is $\frac{125}{9} \cdot \left( \frac{3}{5} \right)^{n-1}$. Plugging in $n=8$, we get $$
\frac{125}{9} \cdot \left( \frac{3}{5} \right)^{7} = \frac{5^3}{3^2} \cdot \frac{3^7}{5^7}
= \frac{3^5}{5^4}
= \boxed{\frac{243}{625}}.
$$


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 consider that in a parallelogram, the diagonals bisect each other, so the midpoint of AC is the same as the midpoint of BD. So if I can find the midpoint of AC, then set that equal to the midpoint of BD, and solve for D.

But first, I need to figure out which points are connected by diagonals. If A, B, C are three vertices, then the diagonals could be AC and BD, or AB and CD, or AD and BC. So depending on which two points are connected by a diagonal, the calculation will differ.

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

Let me try assuming that AC is a diagonal. Then, the midpoint of AC should be the same as the midpoint of BD. Let me compute the midpoint of AC first.

Midpoint of AC: A(5, 3) and C(7, 4). The midpoint is ((5 + 7)/2, (3 + 4)/2) = (12/2, 7/2) = (6, 3.5). So if BD is the other diagonal, then midpoint of BD should also be (6, 3.5). Let me denote B as (6, 8) and D as (x, y). Then midpoint of BD is ((6 + x)/2, (8 + y)/2). Setting this equal to (6, 3.5):

(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 states that x > 7, so this case is invalid. Therefore, AC cannot be a diagonal. So maybe AC is not a diagonal. Let me try another possibility.

Case 2: Suppose AB is a diagonal. Then, the midpoint of AB should be the same as the midpoint of CD. Let me compute midpoint of AB: A(5, 3) and B(6, 8). Midpoint is ((5 + 6)/2, (3 + 8)/2) = (11/2, 11/2) = (5.5, 5.5). Then midpoint of CD should be the same. Let me denote C(7, 4) and D(x, y). Midpoint of CD is ((7 + x)/2, (4 + y)/2). Setting equal to (5.5, 5.5):

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

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

So D would be (4, 7). But x = 4, which is less than 7. Again, the problem says x > 7, so this is invalid. So AB cannot be a diagonal.

Case 3: Suppose BC is a diagonal. Then midpoint of BC should be same as midpoint of AD. Let me compute 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). Then midpoint of AD should be same. A(5, 3) and D(x, y). Midpoint is ((5 + x)/2, (3 + y)/2). Setting equal to (6.5, 6):

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

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

So D would be (8, 9). Here, x = 8, which is greater than 7. So this satisfies the condition x > 7. So maybe this is the correct point? Let me check if this makes sense.

But let me verify if with D(8, 9), the figure is indeed a parallelogram. Let me check the vectors.

If BC is a diagonal, then the other diagonal is AD. But maybe I need to check the sides. Let me think. Alternatively, maybe I can check if the sides are equal and parallel.

Alternatively, let me check if the midpoints are correct. If BC is a diagonal, then midpoint of BC is (6.5, 6), and midpoint of AD is ((5 + 8)/2, (3 + 9)/2) = (13/2, 12/2) = (6.5, 6). So that checks out. So that seems correct.

But let me also check if the sides are equal and parallel. Let me assume that the order of the parallelogram is A, B, C, D. Wait, but if BC is a diagonal, then the order might be different. Let me think.

Alternatively, maybe the order is A, B, D, C? Or some other order. Let me try to figure out the connections.

Alternatively, maybe I can use vectors. Let me consider the vectors between the points.

If D is (8, 9), then let me check the vectors AB, DC, AD, BC.

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

Vector DC is C - D = (7 - 8, 4 - 9) = (-1, -5). Which is the negative of AB, so they are equal in magnitude and opposite in direction. So AB and DC are equal and opposite, which is consistent with being opposite sides of a parallelogram.

Vector AD is D - A = (8 - 5, 9 - 3) = (3, 6)

Vector BC is C - B = (7 - 6, 4 - 8) = (1, -4). Wait, that's not equal to AD. Hmm, so maybe that's not correct?

Wait, maybe I need to check another pair of sides. If AB and DC are one pair of sides, then the other pair should be AD and BC? Wait, but AD is (3, 6) and BC is (1, -4). These are not equal. So that seems inconsistent. So maybe my assumption is wrong?

Wait, but if BC is a diagonal, then maybe the sides are AB and AC? Wait, no. Maybe I need to think differently.

Alternatively, maybe the sides are AB and AC? Wait, but AC is a diagonal. Maybe I need to check the other sides.

Alternatively, maybe the sides are AB and AD? Let me check.

If AB is (1, 5) and AD is (3, 6). Then, if the parallelogram is A, B, C, D, then BC should be equal to AD? Wait, BC is (1, -4). Which is not equal to AD. So that's not matching.

Alternatively, maybe the sides are AB and AC? Wait, AC is from A(5,3) to C(7,4), which is (2, 1). Then, if AB is (1,5), then maybe another side?

This is getting confusing. Maybe I need to check another approach.

Alternatively, since we found D(8,9) by using the midpoint of the diagonals, and that gives x > 7, maybe that is the correct answer? But why does the vector check not match?

Wait, maybe the order of the points is different. Let me try to think of the order of the parallelogram. If the diagonals are BC and AD, then the vertices are connected as A, B, D, C? Or A, D, B, C?

Wait, if diagonals are BC and AD, then the vertices are connected as A, B, C, D? No, diagonals connect opposite vertices. So if BC is a diagonal, then the other diagonal is AD. Therefore, the vertices are A, B, D, C? Or A, D, B, C? Let me try to imagine.

Let me think of the parallelogram as having vertices A, B, D, C. So sides AB, BD, DC, and CA? No, that doesn't make sense. Wait, maybe the order is A, B, C, D? But if BC is a diagonal, then the order would be A, B, D, C? Let me try to draw this mentally.

If A is (5,3), B is (6,8), C is (7,4), and D is (8,9). Let me check the sides:

AB: from A(5,3) to B(6,8)

BC: from B(6,8) to C(7,4)

CD: from C(7,4) to D(8,9)

DA: from D(8,9) to A(5,3)

Wait, but in this case, AB and CD should be opposite sides. Let me check vectors:

AB: (1,5)

CD: (8 - 7, 9 - 4) = (1,5). Oh! So AB and CD are equal and same direction. So that's good.

Then BC and DA should be opposite sides. Vector BC is (1, -4). Vector DA is A - D = (5 - 8, 3 - 9) = (-3, -6). Wait, that's not equal to BC. Wait, but maybe DA is from D to A, which is (-3, -6). But BC is (1, -4). These are not equal. So something is wrong here.

Wait, but if AB and CD are equal, then the other pair of sides should be BC and AD? Wait, vector BC is (1, -4), vector AD is D - A = (8 - 5, 9 - 3) = (3, 6). These are not equal. So that's not matching. So why is the midpoint method giving me a point that doesn't satisfy the side conditions?

Hmm, maybe I made a mistake in assuming which diagonal is which. Let me check again.

If the diagonals are BC and AD, then the midpoint of BC is (6.5, 6) and midpoint of AD is ((5 + 8)/2, (3 + 9)/2) = (6.5, 6). So that's correct. But then why the sides aren't matching?

Wait, 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. Then, AD is (3,6), and BC is (1, -4). These are not equal. So that's not a parallelogram? But according to the midpoint, it should be. Wait, maybe I need to check the other sides?

Wait, if AB and DC are one pair of sides, then the other pair of sides should be BC and AD? But BC is (1, -4) and AD is (3, 6). These are not equal. So that's not a parallelogram. So something is wrong here. Maybe my assumption that BC is a diagonal is wrong?

But the midpoint method gave me a valid midpoint. Wait, but maybe the diagonals are not BC and AD? Maybe the diagonals are BD and AC? Wait, but earlier when I assumed AC was a diagonal, I got x = 6, which was invalid. But maybe I need to check again?

Wait, let me try again. If diagonals are BD and AC. Then midpoint of AC is (6, 3.5) as before. Midpoint of BD should be same. So midpoint of BD is ((6 + x)/2, (8 + y)/2) = (6, 3.5). So solving gives x = 6, y = -1. Which is invalid because x is not greater than 7. So that case is invalid.

Alternatively, if diagonals are AB and CD. Then midpoint of AB is (5.5, 5.5). Midpoint of CD is ((7 + x)/2, (4 + y)/2). Setting equal:

(7 + x)/2 = 5.5 => x = 4, and (4 + y)/2 = 5.5 => y = 7. So D is (4,7). But x = 4 < 7, invalid.

So only case where x > 7 is when diagonals are BC and AD, giving D(8,9). But then why the sides don't match? Maybe I need to check the other sides?

Wait, if AB and DC are equal, and AD and BC are not equal? Wait, but in a parallelogram, both pairs of opposite sides must be equal. So if one pair is equal, the other pair should also be equal. So maybe there is a mistake in my calculation?

Wait, let me check vector BC again. If B is (6,8) and C is (7,4), then vector BC is (7 - 6, 4 - 8) = (1, -4). Vector AD is D - A = (8 - 5, 9 - 3) = (3, 6). These are not equal. But if I check vector DA, which is A - D = (5 - 8, 3 - 9) = (-3, -6). Which is -3 times (1, 2). Wait, but vector BC is (1, -4). Not sure.

Wait, maybe I need to check another pair of sides. If AB is (1,5) and DC is (1,5). Then, the other sides should be BC and AD? Wait, but BC is (1, -4) and AD is (3, 6). Not equal. Alternatively, maybe the sides are AB and AD? Wait, AB is (1,5), AD is (3,6). Then, the other sides would be BC and DC? Wait, BC is (1, -4), DC is (1,5). Not equal. Hmm.

Wait, maybe I need to check if the sides are AB and DC, and BD and AC? Wait, BD is from B(6,8) to D(8,9): (2,1). AC is from A(5,3) to C(7,4): (2,1). So BD and AC are equal vectors. So BD and AC are equal and parallel. So that's another pair of opposite sides? Wait, but BD and AC are diagonals, not sides. Wait, but if BD and AC are diagonals, then they should bisect each other, which they do. But in a parallelogram, the diagonals are not equal unless it's a rectangle. But here, BD is (2,1) and AC is (2,1). Wait, they are equal vectors? Wait, BD is from B to D: (8 - 6, 9 - 8) = (2,1). AC is from A to C: (7 - 5, 4 - 3) = (2,1). So they are equal vectors. So that's interesting. So if diagonals are equal and bisect each other, then the figure is a parallelogram? Wait, no. In a parallelogram, diagonals bisect each other but are not necessarily equal. They are equal only in a rectangle. So if diagonals are equal and bisect each other, then it's a rectangle. But here, if BD and AC are equal and bisect each other, then it's a rectangle. But in that case, the sides should be perpendicular?

Wait, let me check if AB is perpendicular to AD. Vector AB is (1,5), vector AD is (3,6). The dot product is 1*3 + 5*6 = 3 + 30 = 33 ≠ 0. So they are not perpendicular. Therefore, it's not a rectangle. But diagonals are equal? Wait, BD and AC are equal vectors? Wait, BD is from B to D, which is (2,1). AC is from A to C, which is (2,1). So they are equal vectors. But in a parallelogram, the diagonals are not necessarily equal. So this seems contradictory. Wait, but if diagonals are equal and bisect each other, then it's a rectangle. But here, the sides are not perpendicular. Therefore, something is wrong here.

Wait, maybe my mistake is in assuming that BD and AC are diagonals? Wait, if the diagonals are BC and AD, then BD is not a diagonal. Wait, if diagonals are BC and AD, then BD is a side? Wait, no. Let me clarify.

If the vertices are A, B, D, C, then the diagonals would be AC and BD. Wait, if the order is A, B, D, C, then the diagonals are AD and BC? Or AC and BD?

Wait, in a quadrilateral, the diagonals are the connections between opposite vertices. So if the vertices are A, B, D, C, then the diagonals are AD and BC. Wait, but if the order is A, B, C, D, then diagonals are AC and BD. So depending on the order of the vertices, the diagonals change.

So if I have the points A(5,3), B(6,8), C(7,4), D(8,9). Let me check the order. If the order is A, B, C, D, then sides are AB, BC, CD, DA. Then diagonals are AC and BD. Earlier, we saw that AC is (2,1) and BD is (2,1). So diagonals are equal. Which would imply that the figure is a rectangle. But as we saw, the sides are not perpendicular. Therefore, contradiction. Therefore, something is wrong here.

Alternatively, maybe the order is different. Let me try another order. Suppose the order is A, B, D, C. Then sides are AB, BD, DC, CA. Wait, but then BD and DC are sides? That seems odd. Maybe not. Alternatively, order A, D, B, C. Then sides are AD, DB, BC, CA. Not sure.

Alternatively, maybe the order is A, C, B, D? Let me try. Then sides are AC, CB, BD, DA. Not sure.

Alternatively, maybe the problem is that I assumed the diagonals are BC and AD, but actually, the diagonals are AC and BD. Wait, but earlier when I calculated midpoint of AC and BD, they were different. Wait, midpoint of AC is (6, 3.5). Midpoint of BD is ((6 + 8)/2, (8 + 9)/2) = (14/2, 17/2) = (7, 8.5). Which is not equal. Therefore, diagonals AC and BD are not bisecting each other. Therefore, the figure is not a parallelogram? But we found D(8,9) by assuming that BC and AD are diagonals, which have the same midpoint. So if diagonals are BC and AD, then the midpoints are same. So that should be a parallelogram. But then why the sides are not matching?

Wait, maybe I need to check the sides again. If diagonals are BC and AD, then the sides are AB, BD, DC, and CA? Wait, no. Wait, if diagonals are BC and AD, then the vertices are connected as A, B, C, D? Wait, no. Let me think again.

In a parallelogram, the diagonals connect opposite vertices. So if diagonals are BC and AD, then the vertices are A, B, C, D arranged such that B and D are opposite, and A and C are opposite. Therefore, the sides would be AB, BC, CD, DA? Wait, no. If diagonals are BC and AD, then the sides would be AB, BD, DC, and CA? No, that doesn't make sense. Maybe I need to think of the vertices as A, B, D, C. So sides AB, BD, DC, and CA? But that would make BD and CA as sides? Not sure.

Alternatively, maybe the sides are AB, BC, CD, DA. If diagonals are AC and BD. Wait, but earlier, if D is (8,9), then diagonals AC and BD have different midpoints. Wait, midpoint of AC is (6, 3.5), midpoint of BD is (7, 8.5). Not same. Therefore, not a parallelogram. But earlier, when I assumed diagonals BC and AD, their midpoints are same. So if diagonals are BC and AD, then the figure is a parallelogram. But how?

Let me try to think of the vertices as A, B, D, C. So the sides are AB, BD, DC, and CA? Wait, but then BD and CA are diagonals? No, if diagonals are BC and AD, then the sides are AB, BC, CD, DA? Wait, no. This is getting too confusing. Maybe I need to use another method.

Let me try using vectors. Let me suppose that the three points are A(5,3), B(6,8), C(7,4), and D(x,y). In a parallelogram, the vector AB should equal the vector DC, and the vector AD should equal the vector BC.

So, vector AB = B - A = (6 - 5, 8 - 3) = (1, 5)

Vector DC = C - D = (7 - x, 4 - y)

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 x = 6 < 7, which is invalid. So this case is invalid.

Alternatively, maybe vector AB equals vector CD. Vector CD = D - C = (x - 7, y - 4). So AB = (1,5) = CD = (x - 7, y - 4). Therefore:

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 to be the correct point. Then, why earlier when I checked the sides, there was inconsistency?

Wait, if vector AB equals vector CD, then AB and CD are equal and parallel. Then, the other pair of sides should be AD and BC. Let me check vector AD and vector BC.

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

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

These are not equal. So why is this? Because if AB equals CD, then the other pair of sides should be AD and BC? But they are not equal. Wait, but in a parallelogram, both pairs of opposite sides should be equal. So if AB equals CD, then AD should equal BC. But here they are not. So something is wrong here.

Wait, maybe I need to check vector BC and vector AD? Wait, vector BC is (1, -4), vector AD is (3, 6). These are not equal. So why is this happening?

Wait, maybe I need to check another pair. If AB equals CD, then the other pair of sides should be AC and BD? Wait, no. Wait, in a parallelogram, opposite sides are AB and CD, and AD and BC. So if AB equals CD, then AD should equal BC. But they are not. Therefore, this suggests that the figure is not a parallelogram? But according to the vector method, if AB equals CD, then it should be a parallelogram. Wait, maybe I made a mistake in the vector calculation?

Wait, let me check again. If AB equals CD, then AB = CD. So AB is from A to B, CD is from C to D. So if AB = CD, then the figure is a parallelogram. But then AD and BC should also be equal. Wait, but in this case, they are not. So why?

Wait, maybe I need to check the other way. If AB equals DC, then AB = DC. Wait, earlier I considered AB = CD, but maybe it's AB = DC?

Wait, vector AB is (1,5). Vector DC is C - D = (7 - x, 4 - y). If AB = DC, then:

1 = 7 - x => x = 6

5 = 4 - y => y = -1

Which gives D(6, -1). Which is invalid. But if AB = CD, which is D - C = (x - 7, y - 4). So AB = CD:

(1,5) = (x - 7, y - 4) => x = 8, y = 9. So D(8,9). Then, vector AD is (3,6), vector BC is (1, -4). Not equal. So why?

Wait, maybe the other pair of sides is not AD and BC, but something else? Wait, if AB and CD are one pair, then the other pair should be AC and BD? No, AC and BD are diagonals. Wait, no. Wait, in a parallelogram, the sides are AB, BC, CD, DA. So if AB and CD are opposite sides, then BC and DA are the other pair of opposite sides. So vector BC should equal vector DA.

Vector BC is (1, -4). Vector DA is A - D = (5 - 8, 3 - 9) = (-3, -6). Which is not equal to BC. But if we take vector DA as D - A, which is (3, 6). Wait, no. Vector DA is from D to A, which is A - D. But vector BC is from B to C, which is C - B. So if we want vector BC to equal vector DA, then:

C - B = A - D => (1, -4) = (5 - x, 3 - y). Wait, solving:

1 = 5 - x => x = 5 - 1 = 4

-4 = 3 - y => y = 3 + 4 = 7

So D would be (4,7). Which is the same as before, which gives x = 4 < 7. So invalid.

But if we want vector BC to equal vector AD, then:

C - B = D - A => (1, -4) = (x - 5, y - 3). Therefore:

x - 5 = 1 => x = 6

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

Which gives D(6, -1). Again, invalid.

So this suggests that if AB equals CD, then BC does not equal DA. Therefore, the figure is not a parallelogram? But according to the midpoint method, if diagonals BC and AD have the same midpoint, then it should be a parallelogram. So there is a contradiction here.

Wait, maybe my mistake is in assuming which sides are which? Let me try to think differently. If AB equals CD, and AD equals BC, then it's a parallelogram. But in this case, AD is not equal to BC. So maybe there is another way?

Wait, maybe the sides are AB and DC, and AD and BC? Wait, AB equals DC, but AD does not equal BC. So that's not a parallelogram. Therefore, something is wrong here. How come the midpoint method gives a point that doesn't satisfy the side conditions?

Wait, let me check the coordinates again. If D is (8,9), then let me check all sides:

AB: from A(5,3) to B(6,8): length sqrt(1² + 5²) = sqrt(26)

BC: from B(6,8) to C(7,4): sqrt(1² + (-4)²) = sqrt(17)

CD: from C(7,4) to D(8,9): sqrt(1² + 5²) = sqrt(26)

DA: from D(8,9) to A(5,3): sqrt((-3)² + (-6)²) = sqrt(9 + 36) = sqrt(45) = 3*sqrt(5)

So AB = CD, but BC ≠ DA. Therefore, it's not a parallelogram? But according to the midpoint of diagonals BC and AD, they should be a parallelogram. What's going on here?

Wait, maybe the diagonals are not BC and AD? Wait, if diagonals are AC and BD, then their midpoints are different. But if diagonals are BD and AC, then midpoints are different. Wait, but earlier when I assumed diagonals BC and AD, their midpoints are same. So according to that, it should be a parallelogram. But the sides are not equal. So there must be a mistake in my understanding.

Wait, let me recall the property: In a parallelogram, the diagonals bisect each other. So if the midpoints of the diagonals are the same, then it is a parallelogram. So if I have diagonals BC and AD with same midpoint, then it is a parallelogram. Therefore, even if the sides don't seem to match, it should be a parallelogram. So maybe my calculation of the sides is wrong?

Wait, let me check again. If diagonals are BC and AD, then the sides are AB, BD, DC, and CA? Wait, no. Wait, if diagonals are BC and AD, then the vertices are connected as A, B, C, D? Or A, B, D, C?

Wait, suppose the vertices are A, B, C, D arranged such that diagonals are BC and AD. So the sides would be AB, BC, CD, DA? Wait, but then BC is a diagonal, not a side. So that can't be. Therefore, my mistake is in assuming which diagonals are which. If diagonals are BC and AD, then the sides are AB, BD, DC, and CA? No, that doesn't make sense. I think I need to use the correct definition.

In a quadrilateral, the diagonals are the connections between opposite vertices. So if the vertices are A, B, C, D in order, then diagonals are AC and BD. If the vertices are A, B, D, C, then diagonals are AD and BC. So if diagonals are AD and BC, then the sides are AB, BD, DC, and CA? No, that's not correct. The sides should be AB, BC, CD, DA. Wait, if the vertices are A, B, C, D, then sides are AB, BC, CD, DA. Diagonals are AC and BD. If the vertices are A, B, D, C, then sides are AB, BD, DC, CA. Diagonals are AD and BC. So if diagonals are AD and BC, then the sides are AB, BD, DC, CA. But in this case, BD and CA are diagonals? No, BD is a side, and CA is a side? Wait, no. If the vertices are A, B, D, C, then sides are AB, BD, DC, and CA. So BD is a side, and CA is a side. Then, diagonals would be AD and BC. So if diagonals are AD and BC, then they should bisect each other. Which they do, as we saw earlier. So in this case, the sides are AB, BD, DC, and CA. So AB and DC are opposite sides? Wait, AB is from A to B, DC is from D to C. Are they parallel and equal?

Vector AB is (1,5). Vector DC is C - D = (7 - 8, 4 - 9) = (-1, -5). Which is -AB. So they are equal in magnitude and opposite in direction. So AB and DC are equal and parallel. Then, the other pair of sides are BD and CA. Vector BD is D - B = (8 - 6, 9 - 8) = (2,1). Vector CA is A - C = (5 - 7, 3 - 4) = (-2, -1). Which is -BD. So BD and CA are equal and parallel. Therefore, this is a parallelogram! So the sides are AB and DC, and BD and CA. So the sides are AB, BD, DC, and CA. So the figure is a parallelogram with sides AB and DC, and sides BD and CA. Therefore, even though when I checked earlier, I was looking at BC and AD as sides, which were not equal, but actually, the sides are AB, BD, DC, and CA. So the other pair of sides are BD and CA, which are equal and parallel. Therefore, it is indeed a parallelogram. So my earlier confusion was because I was checking the wrong sides. So D(8,9) is correct.

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

But let me just confirm once again. If the sides are AB, BD, DC, and CA, then:

AB: (1,5)

BD: (2,1)

DC: (-1,-5)

CA: (-2,-1)

So AB and DC are negatives of each other, so equal and parallel. BD and CA are negatives of each other, so equal and parallel. Therefore, it is a parallelogram. So yes, D(8,9) is correct. Therefore, x + y = 17.

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

To determine the value of $ x + y $ for the fourth vertex $ D(x, y) $ of a parallelogram, we are given three vertices:
- $ A(5, 3) $
- $ B(6, 8) $
- $ C(7, 4) $

We are also told that $ x > 7 $, and we are to find $ x + y $.

---

### Step 1: Use the Property of Parallelograms

In a parallelogram, **opposite sides are equal and parallel**, and **the diagonals bisect each other**. This means that the **midpoint of one diagonal is the same as the midpoint of the other diagonal**.

Let’s assume that the diagonals are $ AC $ and $ BD $. Then, the midpoint of $ AC $ should be the same as the midpoint of $ BD $.

---

### 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 $ BD $ Equal to That of $ AC $

Let $ D = (x, y) $. Then the midpoint of $ BD $ is:

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

Set this equal to the midpoint of $ AC $:

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

So, $ D = (6, -1) $, but this gives $ x = 6 $, which **violates the condition** $ x > 7 $. So this configuration is **not valid**.

---

### Step 4: Try a Different Pair of Diagonals

Let’s now assume that the diagonals are $ BC $ and $ AD $. Then the midpoint of $ BC $ should equal the midpoint of $ AD $.

Compute the midpoint of $ BC $:

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

Now, the midpoint of $ AD $ is:

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

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

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

So, $ D = (8, 9) $, and since $ x = 8 > 7 $, this satisfies the condition.

---

### Step 5: Verify the Parallelogram

Let’s confirm that the figure with vertices $ A(5,3) $, $ B(6,8) $, $ C(7,4) $, and $ D(8,9) $ is indeed a parallelogram.

- **Vector $ \vec{AB} = (1, 5) $**
- **Vector $ \vec{DC} = (1, 5) $** (since $ \vec{DC} = C - D = (7 - 8, 4 - 9) = (-1, -5) $, which is the negative of $ \vec{AB} $, so they are equal in magnitude and opposite in direction)

- **Vector $ \vec{BD} = (2, 1) $**
- **Vector $ \vec{AC} = (2, 1) $** (since $ \vec{AC} = C - A = (7 - 5, 4 - 3) = (2, 1) $, which is the same as $ \vec{BD} $)

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

---

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

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

[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 consider that in a parallelogram, the diagonals bisect each other, so the midpoint of AC is the same as the midpoint of BD. So if I can find the midpoint of AC, then set that equal to the midpoint of BD, and solve for D.

But first, I need to figure out which points are connected by diagonals. If A, B, C are three vertices, then the diagonals could be AC and BD, or AB and CD, or AD and BC. So depending on which two points are connected by a diagonal, the calculation will differ.

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

Let me try assuming that AC is a diagonal. Then, the midpoint of AC should be the same as the midpoint of BD. Let me compute the midpoint of AC first.

Midpoint of AC: A(5, 3) and C(7, 4). The midpoint is ((5 + 7)/2, (3 + 4)/2) = (12/2, 7/2) = (6, 3.5). So if BD is the other diagonal, then midpoint of BD should also be (6, 3.5). Let me denote B as (6, 8) and D as (x, y). Then midpoint of BD is ((6 + x)/2, (8 + y)/2). Setting this equal to (6, 3.5):

(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 states that x > 7, so this case is invalid. Therefore, AC cannot be a diagonal. So maybe AC is not a diagonal. Let me try another possibility.

Case 2: Suppose AB is a diagonal. Then, the midpoint of AB should be the same as the midpoint of CD. Let me compute midpoint of AB: A(5, 3) and B(6, 8). Midpoint is ((5 + 6)/2, (3 + 8)/2) = (11/2, 11/2) = (5.5, 5.5). Then midpoint of CD should be the same. Let me denote C(7, 4) and D(x, y). Midpoint of CD is ((7 + x)/2, (4 + y)/2). Setting equal to (5.5, 5.5):

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

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

So D would be (4, 7). But x = 4, which is less than 7. Again, the problem says x > 7, so this is invalid. So AB cannot be a diagonal.

Case 3: Suppose BC is a diagonal. Then midpoint of BC should be same as midpoint of AD. Let me compute 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). Then midpoint of AD should be same. A(5, 3) and D(x, y). Midpoint is ((5 + x)/2, (3 + y)/2). Setting equal to (6.5, 6):

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

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

So D would be (8, 9). Here, x = 8, which is greater than 7. So this satisfies the condition x > 7. So maybe this is the correct point? Let me check if this makes sense.

But let me verify if with D(8, 9), the figure is indeed a parallelogram. Let me check the vectors.

If BC is a diagonal, then the other diagonal is AD. But maybe I need to check the sides. Let me think. Alternatively, maybe I can check if the sides are equal and parallel.

Alternatively, let me check if the midpoints are correct. If BC is a diagonal, then midpoint of BC is (6.5, 6), and midpoint of AD is ((5 + 8)/2, (3 + 9)/2) = (13/2, 12/2) = (6.5, 6). So that checks out. So that seems correct.

But let me also check if the sides are equal and parallel. Let me assume that the order of the parallelogram is A, B, C, D. Wait, but if BC is a diagonal, then the order might be different. Let me think.

Alternatively, maybe the order is A, B, D, C? Or some other order. Let me try to figure out the connections.

Alternatively, maybe I can use vectors. Let me consider the vectors between the points.

If D is (8, 9), then let me check the vectors AB, DC, AD, BC.

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

Vector DC is C - D = (7 - 8, 4 - 9) = (-1, -5). Which is the negative of AB, so they are equal in magnitude and opposite in direction. So AB and DC are equal and opposite, which is consistent with being opposite sides of a parallelogram.

Vector AD is D - A = (8 - 5, 9 - 3) = (3, 6)

Vector BC is C - B = (7 - 6, 4 - 8) = (1, -4). Wait, that's not equal to AD. Hmm, so maybe that's not correct?

Wait, maybe I need to check another pair of sides. If AB and DC are one pair of sides, then the other pair should be AD and BC? Wait, but AD is (3, 6) and BC is (1, -4). These are not equal. So that seems inconsistent. So maybe my assumption is wrong?

Wait, but if BC is a diagonal, then maybe the sides are AB and AC? Wait, no. Maybe I need to think differently.

Alternatively, maybe the sides are AB and AC? Wait, but AC is a diagonal. Maybe I need to check the other sides.

Alternatively, maybe the sides are AB and AD? Let me check.

If AB is (1, 5) and AD is (3, 6). Then, if the parallelogram is A, B, C, D, then BC should be equal to AD? Wait, BC is (1, -4). Which is not equal to AD. So that's not matching.

Alternatively, maybe the sides are AB and AC? Wait, AC is from A(5,3) to C(7,4), which is (2, 1). Then, if AB is (1,5), then maybe another side?

This is getting confusing. Maybe I need to check another approach.

Alternatively, since we found D(8,9) by using the midpoint of the diagonals, and that gives x > 7, maybe that is the correct answer? But why does the vector check not match?

Wait, maybe the order of the points is different. Let me try to think of the order of the parallelogram. If the diagonals are BC and AD, then the vertices are connected as A, B, D, C? Or A, D, B, C?

Wait, if diagonals are BC and AD, then the vertices are connected as A, B, C, D? No, diagonals connect opposite vertices. So if BC is a diagonal, then the other diagonal is AD. Therefore, the vertices are A, B, D, C? Or A, D, B, C? Let me try to imagine.

Let me think of the parallelogram as having vertices A, B, D, C. So sides AB, BD, DC, and CA? No, that doesn't make sense. Wait, maybe the order is A, B, C, D? But if BC is a diagonal, then the order would be A, B, D, C? Let me try to draw this mentally.

If A is (5,3), B is (6,8), C is (7,4), and D is (8,9). Let me check the sides:

AB: from A(5,3) to B(6,8)

BC: from B(6,8) to C(7,4)

CD: from C(7,4) to D(8,9)

DA: from D(8,9) to A(5,3)

Wait, but in this case, AB and CD should be opposite sides. Let me check vectors:

AB: (1,5)

CD: (8 - 7, 9 - 4) = (1,5). Oh! So AB and CD are equal and same direction. So that's good.

Then BC and DA should be opposite sides. Vector BC is (1, -4). Vector DA is A - D = (5 - 8, 3 - 9) = (-3, -6). Wait, that's not equal to BC. Wait, but maybe DA is from D to A, which is (-3, -6). But BC is (1, -4). These are not equal. So something is wrong here.

Wait, but if AB and CD are equal, then the other pair of sides should be BC and AD? Wait, vector BC is (1, -4), vector AD is D - A = (8 - 5, 9 - 3) = (3, 6). These are not equal. So that's not matching. So why is the midpoint method giving me a point that doesn't satisfy the side conditions?

Hmm, maybe I made a mistake in assuming which diagonal is which. Let me check again.

If the diagonals are BC and AD, then the midpoint of BC is (6.5, 6) and midpoint of AD is ((5 + 8)/2, (3 + 9)/2) = (6.5, 6). So that's correct. But then why the sides aren't matching?

Wait, 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. Then, AD is (3,6), and BC is (1, -4). These are not equal. So that's not a parallelogram? But according to the midpoint, it should be. Wait, maybe I need to check the other sides?

Wait, if AB and DC are one pair of sides, then the other pair of sides should be BC and AD? But BC is (1, -4) and AD is (3, 6). These are not equal. So that's not a parallelogram. So something is wrong here. Maybe my assumption that BC is a diagonal is wrong?

But the midpoint method gave me a valid midpoint. Wait, but maybe the diagonals are not BC and AD? Maybe the diagonals are BD and AC? Wait, but earlier when I assumed AC was a diagonal, I got x = 6, which was invalid. But maybe I need to check again?

Wait, let me try again. If diagonals are BD and AC. Then midpoint of AC is (6, 3.5) as before. Midpoint of BD should be same. So midpoint of BD is ((6 + x)/2, (8 + y)/2) = (6, 3.5). So solving gives x = 6, y = -1. Which is invalid because x is not greater than 7. So that case is invalid.

Alternatively, if diagonals are AB and CD. Then midpoint of AB is (5.5, 5.5). Midpoint of CD is ((7 + x)/2, (4 + y)/2). Setting equal:

(7 + x)/2 = 5.5 => x = 4, and (4 + y)/2 = 5.5 => y = 7. So D is (4,7). But x = 4 < 7, invalid.

So only case where x > 7 is when diagonals are BC and AD, giving D(8,9). But then why the sides don't match? Maybe I need to check the other sides?

Wait, if AB and DC are equal, and AD and BC are not equal? Wait, but in a parallelogram, both pairs of opposite sides must be equal. So if one pair is equal, the other pair should also be equal. So maybe there is a mistake in my calculation?

Wait, let me check vector BC again. If B is (6,8) and C is (7,4), then vector BC is (7 - 6, 4 - 8) = (1, -4). Vector AD is D - A = (8 - 5, 9 - 3) = (3, 6). These are not equal. But if I check vector DA, which is A - D = (5 - 8, 3 - 9) = (-3, -6). Which is -3 times (1, 2). Wait, but vector BC is (1, -4). Not sure.

Wait, maybe I need to check another pair of sides. If AB is (1,5) and DC is (1,5). Then, the other sides should be BC and AD? Wait, but BC is (1, -4) and AD is (3, 6). Not equal. Alternatively, maybe the sides are AB and AD? Wait, AB is (1,5), AD is (3,6). Then, the other sides would be BC and DC? Wait, BC is (1, -4), DC is (1,5). Not equal. Hmm.

Wait, maybe I need to check if the sides are AB and DC, and BD and AC? Wait, BD is from B(6,8) to D(8,9): (2,1). AC is from A(5,3) to C(7,4): (2,1). So BD and AC are equal vectors. So BD and AC are equal and parallel. So that's another pair of opposite sides? Wait, but BD and AC are diagonals, not sides. Wait, but if BD and AC are diagonals, then they should bisect each other, which they do. But in a parallelogram, the diagonals are not equal unless it's a rectangle. But here, BD is (2,1) and AC is (2,1). Wait, they are equal vectors? Wait, BD is from B to D: (8 - 6, 9 - 8) = (2,1). AC is from A to C: (7 - 5, 4 - 3) = (2,1). So they are equal vectors. So that's interesting. So if diagonals are equal and bisect each other, then the figure is a parallelogram? Wait, no. In a parallelogram, diagonals bisect each other but are not necessarily equal. They are equal only in a rectangle. So if diagonals are equal and bisect each other, then it's a rectangle. But here, if BD and AC are equal and bisect each other, then it's a rectangle. But in that case, the sides should be perpendicular?

Wait, let me check if AB is perpendicular to AD. Vector AB is (1,5), vector AD is (3,6). The dot product is 1*3 + 5*6 = 3 + 30 = 33 ≠ 0. So they are not perpendicular. Therefore, it's not a rectangle. But diagonals are equal? Wait, BD and AC are equal vectors? Wait, BD is from B to D, which is (2,1). AC is from A to C, which is (2,1). So they are equal vectors. But in a parallelogram, the diagonals are not necessarily equal. So this seems contradictory. Wait, but if diagonals are equal and bisect each other, then it's a rectangle. But here, the sides are not perpendicular. Therefore, something is wrong here.

Wait, maybe my mistake is in assuming that BD and AC are diagonals? Wait, if the diagonals are BC and AD, then BD is not a diagonal. Wait, if diagonals are BC and AD, then BD is a side? Wait, no. Let me clarify.

If the vertices are A, B, D, C, then the diagonals would be AC and BD. Wait, if the order is A, B, D, C, then the diagonals are AD and BC? Or AC and BD?

Wait, in a quadrilateral, the diagonals are the connections between opposite vertices. So if the vertices are A, B, D, C, then the diagonals are AD and BC. Wait, but if the order is A, B, C, D, then diagonals are AC and BD. So depending on the order of the vertices, the diagonals change.

So if I have the points A(5,3), B(6,8), C(7,4), D(8,9). Let me check the order. If the order is A, B, C, D, then sides are AB, BC, CD, DA. Then diagonals are AC and BD. Earlier, we saw that AC is (2,1) and BD is (2,1). So diagonals are equal. Which would imply that the figure is a rectangle. But as we saw, the sides are not perpendicular. Therefore, contradiction. Therefore, something is wrong here.

Alternatively, maybe the order is different. Let me try another order. Suppose the order is A, B, D, C. Then sides are AB, BD, DC, CA. Wait, but then BD and DC are sides? That seems odd. Maybe not. Alternatively, order A, D, B, C. Then sides are AD, DB, BC, CA. Not sure.

Alternatively, maybe the order is A, C, B, D? Let me try. Then sides are AC, CB, BD, DA. Not sure.

Alternatively, maybe the problem is that I assumed the diagonals are BC and AD, but actually, the diagonals are AC and BD. Wait, but earlier when I calculated midpoint of AC and BD, they were different. Wait, midpoint of AC is (6, 3.5). Midpoint of BD is ((6 + 8)/2, (8 + 9)/2) = (14/2, 17/2) = (7, 8.5). Which is not equal. Therefore, diagonals AC and BD are not bisecting each other. Therefore, the figure is not a parallelogram? But we found D(8,9) by assuming that BC and AD are diagonals, which have the same midpoint. So if diagonals are BC and AD, then the midpoints are same. So that should be a parallelogram. But then why the sides are not matching?

Wait, maybe I need to check the sides again. If diagonals are BC and AD, then the sides are AB, BD, DC, and CA? Wait, no. Wait, if diagonals are BC and AD, then the vertices are connected as A, B, C, D? Wait, no. Let me think again.

In a parallelogram, the diagonals connect opposite vertices. So if diagonals are BC and AD, then the vertices are A, B, C, D arranged such that B and D are opposite, and A and C are opposite. Therefore, the sides would be AB, BC, CD, DA? Wait, no. If diagonals are BC and AD, then the sides would be AB, BD, DC, and CA? No, that doesn't make sense. Maybe I need to think of the vertices as A, B, D, C. So sides AB, BD, DC, and CA? But that would make BD and CA as sides? Not sure.

Alternatively, maybe the sides are AB, BC, CD, DA. If diagonals are AC and BD. Wait, but earlier, if D is (8,9), then diagonals AC and BD have different midpoints. Wait, midpoint of AC is (6, 3.5), midpoint of BD is (7, 8.5). Not same. Therefore, not a parallelogram. But earlier, when I assumed diagonals BC and AD, their midpoints are same. So if diagonals are BC and AD, then the figure is a parallelogram. But how?

Let me try to think of the vertices as A, B, D, C. So the sides are AB, BD, DC, and CA? Wait, but then BD and CA are diagonals? No, if diagonals are BC and AD, then the sides are AB, BC, CD, DA? Wait, no. This is getting too confusing. Maybe I need to use another method.

Let me try using vectors. Let me suppose that the three points are A(5,3), B(6,8), C(7,4), and D(x,y). In a parallelogram, the vector AB should equal the vector DC, and the vector AD should equal the vector BC.

So, vector AB = B - A = (6 - 5, 8 - 3) = (1, 5)

Vector DC = C - D = (7 - x, 4 - y)

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 x = 6 < 7, which is invalid. So this case is invalid.

Alternatively, maybe vector AB equals vector CD. Vector CD = D - C = (x - 7, y - 4). So AB = (1,5) = CD = (x - 7, y - 4). Therefore:

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 to be the correct point. Then, why earlier when I checked the sides, there was inconsistency?

Wait, if vector AB equals vector CD, then AB and CD are equal and parallel. Then, the other pair of sides should be AD and BC. Let me check vector AD and vector BC.

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

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

These are not equal. So why is this? Because if AB equals CD, then the other pair of sides should be AD and BC? But they are not equal. Wait, but in a parallelogram, both pairs of opposite sides should be equal. So if AB equals CD, then AD should equal BC. But here they are not. So something is wrong here.

Wait, maybe I need to check vector BC and vector AD? Wait, vector BC is (1, -4), vector AD is (3, 6). These are not equal. So why is this happening?

Wait, maybe I need to check another pair. If AB equals CD, then the other pair of sides should be AC and BD? Wait, no. Wait, in a parallelogram, opposite sides are AB and CD, and AD and BC. So if AB equals CD, then AD should equal BC. But they are not. Therefore, this suggests that the figure is not a parallelogram? But according to the vector method, if AB equals CD, then it should be a parallelogram. Wait, maybe I made a mistake in the vector calculation?

Wait, let me check again. If AB equals CD, then AB = CD. So AB is from A to B, CD is from C to D. So if AB = CD, then the figure is a parallelogram. But then AD and BC should also be equal. Wait, but in this case, they are not. So why?

Wait, maybe I need to check the other way. If AB equals DC, then AB = DC. Wait, earlier I considered AB = CD, but maybe it's AB = DC?

Wait, vector AB is (1,5). Vector DC is C - D = (7 - x, 4 - y). If AB = DC, then:

1 = 7 - x => x = 6

5 = 4 - y => y = -1

Which gives D(6, -1). Which is invalid. But if AB = CD, which is D - C = (x - 7, y - 4). So AB = CD:

(1,5) = (x - 7, y - 4) => x = 8, y = 9. So D(8,9). Then, vector AD is (3,6), vector BC is (1, -4). Not equal. So why?

Wait, maybe the other pair of sides is not AD and BC, but something else? Wait, if AB and CD are one pair, then the other pair should be AC and BD? No, AC and BD are diagonals. Wait, no. Wait, in a parallelogram, the sides are AB, BC, CD, DA. So if AB and CD are opposite sides, then BC and DA are the other pair of opposite sides. So vector BC should equal vector DA.

Vector BC is (1, -4). Vector DA is A - D = (5 - 8, 3 - 9) = (-3, -6). Which is not equal to BC. But if we take vector DA as D - A, which is (3, 6). Wait, no. Vector DA is from D to A, which is A - D. But vector BC is from B to C, which is C - B. So if we want vector BC to equal vector DA, then:

C - B = A - D => (1, -4) = (5 - x, 3 - y). Wait, solving:

1 = 5 - x => x = 5 - 1 = 4

-4 = 3 - y => y = 3 + 4 = 7

So D would be (4,7). Which is the same as before, which gives x = 4 < 7. So invalid.

But if we want vector BC to equal vector AD, then:

C - B = D - A => (1, -4) = (x - 5, y - 3). Therefore:

x - 5 = 1 => x = 6

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

Which gives D(6, -1). Again, invalid.

So this suggests that if AB equals CD, then BC does not equal DA. Therefore, the figure is not a parallelogram? But according to the midpoint method, if diagonals BC and AD have the same midpoint, then it should be a parallelogram. So there is a contradiction here.

Wait, maybe my mistake is in assuming which sides are which? Let me try to think differently. If AB equals CD, and AD equals BC, then it's a parallelogram. But in this case, AD is not equal to BC. So maybe there is another way?

Wait, maybe the sides are AB and DC, and AD and BC? Wait, AB equals DC, but AD does not equal BC. So that's not a parallelogram. Therefore, something is wrong here. How come the midpoint method gives a point that doesn't satisfy the side conditions?

Wait, let me check the coordinates again. If D is (8,9), then let me check all sides:

AB: from A(5,3) to B(6,8): length sqrt(1² + 5²) = sqrt(26)

BC: from B(6,8) to C(7,4): sqrt(1² + (-4)²) = sqrt(17)

CD: from C(7,4) to D(8,9): sqrt(1² + 5²) = sqrt(26)

DA: from D(8,9) to A(5,3): sqrt((-3)² + (-6)²) = sqrt(9 + 36) = sqrt(45) = 3*sqrt(5)

So AB = CD, but BC ≠ DA. Therefore, it's not a parallelogram? But according to the midpoint of diagonals BC and AD, they should be a parallelogram. What's going on here?

Wait, maybe the diagonals are not BC and AD? Wait, if diagonals are AC and BD, then their midpoints are different. But if diagonals are BD and AC, then midpoints are different. Wait, but earlier when I assumed diagonals BC and AD, their midpoints are same. So according to that, it should be a parallelogram. But the sides are not equal. So there must be a mistake in my understanding.

Wait, let me recall the property: In a parallelogram, the diagonals bisect each other. So if the midpoints of the diagonals are the same, then it is a parallelogram. So if I have diagonals BC and AD with same midpoint, then it is a parallelogram. Therefore, even if the sides don't seem to match, it should be a parallelogram. So maybe my calculation of the sides is wrong?

Wait, let me check again. If diagonals are BC and AD, then the sides are AB, BD, DC, and CA? Wait, no. Wait, if diagonals are BC and AD, then the vertices are connected as A, B, C, D? Or A, B, D, C?

Wait, suppose the vertices are A, B, C, D arranged such that diagonals are BC and AD. So the sides would be AB, BC, CD, DA? Wait, but then BC is a diagonal, not a side. So that can't be. Therefore, my mistake is in assuming which diagonals are which. If diagonals are BC and AD, then the sides are AB, BD, DC, and CA? No, that doesn't make sense. I think I need to use the correct definition.

In a quadrilateral, the diagonals are the connections between opposite vertices. So if the vertices are A, B, C, D in order, then diagonals are AC and BD. If the vertices are A, B, D, C, then diagonals are AD and BC. So if diagonals are AD and BC, then the sides are AB, BD, DC, and CA? No, that's not correct. The sides should be AB, BC, CD, DA. Wait, if the vertices are A, B, C, D, then sides are AB, BC, CD, DA. Diagonals are AC and BD. If the vertices are A, B, D, C, then sides are AB, BD, DC, CA. Diagonals are AD and BC. So if diagonals are AD and BC, then the sides are AB, BD, DC, CA. But in this case, BD and CA are diagonals? No, BD is a side, and CA is a side? Wait, no. If the vertices are A, B, D, C, then sides are AB, BD, DC, and CA. So BD is a side, and CA is a side. Then, diagonals would be AD and BC. So if diagonals are AD and BC, then they should bisect each other. Which they do, as we saw earlier. So in this case, the sides are AB, BD, DC, and CA. So AB and DC are opposite sides? Wait, AB is from A to B, DC is from D to C. Are they parallel and equal?

Vector AB is (1,5). Vector DC is C - D = (7 - 8, 4 - 9) = (-1, -5). Which is -AB. So they are equal in magnitude and opposite in direction. So AB and DC are equal and parallel. Then, the other pair of sides are BD and CA. Vector BD is D - B = (8 - 6, 9 - 8) = (2,1). Vector CA is A - C = (5 - 7, 3 - 4) = (-2, -1). Which is -BD. So BD and CA are equal and parallel. Therefore, this is a parallelogram! So the sides are AB and DC, and BD and CA. So the sides are AB, BD, DC, and CA. So the figure is a parallelogram with sides AB and DC, and sides BD and CA. Therefore, even though when I checked earlier, I was looking at BC and AD as sides, which were not equal, but actually, the sides are AB, BD, DC, and CA. So the other pair of sides are BD and CA, which are equal and parallel. Therefore, it is indeed a parallelogram. So my earlier confusion was because I was checking the wrong sides. So D(8,9) is correct.

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

But let me just confirm once again. If the sides are AB, BD, DC, and CA, then:

AB: (1,5)

BD: (2,1)

DC: (-1,-5)

CA: (-2,-1)

So AB and DC are negatives of each other, so equal and parallel. BD and CA are negatives of each other, so equal and parallel. Therefore, it is a parallelogram. So yes, D(8,9) is correct. Therefore, x + y = 17.

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

To determine the value of $ x + y $ for the fourth vertex $ D(x, y) $ of a parallelogram, we are given three vertices:
- $ A(5, 3) $
- $ B(6, 8) $
- $ C(7, 4) $

We are also told that $ x > 7 $, and we are to find $ x + y $.

---

### Step 1: Use the Property of Parallelograms

In a parallelogram, **opposite sides are equal and parallel**, and **the diagonals bisect each other**. This means that the **midpoint of one diagonal is the same as the midpoint of the other diagonal**.

Let’s assume that the diagonals are $ AC $ and $ BD $. Then, the midpoint of $ AC $ should be the same as the midpoint of $ BD $.

---

### 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 $ BD $ Equal to That of $ AC $

Let $ D = (x, y) $. Then the midpoint of $ BD $ is:

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

Set this equal to the midpoint of $ AC $:

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

So, $ D = (6, -1) $, but this gives $ x = 6 $, which **violates the condition** $ x > 7 $. So this configuration is **not valid**.

---

### Step 4: Try a Different Pair of Diagonals

Let’s now assume that the diagonals are $ BC $ and $ AD $. Then the midpoint of $ BC $ should equal the midpoint of $ AD $.

Compute the midpoint of $ BC $:

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

Now, the midpoint of $ AD $ is:

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

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

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

So, $ D = (8, 9) $, and since $ x = 8 > 7 $, this satisfies the condition.

---

### Step 5: Verify the Parallelogram

Let’s confirm that the figure with vertices $ A(5,3) $, $ B(6,8) $, $ C(7,4) $, and $ D(8,9) $ is indeed a parallelogram.

- **Vector $ \vec{AB} = (1, 5) $**
- **Vector $ \vec{DC} = (1, 5) $** (since $ \vec{DC} = C - D = (7 - 8, 4 - 9) = (-1, -5) $, which is the negative of $ \vec{AB} $, so they are equal in magnitude and opposite in direction)

- **Vector $ \vec{BD} = (2, 1) $**
- **Vector $ \vec{AC} = (2, 1) $** (since $ \vec{AC} = C - A = (7 - 5, 4 - 3) = (2, 1) $, which is the same as $ \vec{BD} $)

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

---

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

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

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]