Find the integer solution of the following Linear Diophantine Equation

sagutan nang tama at simple.

1. 4x + 3y = 7

2. 15x + 25y = 10

3. 6x + 9y = 3s

Let's solve the given Linear Diophantine Equations for integer solutions. 

### 1. \( 4x + 3y = 7 \)
We need to find integer values of \( x \) and \( y \) that satisfy this equation.

First, we can express \( y \) in terms of \( x \):

\[
y = \frac{7 - 4x}{3}
\]

For \( y \) to be an integer, \( 7 - 4x \) must be divisible by 3. Let's test integer values for \( x \):

- For \( x = 1 \):

\[
y = \frac{7 - 4(1)}{3} = \frac{7 - 4}{3} = \frac{3}{3} = 1
\]

So, \( x = 1 \) and \( y = 1 \) is a solution.

**Solution**: \( x = 1, y = 1 \) is an integer solution.

### 2. \( 15x + 25y = 10 \)
This equation can be simplified by dividing the entire equation by 5:

\[
3x + 5y = 2
\]

Now, express \( y \) in terms of \( x \):

\[
y = \frac{2 - 3x}{5}
\]

For \( y \) to be an integer, \( 2 - 3x \) must be divisible by 5. Let's test integer values for \( x \):

- For \( x = 4 \):

\[
y = \frac{2 - 3(4)}{5} = \frac{2 - 12}{5} = \frac{-10}{5} = -2
\]

So, \( x = 4 \) and \( y = -2 \) is a solution.

**Solution**: \( x = 4, y = -2 \) is an integer solution.

### 3. \( 6x + 9y = 3s \)
We need to find integer values of \( x \), \( y \), and \( s \) that satisfy this equation.

We can simplify the equation by dividing the entire equation by 3:

\[
2x + 3y = s
\]

Now, \( s \) can be any integer that satisfies this equation. We can choose \( s \) such that \( x \) and \( y \) are integers.

Let's express \( y \) in terms of \( x \) and \( s \):

\[
y = \frac{s - 2x}{3}
\]

For \( y \) to be an integer, \( s - 2x \) must be divisible by 3. Since \( s \) can be any integer, this equation can have infinite integer solutions depending on the choice of \( s \).

For example, if \( s = 1 \):

\[
y = \frac{1 - 2x}{3}
\]

Testing \( x = 1 \):

\[
y = \frac{1 - 2(1)}{3} = \frac{1 - 2}{3} = \frac{-1}{3} \quad \text{(not an integer, so not a solution)}
\]

Testing \( x = 2 \):

\[
y = \frac{1 - 2(2)}{3} = \frac{1 - 4}{3} = \frac{-3}{3} = -1
\]

So, \( x = 2 \), \( y = -1 \), and \( s = 1 \) is a solution.

**Solution**: One solution is \( x = 2 \), \( y = -1 \), \( s = 1 \).

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### Related Problems:
1. Find the integer solutions of \( 7x + 5y = 10 \).
2. Solve the Diophantine equation \( 2x + 3y = 1 \).
3. Determine the integer solution of \( 8x + 6y = 14 \).
4. Find all integer solutions of \( 9x + 4y = 20 \).
5. Solve for integer solutions \( 5x + 7y = 12 \).

**Tip:** For Linear Diophantine Equations of the form \( ax + by = c \), always check if the greatest common divisor (GCD) of \( a \) and \( b \) divides \( c \). If not, there are no integer solutions.

Explore how to find integer solutions for linear Diophantine equations with step-by-step examples. Learn key concepts including linear Diophantine equations and methods to determine integer solutions. Suitable for high school students studying algebra and number theory.

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