`(x)/(8)+(y)/(5)=(31)/(40)`
In the equation above, if x and y are positive integers, what is the value of x+y?

Answer

To solve the equation \(\frac{x}{8} + \frac{y}{5} = \frac{31}{40}\), we will follow these steps:

### Step 1: Simplify the Equation
First, we will eliminate the fractions by finding a common denominator. The least common multiple (LCM) of 8 and 5 is 40. We can rewrite the equation as follows:

\[ \frac{x}{8} = \frac{5x}{40} \quad \text{and} \quad \frac{y}{5} = \frac{8y}{40} \]

Thus, the equation becomes:

\[ \frac{5x + 8y}{40} = \frac{31}{40} \]

### Step 2: Remove the Denominator
Next, we can multiply both sides of the equation by 40 to eliminate the denominator:

\[ 5x + 8y = 31 \]

### Step 3: Solve for y in terms of x
Now, we can express \(y\) in terms of \(x\):

\[ 8y = 31 - 5x \]
\[ y = \frac{31 - 5x}{8} \]

### Step 4: Find Integer Solutions
Since \(x\) and \(y\) must be positive integers, \(31 - 5x\) must be a positive multiple of 8. This means \(31 - 5x > 0\) or \(x < \frac{31}{5} = 6.2\). Therefore, \(x\) can take integer values from 1 to 6.

We will test these integer values for \(x\):

1. **If \(x = 1\)**:
\[ y = \frac{31 - 5 \cdot 1}{8} = \frac{26}{8} = 3.25 \quad \text{(not an integer)} \]

2. **If \(x = 2\)**:
\[ y = \frac{31 - 5 \cdot 2}{8} = \frac{21}{8} = 2.625 \quad \text{(not an integer)} \]

3. **If \(x = 3\)**:
\[ y = \frac{31 - 5 \cdot 3}{8} = \frac{16}{8} = 2 \quad \text{(integer)} \]

4. **If \(x = 4\)**:
\[ y = \frac{31 - 5 \cdot 4}{8} = \frac{11}{8} = 1.375 \quad \text{(not an integer)} \]

5. **If \(x = 5\)**:
\[ y = \frac{31 - 5 \cdot 5}{8} = \frac{6}{8} = 0.75 \quad \text{(not an integer)} \]

6. **If \(x = 6\)**:
\[ y = \frac{31 - 5 \cdot 6}{8} = \frac{1}{8} = 0.125 \quad \text{(not an integer)} \]

### Step 5: Conclusion
The only pair of positive integers \((x, y)\) that satisfies the equation is \((3, 2)\). Therefore, we can find \(x + y\):

\[ x + y = 3 + 2 = 5 \]

### Final Answer
The value of \(x + y\) is \(5\).

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