If the system of linear equations `2x- 3y= gamma +5, alpha x + 5y= beta +1`, where `alpha, beta, gamma in R` has infinitely many solutions, then the value of `|9 alpha +3 beta+ 5 gamma|` is equal to _____

Answer

To solve the given problem, we need to determine the value of \(|9\alpha + 3\beta + 5\gamma|\) under the condition that the system of linear equations has infinitely many solutions.

### Step-by-Step Solution:

1. **Identify the equations**:
We have the following two equations:
\[ 2x - 3y = \gamma + 5 \quad (1) \]
\[ \alpha x + 5y = \beta + 1 \quad (2) \]

2. **Condition for infinitely many solutions**:
For the system to have infinitely many solutions, the ratios of the coefficients of \(x\), \(y\), and the constants must be equal:
\[ \frac{2}{\alpha} = \frac{-3}{5} = \frac{\gamma + 5}{\beta + 1} \]

3. **Finding \(\alpha\)**:
From the first part of the ratio:
\[ \frac{2}{\alpha} = \frac{-3}{5} \]
Cross-multiplying gives:
\[ 2 \cdot 5 = -3 \cdot \alpha \implies 10 = -3\alpha \implies \alpha = -\frac{10}{3} \]

4. **Finding the relationship between \(\beta\) and \(\gamma\)**:
Now, using the second part of the ratio:
\[ \frac{-3}{5} = \frac{\gamma + 5}{\beta + 1} \]
Cross-multiplying gives:
\[ -3(\beta + 1) = 5(\gamma + 5) \]
Expanding both sides:
\[ -3\beta - 3 = 5\gamma + 25 \implies 5\gamma + 3\beta = -28 \quad (3) \]

5. **Substituting \(\alpha\) into the expression**:
We need to find \(9\alpha + 3\beta + 5\gamma\):
\[ 9\alpha = 9 \left(-\frac{10}{3}\right) = -30 \]
Now substituting into the expression:
\[ 9\alpha + 3\beta + 5\gamma = -30 + 3\beta + 5\gamma \]

6. **Using equation (3)**:
From equation (3), we know that:
\[ 3\beta + 5\gamma = -28 \]
Therefore:
\[ 9\alpha + 3\beta + 5\gamma = -30 - 28 = -58 \]

7. **Finding the absolute value**:
Finally, we take the absolute value:
\[ |9\alpha + 3\beta + 5\gamma| = |-58| = 58 \]

### Final Answer:
The value of \(|9\alpha + 3\beta + 5\gamma|\) is \(58\).