Find the value of k for which the system of equations
` " 5x - 3y = 0 , 2x + ky = 0 "` has a nonzero solution.

Answer

To find the value of \( k \) for which the system of equations

\[ 5x - 3y = 0 \]
and
\[ 2x + ky = 0 \]

has a non-zero solution, we can use the condition for the existence of non-trivial solutions for a system of linear equations. The condition states that the ratios of the coefficients of \( x \) and \( y \) must be equal.

### Step-by-Step Solution:

1. **Identify the coefficients**:
From the first equation \( 5x - 3y = 0 \), we identify:
- \( a_1 = 5 \)
- \( b_1 = -3 \)

From the second equation \( 2x + ky = 0 \), we identify:
- \( a_2 = 2 \)
- \( b_2 = k \)

2. **Set up the ratio condition**:
According to the condition for non-zero solutions, we have:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \]
Substituting the values we identified:
\[ \frac{5}{2} = \frac{-3}{k} \]

3. **Cross-multiply to solve for \( k \)**:
Cross-multiplying gives:
\[ 5k = -6 \]

4. **Isolate \( k \)**:
Dividing both sides by 5:
\[ k = \frac{-6}{5} \]

Thus, the value of \( k \) for which the system of equations has a non-zero solution is
\[ \boxed{-\frac{6}{5}}. \]