Find k, such that the system 3x + 5y = 0 and kx + 10y = 0 has a non-zero solution.

Answer

To find the value of \( k \) such that the system of equations

1. \( 3x + 5y = 0 \)
2. \( kx + 10y = 0 \)

has a non-zero solution, we can use the condition for the existence of non-zero solutions in a system of linear equations. This condition states that the ratios of the coefficients of the variables must be equal. Specifically, for the equations \( a_1x + b_1y = 0 \) and \( a_2x + b_2y = 0 \), the following must hold true:

\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

In our case, we can identify the coefficients as follows:

- From the first equation \( 3x + 5y = 0 \), we have \( a_1 = 3 \), \( b_1 = 5 \), and \( c_1 = 0 \).
- From the second equation \( kx + 10y = 0 \), we have \( a_2 = k \), \( b_2 = 10 \), and \( c_2 = 0 \).

Now, we can set up the ratios:

\[ \frac{3}{k} = \frac{5}{10} \]

Next, we simplify the right side of the equation:

\[ \frac{5}{10} = \frac{1}{2} \]

So, we have:

\[ \frac{3}{k} = \frac{1}{2} \]

Now, we can cross-multiply to solve for \( k \):

\[ 3 \cdot 2 = 1 \cdot k \]

This simplifies to:

\[ 6 = k \]

Thus, the value of \( k \) that allows the system to have a non-zero solution is:

\[ \boxed{6} \]