`(1)/(3)r+4s=1`
`kr+6s=-5`
In the system of equations above, k and s are nonzero constant. If the system has no solutions, what is the value of k?

Answer

To solve the system of equations given by:

1. \(\frac{1}{3}r + 4s = 1\)
2. \(kr + 6s = -5\)

we need to determine the value of \(k\) such that the system has no solutions. This occurs when the two lines represented by the equations are parallel, which means their slopes must be equal.

### Step 1: Rewrite the first equation in slope-intercept form (y = mx + b)
Starting with the first equation:

\[ \frac{1}{3}r + 4s = 1 \]

We can isolate \(r\):

\[ \frac{1}{3}r = 1 - 4s \]

Now, multiply both sides by 3 to eliminate the fraction:

\[ r = 3 - 12s \]

This can be rewritten as:

\[ r = -12s + 3 \]

From this, we can identify the slope \(m_1 = -12\).

### Step 2: Rewrite the second equation in slope-intercept form
Now, let's rewrite the second equation:

\[ kr + 6s = -5 \]

Isolate \(r\):

\[ kr = -5 - 6s \]

Now, divide by \(k\) (assuming \(k \neq 0\)):

\[ r = -\frac{6}{k}s - \frac{5}{k} \]

From this, we can identify the slope \(m_2 = -\frac{6}{k}\).

### Step 3: Set the slopes equal for parallel lines
Since the lines are parallel, we set the slopes equal to each other:

\[ -12 = -\frac{6}{k} \]

### Step 4: Solve for \(k\)
To solve for \(k\), we can multiply both sides by \(-k\) (again, assuming \(k \neq 0\)):

\[ 12k = 6 \]

Now, divide both sides by 12:

\[ k = \frac{6}{12} = \frac{1}{2} \]

### Conclusion
The value of \(k\) for which the system of equations has no solutions is:

\[ \boxed{\frac{1}{2}} \]