`(1)/(2)x-(5)/(6)y=5`
`-2x+ky=3`
In the system of linear equations above, k is a constant. If the system has no solution, what is the value of k?

Options

A. `(5)/(3)`
B. `(5)/(2)`
C. `(10)/(3)`
D. `(15)/(2)`

Answer

To find the value of \( k \) such that the system of linear equations has no solution, we need to determine the conditions under which the two lines represented by the equations are parallel.

The given equations are:

1. \( \frac{1}{2}x - \frac{5}{6}y = 5 \)
2. \( -2x + ky = 3 \)

### Step 1: Rewrite the equations in slope-intercept form (y = mx + b)

**For the first equation:**

Starting with:
\[ \frac{1}{2}x - \frac{5}{6}y = 5 \]

We can isolate \( y \):
\[ -\frac{5}{6}y = -\frac{1}{2}x + 5 \]
Multiplying through by \(-1\):
\[ \frac{5}{6}y = \frac{1}{2}x - 5 \]
Now, multiply everything by \(\frac{6}{5}\) to solve for \( y \):
\[ y = \frac{6}{5} \cdot \frac{1}{2}x - \frac{6}{5} \cdot 5 \]
\[ y = \frac{3}{5}x - 6 \]

The slope \( m_1 \) of the first line is \( \frac{3}{5} \).

**For the second equation:**

Starting with:
\[ -2x + ky = 3 \]

Isolating \( y \):
\[ ky = 2x + 3 \]
Now, divide everything by \( k \):
\[ y = \frac{2}{k}x + \frac{3}{k} \]

The slope \( m_2 \) of the second line is \( \frac{2}{k} \).

### Step 2: Set the slopes equal for parallel lines

For the lines to be parallel, their slopes must be equal:
\[ \frac{3}{5} = \frac{2}{k} \]

### Step 3: Solve for \( k \)

Cross-multiplying gives:
\[ 3k = 10 \]
Now, divide both sides by 3:
\[ k = \frac{10}{3} \]

### Conclusion

The value of \( k \) such that the system of equations has no solution (the lines are parallel) is:
\[ \boxed{\frac{10}{3}} \]