`{((3)/(4)x-(1)/(2)y=12), (kx-2y=22):}`
If the system of linear equations above has no solution, and k is a constant, what is the value of k?

Options

A. `-(4)/(3)`
B. `-(3)/(4)`
C. `3`
D. `4`

Answer

To determine the value of \( k \) such that the system of linear equations has no solution, we need to analyze the given equations:

1. **Equations**:
\[ \frac{3}{4}x - \frac{1}{2}y = 12 \quad \text{(Equation 1)} \]
\[ kx - 2y = 22 \quad \text{(Equation 2)} \]

2. **Understanding No Solution**:
A system of linear equations has no solution if the lines represented by the equations are parallel. This occurs when the slopes of the lines are equal but the y-intercepts are different.

3. **Rearranging Equation 1**:
To find the slope of the first equation, we can rearrange it into the slope-intercept form \( y = mx + b \):
\[ -\frac{1}{2}y = -\frac{3}{4}x + 12 \]
Multiplying through by -2 to isolate \( y \):
\[ y = \frac{3}{2}x - 24 \]
Here, the slope \( m_1 \) of the first equation is \( \frac{3}{2} \).

4. **Rearranging Equation 2**:
Now, we rearrange the second equation into slope-intercept form:
\[ -2y = -kx + 22 \]
Dividing through by -2 gives:
\[ y = \frac{k}{2}x - 11 \]
Here, the slope \( m_2 \) of the second equation is \( \frac{k}{2} \).

5. **Setting the Slopes Equal**:
Since the lines are parallel, we set the slopes equal to each other:
\[ \frac{3}{2} = \frac{k}{2} \]

6. **Solving for \( k \)**:
To solve for \( k \), we can cross-multiply:
\[ 3 \cdot 2 = k \cdot 2 \]
This simplifies to:
\[ 3 = k \]

7. **Conclusion**:
Therefore, the value of \( k \) that makes the system of equations have no solution is:
\[ k = 3 \]