`(k-1)x+(1)/(3)y=4`
`k(x+2y)=7`
In the system of linear equations above, k is a constant. If the system has no system, what is the value of k?

Answer

To solve the problem, we need to find the value of \( k \) for which the system of equations has no solution. The given equations are:

1. \( (k-1)x + \frac{1}{3}y = 4 \)
2. \( k(x + 2y) = 7 \)

### Step 1: Rewrite the second equation in standard form
We can rewrite the second equation as:
\[ kx + 2ky = 7 \]

### Step 2: Identify coefficients
From the equations, we identify the coefficients:
- For the first equation:
- \( a_1 = k - 1 \)
- \( b_1 = \frac{1}{3} \)
- \( c_1 = 4 \)

- For the second equation:
- \( a_2 = k \)
- \( b_2 = 2k \)
- \( c_2 = 7 \)

### Step 3: Apply the condition for no solution
For the system of equations to have no solution, the following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \quad \text{and} \quad \frac{a_1}{a_2} \neq \frac{c_1}{c_2} \]

Substituting the coefficients:
\[ \frac{k - 1}{k} = \frac{\frac{1}{3}}{2k} \]

### Step 4: Cross-multiply and simplify
Cross-multiplying gives:
\[ 3(k - 1) = k \cdot 2k \]
This simplifies to:
\[ 3k - 3 = 2k^2 \]
Rearranging gives:
\[ 2k^2 - 3k + 3 = 0 \]

### Step 5: Solve the quadratic equation
To solve \( 2k^2 - 3k + 3 = 0 \), we can use the quadratic formula:
\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = -3, c = 3 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-3)^2 - 4 \cdot 2 \cdot 3 = 9 - 24 = -15 \]
Since the discriminant is negative, this means there are no real solutions for \( k \) from this equation.

### Step 6: Check the condition for no solution
Next, we need to check the condition for no solution:
\[ \frac{a_1}{a_2} \neq \frac{c_1}{c_2} \]
Substituting the values:
\[ \frac{k - 1}{k} \neq \frac{4}{7} \]

### Step 7: Solve for k
Setting up the equation:
\[ \frac{k - 1}{k} = \frac{4}{7} \]
Cross-multiplying gives:
\[ 7(k - 1) = 4k \]
Expanding and simplifying:
\[ 7k - 7 = 4k \implies 3k = 7 \implies k = \frac{7}{3} \]

### Conclusion
Thus, the value of \( k \) for which the system of equations has no solution is:
\[ \boxed{\frac{7}{3}} \]