Find the value of k for which the system of equations `kx - y = 2, 6x - 2y = 4` has infinitely many solutions

Options

A. 1
B. 3
C. `-1`
D. no vlaue

Answer

To find the value of \( k \) for which the system of equations

1. \( kx - y = 2 \)
2. \( 6x - 2y = 4 \)

has infinitely many solutions, we need to analyze the conditions under which two linear equations represent the same line. This occurs when the ratios of the coefficients of \( x \), \( y \), and the constant terms are equal.

### Step 1: Rewrite the equations in standard form
We can rewrite the first equation as:
\[ kx - y - 2 = 0 \]
This gives us:
- \( a_1 = k \)
- \( b_1 = -1 \)
- \( c_1 = -2 \)

For the second equation, we can rewrite it as:
\[ 6x - 2y - 4 = 0 \]
This gives us:
- \( a_2 = 6 \)
- \( b_2 = -2 \)
- \( c_2 = -4 \)

### Step 2: Set up the ratios for infinitely many solutions
For the system to have infinitely many solutions, the following must hold true:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

Substituting the values we found:
\[ \frac{k}{6} = \frac{-1}{-2} = \frac{-2}{-4} \]

### Step 3: Simplify the ratios
Calculating the right side:
\[ \frac{-1}{-2} = \frac{1}{2} \]
\[ \frac{-2}{-4} = \frac{1}{2} \]

Thus, we have:
\[ \frac{k}{6} = \frac{1}{2} \]

### Step 4: Solve for \( k \)
To find \( k \), we can cross-multiply:
\[ k \cdot 2 = 6 \cdot 1 \]
\[ 2k = 6 \]
Dividing both sides by 2:
\[ k = 3 \]

### Conclusion
The value of \( k \) for which the system of equations has infinitely many solutions is \( k = 3 \).

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