For what value of k, the following pair of linear equations have infinitely many solution?
`kx + 4y + 6 = 0, 3x + 8y + 12 = 0`.

Options

A. `k = 0`
B. `k = 3`
C. `k = 2`
D. `k = 1.5`

Answer

To determine the value of \( k \) for which the given pair of linear equations has infinitely many solutions, we need to use the condition that the ratios of the coefficients of \( x \), \( y \), and the constant terms must be equal.

The given equations are:

1. \( kx + 4y + 6 = 0 \)
2. \( 3x + 8y + 12 = 0 \)

We can rewrite these equations in the standard form \( Ax + By + C = 0 \):

1. \( kx + 4y + 6 = 0 \) can be written as \( kx + 4y = -6 \)
2. \( 3x + 8y + 12 = 0 \) can be written as \( 3x + 8y = -12 \)

Now, we can identify the coefficients:

- For the first equation, \( A_1 = k \), \( B_1 = 4 \), and \( C_1 = 6 \).
- For the second equation, \( A_2 = 3 \), \( B_2 = 8 \), and \( C_2 = 12 \).

For the two equations to have infinitely many solutions, the following condition must hold:

\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]

Substituting the coefficients into the equation gives us:

\[ \frac{k}{3} = \frac{4}{8} = \frac{6}{12} \]

Now, we simplify the ratios:

1. \( \frac{4}{8} = \frac{1}{2} \)
2. \( \frac{6}{12} = \frac{1}{2} \)

Thus, we have:

\[ \frac{k}{3} = \frac{1}{2} \]

To find \( k \), we can cross-multiply:

\[ 2k = 3 \cdot 1 \]

This simplifies to:

\[ 2k = 3 \]

Now, divide both sides by 2:

\[ k = \frac{3}{2} \]

Thus, the value of \( k \) for which the given pair of linear equations has infinitely many solutions is:

\[ \boxed{\frac{3}{2}} \]