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=6`
B. `k=3`
C. `k=2`
D. `k=1.5`

Answer

To find the value of \( k \) for which the given pair of linear equations has infinitely many solutions, we need to ensure that the equations are equivalent. This means that the coefficients of \( x \) and \( y \) in both equations must be proportional, as well as the constant terms.

The given equations are:
1. \( kx + 4y + 6 = 0 \) (which we can rewrite as \( kx + 4y = -6 \))
2. \( 3x + 8y + 12 = 0 \) (which we can rewrite as \( 3x + 8y = -12 \))

For these two equations to have infinitely many solutions, the following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]
where \( a_1, b_1, c_1 \) are the coefficients from the first equation and \( a_2, b_2, c_2 \) are the coefficients from the second equation.

From the equations, we identify:
- \( a_1 = k \), \( b_1 = 4 \), \( c_1 = -6 \)
- \( a_2 = 3 \), \( b_2 = 8 \), \( c_2 = -12 \)

Now, we set up the ratios:
1. From \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \):
\[ \frac{k}{3} = \frac{4}{8} \]
Simplifying \( \frac{4}{8} \) gives \( \frac{1}{2} \):
\[ \frac{k}{3} = \frac{1}{2} \]

2. Cross-multiplying gives:
\[ 2k = 3 \]
Therefore,
\[ k = \frac{3}{2} \]

3. Now, we need to check the ratio of the constants:
\[ \frac{c_1}{c_2} = \frac{-6}{-12} = \frac{1}{2} \]
Since we already have \( \frac{k}{3} = \frac{1}{2} \) and \( \frac{c_1}{c_2} = \frac{1}{2} \), the condition for infinitely many solutions is satisfied.

Thus, the value of \( k \) for which the equations have infinitely many solutions is:
\[ \boxed{\frac{3}{2}} \]