Question

If the system of equations 3x-2y-7=0 and kx+2y+11=0 has unique solution then

• A. \(k\ne3\)
• B. \(k\ne-3\)
• C. \(k=3\)
• D. \(k=-3\)

Answer 1

The Correct Option is B

To solve the problem, we need to determine the condition under which the given system of linear equations has a unique solution. The system of equations is:

$$ 3x - 2y - 7 = 0 $$

$$ kx + 2y + 11 = 0 $$

1. Understanding the Condition for a Unique Solution:
A system of linear equations has a unique solution if and only if the lines represented by the equations are not parallel. This means the slopes of the two lines must be different.

2. Rewriting the Equations in Slope-Intercept Form:
We rewrite each equation in the form \( y = mx + c \), where \( m \) is the slope.

For the first equation \( 3x - 2y - 7 = 0 \):

$$ 3x - 2y = 7 $$

$$ -2y = -3x + 7 $$

$$ y = \frac{3}{2}x - \frac{7}{2} $$

So, the slope of the first line is \( m_1 = \frac{3}{2} \).

For the second equation \( kx + 2y + 11 = 0 \):

$$ kx + 2y = -11 $$

$$ 2y = -kx - 11 $$

$$ y = -\frac{k}{2}x - \frac{11}{2} $$

So, the slope of the second line is \( m_2 = -\frac{k}{2} \).

3. Ensuring the Lines Are Not Parallel:
For the system to have a unique solution, the slopes \( m_1 \) and \( m_2 \) must be different:

$$ m_1 \neq m_2 $$

Substituting the values of \( m_1 \) and \( m_2 \):

$$ \frac{3}{2} \neq -\frac{k}{2} $$

4. Solving for \( k \):
Multiply both sides by 2 to eliminate the denominators:

$$ 3 \neq -k $$

$$ k \neq -3 $$

5. Conclusion:
The condition for the system to have a unique solution is that \( k \neq -3 \).

Final Answer:
The correct option is \( {k \neq -3} \).