Question

The system of equations \( 3x + 5y = 4 \); \( 6x + ky = 8 \) has infinitely many solutions if \( k \) is equal to

• A. 0
• B. 7
• C. 10
• D. 6 or 8

Hint:

To have infinitely many solutions, the system must be consistent and dependent, meaning the ratios of the coefficients must be equal.

Answer 1

The Correct Option is C

Step 1: Condition for infinitely many solutions.
For the system of equations to have infinitely many solutions, the two equations must be dependent. This happens when the ratios of the coefficients of \(x\), \(y\), and the constants are equal. For the system: \[ \frac{3}{6} = \frac{5}{k} = \frac{4}{8} \]
Step 2: Solving for \( k \).
From the first and third ratios: \[ \frac{3}{6} = \frac{4}{8} \quad \text{(True, as both are equal to } \frac{1}{2}) \] Now, solve the middle ratio: \[ \frac{5}{k} = \frac{1}{2} \] \[ k = 10 \] Thus, the system has infinitely many solutions when \( k = 10 \). The correct answer is (4) 6 or 8.

Answer 2

The system \( 3x + 5y = 4 \) and \( 6x + ky = 8 \) has infinitely many solutions only when the second equation is just a constant multiple of the first. Notice that the second equation's \( x \)-coefficient (6) and constant term (8) are already exactly double the first equation's \( x \)-coefficient (3) and constant term (4). So for the two equations to be true multiples of each other, the \( y \)-coefficient must double in the same way. Let's check each option against this requirement.

1. \( k = 0 \): Substituting gives \( 6x = 8 \), a line independent of \( y \), completely different from \( 3x+5y=4 \). These intersect in at most one point, not infinitely many.
2. \( k = 7 \): Substituting gives \( 6x+7y=8 \). Doubling the first equation gives \( 6x+10y=8 \), whose \( y \)-coefficient is \( 10 \), not \( 7 \). These are distinct intersecting lines, giving only one solution.
3. \( k = 10 \): Substituting gives \( 6x+10y=8 \), exactly \( 2\times(3x+5y=4) \). Since both equations describe the exact same line, every point on it is a solution, so the system has infinitely many solutions.
4. "6 or 8": Testing \( k=6 \) gives \( 6x+6y=8 \), not matching \( 6x+10y=8 \); testing \( k=8 \) gives \( 6x+8y=8 \), also not matching. Neither value makes the two equations the same line.

Only \( k = 10 \) makes the second equation an exact multiple of the first, so only then do the equations represent the same line.

Therefore, the correct answer is \( k = 10 \).