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: For \( 3x+5y=4 \) and \( 6x+ky=8 \) to have infinitely many solutions, the coefficient matrix and augmented matrix must both have rank \( 1 \), i.e. every row must be a scalar multiple of every other row.
Step 2: The ratio of \( x \)-coefficients is \( \frac{6}{3} = 2 \), and the ratio of constants is \( \frac{8}{4} = 2 \) — both already equal \( 2 \).
Step 3: For full proportionality, the \( y \)-coefficients must share this ratio: \( \frac{k}{5} = 2 \), so \( k = 10 \).
\[ \boxed{k = 10} \]

Answer 2

Another way to find when this system has infinitely many solutions is to eliminate one variable directly and see what condition on \( k \) makes the resulting equation vanish identically. Multiply \( 3x+5y=4 \) by \( 2 \): \( 6x+10y=8 \). Subtracting \( 6x+ky=8 \) from this gives \( (10-k)y = 0 \), which must hold for every \( y \) only when the coefficient of \( y \) itself is zero.

1. \( k = 0 \): Then \( (10-0)y=10y \), zero only when \( y=0 \), a single value.
2. \( k = 7 \): Then \( (10-7)y=3y \), again zero only at \( y=0 \).
3. \( k = 10 \): Then \( (10-10)y=0 \), true for every \( y \), meaning the two original equations coincide, giving infinitely many solutions.
4. "6 or 8": \( k=6 \) gives \( (10-6)y=4y \), zero only at \( y=0 \); \( k=8 \) gives \( (10-8)y=2y \), also zero only at \( y=0 \). Neither makes the coefficient vanish identically.

Only \( k = 10 \) makes the coefficient of \( y \) vanish completely, exactly the condition needed for infinitely many solutions.

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