For what value of k, will the equations `2x+3y-5=0, 6x+ky-15=0` have an infinite number of solutions?

Options

A. 0
B. 9
C. 2
D. 3

Answer

To determine the value of \( k \) for which the equations \( 2x + 3y - 5 = 0 \) and \( 6x + ky - 15 = 0 \) have an infinite number of solutions, we can follow these steps:

### Step 1: Identify the coefficients
The first equation can be written in the standard form \( A_1x + B_1y + C_1 = 0 \):
- For the first equation \( 2x + 3y - 5 = 0 \):
- \( A_1 = 2 \)
- \( B_1 = 3 \)
- \( C_1 = -5 \)

The second equation can also be written in the same form:
- For the second equation \( 6x + ky - 15 = 0 \):
- \( A_2 = 6 \)
- \( B_2 = k \)
- \( C_2 = -15 \)

### Step 2: Set up the condition for infinite solutions
For two linear equations to have an infinite number of solutions, the ratios of the coefficients must be equal:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]

### Step 3: Write the ratios
Substituting the coefficients into the ratios gives:
\[ \frac{2}{6} = \frac{3}{k} = \frac{-5}{-15} \]

### Step 4: Simplify the ratios
We can simplify the ratios:
\[ \frac{2}{6} = \frac{1}{3} \quad \text{and} \quad \frac{-5}{-15} = \frac{1}{3} \]
Thus, we have:
\[ \frac{1}{3} = \frac{3}{k} \]

### Step 5: Cross-multiply to solve for \( k \)
Cross-multiplying gives:
\[ 1 \cdot k = 3 \cdot 3 \]
This simplifies to:
\[ k = 9 \]

### Conclusion
The value of \( k \) for which the equations have an infinite number of solutions is \( k = 9 \).

---