The value of 'k' so that the system of equations `3x-y-5 =0 and 6x-2y-k=0` has infinite solutions

Options

A. `k=-10`
B. `k=10`
C. `k=-8`
D. `k=8`

Answer

To find the value of 'k' so that the system of equations \(3x - y - 5 = 0\) and \(6x - 2y - k = 0\) has infinite solutions, we need to analyze the conditions for infinite solutions in a system of linear equations.

### Step-by-Step Solution:

1. **Identify the Coefficients**:
The first equation can be rewritten in the standard form \(a_1x + b_1y + c_1 = 0\):
- For \(3x - y - 5 = 0\), we have:
- \(a_1 = 3\)
- \(b_1 = -1\)
- \(c_1 = -5\)

The second equation can also be rewritten:
- For \(6x - 2y - k = 0\), we have:
- \(a_2 = 6\)
- \(b_2 = -2\)
- \(c_2 = -k\)

2. **Set Up the Condition for Infinite Solutions**:
For the two equations to have infinite solutions, the following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

3. **Calculate the Ratios**:
- Calculate \(\frac{a_1}{a_2}\):
\[ \frac{3}{6} = \frac{1}{2} \]
- Calculate \(\frac{b_1}{b_2}\):
\[ \frac{-1}{-2} = \frac{1}{2} \]
- Now, set \(\frac{c_1}{c_2}\):
\[ \frac{-5}{-k} = \frac{5}{k} \]

4. **Equate the Ratios**:
Since all ratios must be equal, we can set:
\[ \frac{1}{2} = \frac{5}{k} \]

5. **Cross-Multiply**:
To solve for \(k\), cross-multiply:
\[ 1 \cdot k = 2 \cdot 5 \]
\[ k = 10 \]

### Conclusion:
The value of \(k\) for which the system of equations has infinite solutions is \(k = 10\).