Read the statements carefully and select the correct option
Statement - I : The graphical representation of 2 x + y = 6 and 2x - y - 2 = 0 will be a pair of parallel lines
Statements - II : When k = - 1 then linear equations 5x + ky = 4 and 15 x + 3y = 12 have infinitely many solutions

Options

A. Both Statement-I and Statement-II are true.
B. Statement-I is true but Statement-II is false.
C. Statement-I is false but Statement-II is true
D. Both Statement-I and Statement-II are false

Answer

To determine whether the statements are correct, we will analyze each statement step by step.

### Step 1: Analyze Statement I
We need to check if the lines represented by the equations \(2x + y = 6\) and \(2x - y - 2 = 0\) are parallel.

1. **Convert both equations to the standard form**:
- The first equation can be rewritten as:
\[ 2x + y - 6 = 0 \]
- The second equation can be rewritten as:
\[ 2x - y - 2 = 0 \]

2. **Identify coefficients**:
- For the first line \(2x + y - 6 = 0\), we have:
- \(a_1 = 2\), \(b_1 = 1\), \(c_1 = -6\)
- For the second line \(2x - y - 2 = 0\), we have:
- \(a_2 = 2\), \(b_2 = -1\), \(c_2 = -2\)

3. **Check the condition for parallel lines**:
- The condition for two lines to be parallel is:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \quad \text{and} \quad \frac{c_1}{c_2} \neq \frac{a_1}{a_2} \]
- Calculate \(\frac{a_1}{a_2}\) and \(\frac{b_1}{b_2}\):
\[ \frac{a_1}{a_2} = \frac{2}{2} = 1 \]
\[ \frac{b_1}{b_2} = \frac{1}{-1} = -1 \]
- Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), the lines are not parallel.

### Conclusion for Statement I:
**Statement I is false**.

### Step 2: Analyze Statement II
We need to check if the equations \(5x + ky = 4\) and \(15x + 3y = 12\) have infinitely many solutions when \(k = -1\).

1. **Convert the second equation to the standard form**:
- The second equation can be simplified:
\[ 15x + 3y - 12 = 0 \quad \Rightarrow \quad 5x + y - 4 = 0 \]

2. **Identify coefficients**:
- For the first line \(5x + ky - 4 = 0\):
- \(a_1 = 5\), \(b_1 = k\), \(c_1 = -4\)
- For the second line \(5x + y - 4 = 0\):
- \(a_2 = 5\), \(b_2 = 1\), \(c_2 = -4\)

3. **Substituting \(k = -1\)**:
- The first equation becomes:
\[ 5x - y - 4 = 0 \]
- Now we have:
- \(a_1 = 5\), \(b_1 = -1\), \(c_1 = -4\)

4. **Check the condition for infinitely many solutions**:
- The condition for infinitely many solutions is:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]
- Calculate:
\[ \frac{a_1}{a_2} = \frac{5}{5} = 1 \]
\[ \frac{b_1}{b_2} = \frac{-1}{1} = -1 \]
\[ \frac{c_1}{c_2} = \frac{-4}{-4} = 1 \]
- Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), the lines do not have infinitely many solutions.

### Conclusion for Statement II:
**Statement II is false**.

### Final Conclusion:
Both statements are false.

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