Number of solutions of the two equations ` 4 x - y = 2 and 2y -8x + 4 = 0 ` is

Options

A. zero
B. one
C. two
D. infinitely many

Answer

To determine the number of solutions for the two equations \( 4x - y = 2 \) and \( 2y - 8x + 4 = 0 \), we will analyze the equations step by step.

### Step 1: Rewrite the equations in standard form
The first equation is already in standard form:
\[ 4x - y - 2 = 0 \]

The second equation can be rearranged as follows:
\[ 2y - 8x + 4 = 0 \]
This can be rewritten as:
\[ -8x + 2y + 4 = 0 \]
or
\[ 2y - 8x + 4 = 0 \]

### Step 2: Identify coefficients
From the first equation \( 4x - y - 2 = 0 \), we can identify:
- \( a_1 = 4 \)
- \( b_1 = -1 \)
- \( c_1 = -2 \)

From the second equation \( -8x + 2y + 4 = 0 \), we can identify:
- \( a_2 = -8 \)
- \( b_2 = 2 \)
- \( c_2 = 4 \)

### Step 3: Check the conditions for the number of solutions
We will use the following conditions:
1. If \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), then there is exactly one solution.
2. If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) and \( \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), then there are infinitely many solutions.
3. If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), then there are no solutions.

### Step 4: Calculate the ratios
Now we compute the ratios:
- \( \frac{a_1}{a_2} = \frac{4}{-8} = -\frac{1}{2} \)
- \( \frac{b_1}{b_2} = \frac{-1}{2} \)
- \( \frac{c_1}{c_2} = \frac{-2}{4} = -\frac{1}{2} \)

### Step 5: Analyze the results
We find:
- \( \frac{a_1}{a_2} = -\frac{1}{2} \)
- \( \frac{b_1}{b_2} = -\frac{1}{2} \)
- \( \frac{c_1}{c_2} = -\frac{1}{2} \)

Since all three ratios are equal:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

### Conclusion
This indicates that the two equations represent the same line, which means they have infinitely many solutions.

### Final Answer
The number of solutions of the two equations is **infinite**.

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