Separate equations of lines, whose combined equation is `4x^(2)-y^(2)+2x+y=0` are

Options

A. `2x+y+1=0, 2x+y=0`
B. `2x-y=0,2x+y+1=0`
C. `x-2y+1=0,x+2y=0`
D. `2x-y+1=0,x-2y=0`

Answer

To find the separate equations of lines from the combined equation \( 4x^2 - y^2 + 2x + y = 0 \), we can follow these steps:

### Step 1: Rewrite the equation
Start with the given combined equation:
\[ 4x^2 - y^2 + 2x + y = 0 \]

### Step 2: Group the terms
Notice that the first two terms \( 4x^2 - y^2 \) can be recognized as a difference of squares:
\[ (2x)^2 - (y)^2 \]
This can be factored using the difference of squares formula \( a^2 - b^2 = (a - b)(a + b) \).

### Step 3: Factor the difference of squares
Applying the difference of squares to \( (2x)^2 - (y)^2 \):
\[ (2x - y)(2x + y) \]
Now, we can rewrite the equation as:
\[ (2x - y)(2x + y) + 2x + y = 0 \]

### Step 4: Factor out common terms
Next, we can factor out \( 2x + y \):
\[ (2x + y)(2x - y + 1) = 0 \]

### Step 5: Set each factor to zero
Now, we can set each factor equal to zero to find the separate equations of the lines:
1. \( 2x + y = 0 \)
2. \( 2x - y + 1 = 0 \)

### Step 6: Write the final equations
The separate equations of the lines are:
1. \( y = -2x \)
2. \( y = 2x + 1 \)

### Final Answer:
The separate equations of the lines are:
1. \( 2x + y = 0 \)
2. \( 2x - y + 1 = 0 \)

---