If a and b are solutions to the equation `(2x)/(2x+1)+(8)/(x-3)=0, and altb`, what is the value of a?

Options

A. `-4`
B. `-1`
C. `1`
D. `4`

Explanation

To solve the equation \(\frac{2x}{2x+1} + \frac{8}{x-3} = 0\) and find the value of \(a\) (the smaller solution), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \frac{2x}{2x+1} + \frac{8}{x-3} = 0 \]
To eliminate the fractions, we can rewrite it as:
\[ \frac{2x}{2x+1} = -\frac{8}{x-3} \]

### Step 2: Cross-multiply
Cross-multiplying gives us:
\[ 2x \cdot (x - 3) = -8 \cdot (2x + 1) \]
Expanding both sides:
\[ 2x^2 - 6x = -16x - 8 \]

### Step 3: Rearrange the equation
Now, we will move all terms to one side of the equation:
\[ 2x^2 - 6x + 16x + 8 = 0 \]
This simplifies to:
\[ 2x^2 + 10x + 8 = 0 \]

### Step 4: Factor out the common term
We notice that all terms are divisible by 2:
\[ x^2 + 5x + 4 = 0 \]

### Step 5: Factor the quadratic
Next, we need to factor the quadratic:
\[ (x + 4)(x + 1) = 0 \]

### Step 6: Find the solutions
Setting each factor to zero gives us the solutions:
\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

### Step 7: Identify the smaller solution
Since we need the smaller solution \(a\), we compare the two solutions:
\(-4\) and \(-1\). Clearly, \(-4\) is smaller than \(-1\).

Thus, the value of \(a\) is:
\[ \boxed{-4} \]

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