Solve: `(8)/(x+3)-2= (3)/(2-x)`

Answer

To solve the equation \(\frac{8}{x+3} - 2 = \frac{3}{2-x}\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \frac{8}{x+3} - 2 = \frac{3}{2-x} \]

### Step 2: Move the constant to the right side
Add \(2\) to both sides:
\[ \frac{8}{x+3} = \frac{3}{2-x} + 2 \]

### Step 3: Find a common denominator on the right side
The common denominator for the right side is \(2-x\):
\[ \frac{8}{x+3} = \frac{3 + 2(2-x)}{2-x} \]
This simplifies to:
\[ \frac{8}{x+3} = \frac{3 + 4 - 2x}{2-x} = \frac{7 - 2x}{2-x} \]

### Step 4: Cross-multiply
Cross-multiplying gives us:
\[ 8(2-x) = (7 - 2x)(x + 3) \]

### Step 5: Expand both sides
Expanding the left side:
\[ 16 - 8x \]
Expanding the right side:
\[ (7 - 2x)(x + 3) = 7x + 21 - 2x^2 - 6x = -2x^2 + x + 21 \]
Thus, we have:
\[ 16 - 8x = -2x^2 + x + 21 \]

### Step 6: Rearrange the equation
Bringing all terms to one side:
\[ 2x^2 - 9x + 16 - 21 = 0 \]
This simplifies to:
\[ 2x^2 - 9x - 5 = 0 \]

### Step 7: Factor the quadratic equation
To factor \(2x^2 - 9x - 5\), we look for two numbers that multiply to \(2 \times -5 = -10\) and add to \(-9\). The numbers are \(-10\) and \(1\):
\[ 2x^2 - 10x + x - 5 = 0 \]
Factoring by grouping:
\[ 2x(x - 5) + 1(x - 5) = 0 \]
This gives:
\[ (2x + 1)(x - 5) = 0 \]

### Step 8: Solve for \(x\)
Setting each factor to zero:
1. \(2x + 1 = 0 \Rightarrow x = -\frac{1}{2}\)
2. \(x - 5 = 0 \Rightarrow x = 5\)

Thus, the solutions are:
\[ x = 5 \quad \text{and} \quad x = -\frac{1}{2} \]