https://file.aiquickdraw.com/mathbot/file/1727114259175tbiiw4ws.jpg

The equation in the image is:

\[
\frac{1}{x+3} + \frac{1}{x-5} = \frac{1 - x}{x+3}
\]

To solve this equation, let's proceed step by step:

### Step 1: Eliminate denominators by multiplying the entire equation by the least common denominator (LCD).
The LCD of \(x+3\) and \(x-5\) is \((x+3)(x-5)\).

Multiply every term by \((x+3)(x-5)\):

\[
(x+3)(x-5) \cdot \frac{1}{x+3} + (x+3)(x-5) \cdot \frac{1}{x-5} = (x+3)(x-5) \cdot \frac{1 - x}{x+3}
\]

Simplify each term:

\[
(x-5) + (x+3) = (x-5)(1-x)
\]

### Step 2: Simplify both sides.
On the left side:
\[
x - 5 + x + 3 = 2x - 2
\]

On the right side:
\[
(x-5)(1-x) = x(1-x) - 5(1-x) = x - x^2 - 5 + 5x = -x^2 + 6x - 5
\]

Thus, the equation becomes:

\[
2x - 2 = -x^2 + 6x - 5
\]

### Step 3: Rearrange into a standard quadratic form.
Move all terms to one side:

\[
x^2 - 4x + 3 = 0
\]

### Step 4: Factor the quadratic equation.
\[
(x - 3)(x - 1) = 0
\]

### Step 5: Solve for \(x\).
Setting each factor to zero:

\[
x - 3 = 0 \quad \text{or} \quad x - 1 = 0
\]
\[
x = 3 \quad \text{or} \quad x = 1
\]

### Step 6: Check for extraneous solutions.
Since the original equation has denominators with \(x + 3\) and \(x - 5\), we must ensure that neither solution makes these denominators zero.

- For \(x = 3\), none of the denominators become zero.
- For \(x = 1\), none of the denominators become zero.

Thus, both \(x = 3\) and \(x = 1\) are valid solutions.

### Final Answer:
\[
x = 3 \quad \text{or} \quad x = 1
\]

Would you like further details or explanations? Here are some related questions:

1. How is the least common denominator determined in rational equations?
2. Why do we need to check for extraneous solutions in rational equations?
3. Can quadratic equations always be factored easily?
4. What happens if one of the solutions makes the denominator zero?
5. How do you verify the solutions in rational equations?

**Tip:** Always verify that none of the solutions make the denominator zero in rational equations to avoid extraneous solutions.

Solve the equation: 1/(x+3) + 1/(x-5) = (1-x)/(x+3)

Step-by-step solution of the rational equation 1/(x+3) + 1/(x-5) = (1-x)/(x+3), involving key algebraic techniques like finding the least common denominator (LCD), simplifying quadratic equations, and checking for extraneous solutions. Suitable for Grades 10-12.

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