Question

Do the Factor, Restrict, and solve method for these rational expressions : (x+1)/(x-1) - (x+2)/(x-1) and x/(x+2) + 3/(x+4)

Answer 1

Let's tackle each rational expression using the Factor, Restrict, and Solve method.

### Expression 1:
\[
\frac{x+1}{x-1} - \frac{x+2}{x-1}
\]

**Step 1: Factor**
Since both fractions have the same denominator, we can combine them directly:
\[
\frac{(x+1) - (x+2)}{x-1} = \frac{x+1 - x - 2}{x-1}
\]
This simplifies to:
\[
\frac{-1}{x-1}
\]

**Step 2: Restrict**
To restrict the expression, we need to identify values that make the original denominators zero. The denominator in this case is \(x - 1\):
\[
x - 1 \neq 0 \implies x \neq 1
\]

**Step 3: Solve**
Since the expression simplifies to \(-\frac{1}{x - 1}\), it is already solved. The final answer is:
\[
\frac{-1}{x-1}, \quad x \neq 1
\]

---

### Expression 2:
\[
\frac{x}{x+2} + \frac{3}{x+4}
\]

**Step 1: Factor**
To add these fractions, we need a common denominator, which is \((x+2)(x+4)\). We rewrite each fraction:
\[
\frac{x}{x+2} = \frac{x(x+4)}{(x+2)(x+4)} = \frac{x^2 + 4x}{(x+2)(x+4)}
\]
\[
\frac{3}{x+4} = \frac{3(x+2)}{(x+4)(x+2)} = \frac{3x + 6}{(x+2)(x+4)}
\]

Now, combine the two fractions:
\[
\frac{x^2 + 4x + 3x + 6}{(x+2)(x+4)} = \frac{x^2 + 7x + 6}{(x+2)(x+4)}
\]

**Step 2: Restrict**
Next, we find restrictions by determining when the denominators are zero:
- For \(x + 2 = 0\):
\[
x \neq -2
\]
- For \(x + 4 = 0\):
\[
x \neq -4
\]

**Step 3: Solve**
The expression simplifies to:
\[
\frac{x^2 + 7x + 6}{(x+2)(x+4)}
\]

Next, we can factor the numerator \(x^2 + 7x + 6\):
\[
x^2 + 7x + 6 = (x + 1)(x + 6)
\]
Thus,
\[
\frac{(x+1)(x+6)}{(x+2)(x+4)}
\]

The final simplified result is:
\[
\frac{(x+1)(x+6)}{(x+2)(x+4)}, \quad x \neq -2, -4
\]

### Summary of Results:
1. \(\frac{-1}{x-1}, \quad x \neq 1\)
2. \(\frac{(x+1)(x+6)}{(x+2)(x+4)}, \quad x \neq -2, -4\)