Question

day 2
add or subtract the given expressions, simplifying each result and noting the combined excluded values.

1. $\frac{1}{1+x}+\frac{1-x}{x}$
2. $\frac{x+4}{x^2-4}+\frac{-2x-2}{x^2-4}$

Explanation:

For Problem 15:

Step1: Find common denominator

The common denominator of $\frac{1}{1+x}$ and $\frac{1-x}{x}$ is $x(1+x)$.

Step2: Rewrite fractions with LCD

$$\frac{1 \cdot x}{x(1+x)} + \frac{(1-x)(1+x)}{x(1+x)}$$

Step3: Expand numerators

$$\frac{x}{x(1+x)} + \frac{1-x^2}{x(1+x)}$$

Step4: Add numerators

$$\frac{x + 1 - x^2}{x(1+x)}$$

Step5: Identify excluded values

Denominators cannot be 0: $x
eq 0$, $1+x
eq 0 \implies x
eq -1$

Step1: Add numerators (same denominator)

Since denominators are equal, combine numerators:
$$\frac{(x+4) + (-2x-2)}{x^2-4}$$

Step2: Simplify the numerator

$$\frac{x+4-2x-2}{x^2-4} = \frac{-x+2}{x^2-4}$$

Step3: Factor denominator

Factor $x^2-4$ as $(x-2)(x+2)$:
$$\frac{-(x-2)}{(x-2)(x+2)}$$

Step4: Cancel common factors

Cancel $(x-2)$ (where $x
eq 2$):
$$\frac{-1}{x+2}$$

Step5: Identify excluded values

Original denominator $x^2-4
eq 0 \implies (x-2)(x+2)
eq 0$, so $x
eq 2, x
eq -2$

Answer:

Simplified expression: $\frac{-x^2 + x + 1}{x(x+1)}$; Excluded values: $x
eq -1, 0$

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For Problem 16: