Question

find the sum:
\\(\frac{x - 2}{x^2 + 1} + \frac{x + 3}{x^2 + 1}\\)
\\(\frac{2x + 1}{2x^2 + 2}\\)
\\(\frac{1}{x^2 + 1}\\)
\\(\frac{2x + 1}{x^2 + 1}\\)
\\(\frac{2}{x}\\)
find the difference:
\\(\frac{x + 1}{x - 1} - \frac{x - 8}{x - 1}\\)
\\(\frac{9}{x - 1}\\)
\\(\frac{2x - 7}{x - 1}\\)
\\(\frac{-7}{x - 1}\\)
\\(\frac{2x + 9}{x - 1}\\)

Explanation:

Step1: Add numerators (common denominator)

$\frac{(x-2)+(x+3)}{x^2+1}$

Step2: Simplify numerator

$\frac{x-2+x+3}{x^2+1} = \frac{2x+1}{x^2+1}$

Step3: Subtract numerators (common denominator)

$\frac{(x+1)-(x-8)}{x-1}$

Step4: Simplify numerator

$\frac{x+1-x+8}{x-1} = \frac{9}{x-1}$

Answer:

Sum: $\frac{2x+1}{x^2+1}$
Difference: $\frac{9}{x-1}$