What is the sum of [tex]\frac{7x}{x^2-4}[/tex] and [tex]\frac{2}{x+2}[/tex]?

A. [tex]\frac{7x+2}{x^2-4}[/tex]
B. [tex]\frac{7x+2}{x^2+x-2}[/tex]
C. [tex]\frac{9}{x}[/tex]
D. [tex]\frac{9x-4}{x^2-4}[/tex]



Answer :

To find the sum of [tex]\(\frac{7x}{x^2 - 4}\)[/tex] and [tex]\(\frac{2}{x + 2}\)[/tex], we can follow these steps:

1. Simplify the Denominators:
- Notice that [tex]\(x^2 - 4\)[/tex] is a difference of squares and can be factored:
[tex]\[ x^2 - 4 = (x - 2)(x + 2) \][/tex]

2. Rewrite the Expressions:
- The first fraction is already expressed with the factored denominator:
[tex]\[ \frac{7x}{x^2 - 4} = \frac{7x}{(x - 2)(x + 2)} \][/tex]
- Rewrite [tex]\(\frac{2}{x + 2}\)[/tex] over the common denominator [tex]\((x - 2)(x + 2)\)[/tex]:
[tex]\[ \frac{2}{x + 2} = \frac{2(x - 2)}{(x + 2)(x - 2)} = \frac{2x - 4}{(x + 2)(x - 2)} \][/tex]

3. Add the Fractions:
- Now, both fractions have the same denominator [tex]\((x - 2)(x + 2)\)[/tex]. We can add the numerators:
[tex]\[ \frac{7x}{(x - 2)(x + 2)} + \frac{2x - 4}{(x - 2)(x + 2)} = \frac{7x + (2x - 4)}{(x - 2)(x + 2)} \][/tex]
- Combine the terms in the numerator:
[tex]\[ \frac{7x + 2x - 4}{(x - 2)(x + 2)} = \frac{9x - 4}{(x - 2)(x + 2)} \][/tex]

4. Simplified Result:
- The final simplified expression is:
[tex]\[ \frac{9x - 4}{x^2 - 4} \][/tex]

Thus, the sum of [tex]\(\frac{7x}{x^2 - 4}\)[/tex] and [tex]\(\frac{2}{x + 2}\)[/tex] is:
[tex]\[ \boxed{\frac{9x - 4}{x^2 - 4}} \][/tex]