Answer:
C. [tex]\frac{13^{2} + 4x - 2}{3x(x+2)}[/tex]
Explanation:
Since both fractions don't have the same denominator, we have to make them with the same denominator before we can actually do the addition.
So, here's the initial problem:
[tex]\frac{4x + 1}{x + 2} + \frac{x - 1}{3x}[/tex]
To make them with the same denominator, we'll multiply each side by 1, using the denominator of the other part... like this:
[tex]\frac{4x + 1}{x + 2} * \frac{3x}{3x} + \frac{x - 1}{3x} * \frac{x + 2}{x + 2}[/tex]
Since we multiply by 1 (expressed differently, but still just 1), the numbers aren't really changed.
Then we make the multiplications, first part:
[tex]\frac{4x + 1}{x + 2} * \frac{3x}{3x} = \frac{(12x^{2} + 3x) }{3x^{2} + 6}[/tex]
Now the second part:
[tex]\frac{x - 1}{3x} * \frac{x + 2}{x + 2} = \frac{x^{2} + x -2}{3x^{2} + 6}[/tex]
We now have both sides of the addition with the same denominator, and we can add the numerators.
[tex]\frac{(12x^{2} + 3x) }{3x^{2} + 6} + \frac{x^{2} + x -2}{3x^{2} + 6} = \frac{13x^{2} +4x -2}{3 (x^{2} + 2)}[/tex]
We also simplified the denominator at the end, by extracting the 3 which was common.
