How do you solve $\frac{2 x + 3}{x^{2} - 9} + \frac{x}{x - 3}$ $(2x+3)/(x^2-9) + x/(x-3)$ ?

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Answer 1 · 2018-04-03T16:03:16

$\frac{x^{2} + 5 x + 3}{x^{2} - 9}$ $(x^2+5x+3)/(x^2-9)$

Add them like fractions (make sure their denominators are the same)

The two denominators are $x^{2} - 9$ $x^2-9$ are $x - 3$ $x-3$ . Find the least common denominator.

$x^{2} - 9 = (x - 3) (x + 3) \to$ $x^2-9=(x-3)(x+3) rarr$ One of the denominators is a factor of the other

Multiply the second fraction's numerator and denominator by $x + 3$ $x+3$ to get the same denominator as the first fraction

$\frac{x}{x - 3} \cdot \frac{x + 3}{x + 3}$ $x/(x-3)*(x+3)/(x+3)$

$\frac{x^{2} + 3 x}{x^{2} - 9}$ $(x^2+3x)/(x^2-9)$

Now add the fractions together

$\frac{2 x + 3 + x^{2} + 3 x}{x^{2} - 9}$ $(2x+3+x^2+3x)/(x^2-9)$

Combine like terms:

$\frac{x^{2} + 5 x + 3}{x^{2} - 9}$ $(x^2+5x+3)/(x^2-9)$

How do you solve (2x+3)/(x^2-9) + x/(x-3)2x+3x2−9+xx−3(2x+3)/(x^2-9) + x/(x-3)?