# How do solve 3/(x-2)<=3/(x+3) algebraically?

Oct 5, 2016

$- 3 < x < 2$

#### Explanation:

Slower way:

• Bring everything to the right: $\frac{3}{x - 2} - \frac{3}{x + 3} \setminus \le q 0$

• Lowest common denominator: $\frac{3 \left(x + 3\right) - 3 \left(x - 2\right)}{\left(x - 2\right) \left(x + 3\right)} \setminus \le q 0$

• Expand the numerator: $\frac{3 x + 9 - 3 x + 6}{\left(x - 2\right) \left(x + 3\right)} \setminus \le q 0$

• $\frac{9 + 6}{\left(x - 2\right) \left(x + 3\right)} \setminus \le q 0$

• $\frac{15}{\left(x - 2\right) \left(x + 3\right)} \setminus \le q 0$

• Since $15$ is always positive, the sign of the fraction is decided by the sign of the denominator. $\left(x - 2\right) \left(x + 3\right)$ represents a parabola with zeros in $x = - 3$ and $x = 2$. Such a parabola is negative between its solutions, as you can see in the graph here:
graph{(x-3)(x+2) [-4.96, 6.146, -2.933, 2.614]}