For what x is 3/(x+3)>3/(x-2)?

Oct 22, 2015

$x \in \left(- 3 , 2\right)$

Explanation:

First divide both sides by $3$ to get:

$\frac{1}{x + 3} > \frac{1}{x - 2}$

We can ignore the values $x = - 3$ and $x = 2$ since one or the other side of the inequality is undefined for these values. That leaves $\left(- \infty , - 3\right) \cup \left(- 3 , 2\right) \cup \left(2 , \infty\right)$

It is probably easiest to split this into cases as follows:

Case $\boldsymbol{x \in \left(- \infty , - 3\right)}$

$\left(x + 3\right) < 0$ and $\left(x - 2\right) < 0$ so $\left(x + 3\right) \left(x - 2\right) > 0$

Multiply both sides of the inequality by $\left(x + 3\right) \left(x - 2\right)$ to get:

$x - 2 > x + 3$

Subtract $x$ from both sides to get $- 2 > 3$, which is false.
So there is no solution for $x \in \left(- \infty , - 3\right)$

Case $\boldsymbol{x \in \left(- 3 , 2\right)}$

$\left(x + 3\right) > 0$ and $\left(x - 2\right) < 0$ so $\left(x + 3\right) \left(x - 2\right) < 0$

Multiply both sides of the inequality by $\left(x + 3\right) \left(x - 2\right)$ and reverse the inequality (since we're multiplying by a negative value) to get:

$x - 2 < x + 3$

Subtract $x$ from both sides to get $- 2 < 3$, which is true.
So the inequality is true for all $x \in \left(- 3 , 2\right)$

Case $\boldsymbol{x \in \left(2 , \infty\right)}$

$\left(x + 3\right) > 0$ and $\left(x - 2\right) > 0$ so $\left(x + 3\right) \left(x - 2\right) > 0$

Multiply both sides of the inequality by $\left(x + 3\right) \left(x - 2\right)$ to get:

$x - 2 > x + 3$

Subtract $x$ from both sides to get $- 2 > 3$, which is false.
So there is no solution with $x \in \left(2 , \infty\right)$