Question

In the given problem, solve and graph. Express the answer in both inequality and interval notation. $\frac{3}{x-3}\le \frac{2}{x+2}$

Answer 1

$\frac{3}{x-3}\le \frac{2}{x+2}$

$\frac{3}{x-3}-\frac{2}{x+2}\le 0$

$\frac{x+12}{(x-3)(x+2)}\le 0$

	x<-12	x=- 1	- 12<x<-2	x=-2	-2<x<3	x=3	x>3
x+12	-	0	+	+	+	+	+
x-3	-	-	-	-	-	0	+
x+2	-	-	-	0	+	+	+
$\frac{x+12}{(x-3)(x+2)}$	-	0	+	undefined	-	undefined	+

$\left[\begin{array}{cc}Solution:& x\le -12or-2<x<3\\ IntervalNotation:& (-\infty ,-12]\cup (-2,3)\end{array}\right]$