Question

Solve the following inequality: $\frac{x+7}{x-3}>0$

Answer 1

	x<-7	x=-7	-7<x<3	x=3	x>3
x+7	-	0	+	+	+
x-3	-	-	-	0	+
$\frac{x+7}{x-3}$	+	0	-	undefined	+

$\frac{x+7}{x-3}>0$

$x<-7,x>3$

Interval notation $(-\infty ,-7)\cup (3,\infty )$