Question

In the given problem, solve and graph.

Express answers in both inequality and interval notation.

$\frac{3x+1}{x+4}\le 1$

Answer 1

$\frac{3x+1}{x+4}\le 1$

$\frac{3x+1}{x+4}-1\le 0$

$\frac{2x-3}{x+4}\le 0$

	x<-4	x=-4	-4<x<$\frac{3}{2}$	x=$\frac{3}{2}$	x>$\frac{3}{2}$
2x-3	-	-	-	0	+
x+4	-	0	+	+	+
$\frac{2x-3}{x+4}$	+	undefined	-	0	+

$\left[\begin{array}{cc}Solution:& -4<x\le \frac{3}{2}\\ Decimal:& -4<x\le 1.5\\ IntervalNotation:& (-4,\frac{3}{2}]\end{array}\right]$