Answer
$(-∞, 0)$
Work Step by Step
$(x^2+4)/3x \leq1$
Numerator is zero when $x^2+4=0$.
The numerator is never zero. Otherwise, we would have $x^2=-4$, and we can't take the square root of a negative number.
Denominator is zero when $3x=0$.
$3x=0$
$3x/3=0/3$
$x=0$
Two regions to test: $(-∞, 0)$, $(0, ∞)$
We pick one value of $x$ to test in each region. Let $x=-1$ and $x=1$
$x=-1$
$(x^2+4)/3x \leq1$
$((-1)^2+4)/3*(-1) \leq1$
$(1+4)/-3 \leq1$
$5/-3 \leq1$
$-5/3 \leq 1$ (true, so this is part of the solution set)
$x=1$
$(x^2+4)/3x \leq1$
$((1)^2+4)/3*(1) \leq1$
$(1+4)/3 \leq1$
$5/3 \leq1$ (false, so this is not part of the solution set)
Thus, the solution set is $(-∞, 0)$.
