Answer
The inequality is valid for values less than 0 and values between 3 and 4 (not including them) i.e. $(-\infty,0), (3,4)$.
Work Step by Step
First, we are going to move everything to the left side and simplify:
$x+\dfrac{12}{x}<7$
$x+\dfrac{12}{x}-7<0$
$(x+\dfrac{12}{x}-7)\cdot x<0 \cdot x$
$x^2+12-7x<0$
Since we eliminated x and there is the restriction $x\ne0$, we'll take it into account as a critical point.
$x^2-7x+12<0$
Now, we find critical points by factoring and equating to zero:
$x^2-7x+12=0$
$(x-3)(x-4)=0$
There are three critical points:
$x_1=0$
$x_2-3=0\rightarrow x_2=3$
$x_3-4=0\rightarrow x_3=4$
Next, we are going to take four values: one less than 0; one between 0 and 3; one between 3 and 4; and one more than 4 to test in the original equation and check if the inequality is true or not:
First test with a value less than 0:
$-1+\dfrac{12}{-1}<7$
$-1-12<7$
$-13<7 \rightarrow \text{ TRUE}$
Second test with a value between 0 and 3:
$1+\dfrac{12}{1}<7$
$1+12<7$
$13<7 \rightarrow \text{ FALSE}$
Third test with a value between 3 and 4:
$3.2+\dfrac{12}{3.2}<7$
$3.2+3.75<7$
$6.95<7 \rightarrow \text{ TRUE}$
Fourth test with a value more than 4:
$6+\dfrac{12}{6}<7$
$6+2<7$
$8<7 \rightarrow \text{ FALSE}$
These tests show that the inequality $x+\dfrac{12}{x}<7$ is valid for values less than 0 and values between 3 and 4 (not including them) i.e. $(-\infty,0)\cap (3,4)$.
