Solving the inequality. ${3x + 5 \over x} \gt 0$ I have a question similar to this:
$${3x + 5 \over x} \gt 0$$
I am not sure why but the 0 is throwing me off as I want to $0 \over x$ to balance it and Ill just get 0 again and just seems wrong.
 A: Here are two different keys that points to two different methods.
Hint 1: $$\dfrac{3x+5}{x}\gt0 \iff \dfrac{3x}x+\dfrac5x\gt0\iff \dfrac5x+3\gt0.$$
Hint 2: Study where the polynomials $x$ and $3x+5$ are both strictly positive and both strictly negative, since only then, their ratio would have been positive.
A: Hint: If $uv > 0$, either $u > 0$ AND $v > 0$ or $u < 0 $ AND $v < 0$. In this case, $u = 3x+5$ and $v = 1/x$
A: In this instance it would be safe to multiply both sides of the inequality by $x^2$ (its okay to do this because $x\neq0$ and because $x^2$ is positive so it does not change the direction of the inequality)
This gives you 
$x(x+5)>0$ drawing the sketch of $y=x(x+5)$ and using the roots we can see that $y>0$ when 
$x>0$ or $x<-5$ which are the required solutions
A: My teacher always said to make a table that include all multiplications(divisons) and check the sign.You add zeroes of the polynomials $$3x+5=0\\x=-\frac{5}{3} \land x=0$$
Now you make a table with columns marked by intervals from ($-\infty,-\frac{5}{3})$ then from ($\frac{-5}{3},0$) then from ($0,+\infty$),you have 3 rows,first one is $3x+5$,second one is $x$ and third $(\frac{3x+5}{x})$,now you fill the table with $+$ and $-$,if the expression(polynomial) is positive in that interval you add a $+$ sign and if it's negative you add $-$,here $3x+5>0$ for $x>-\frac{5}{3}$ you fill the row with $-$,$+$,$+$ and $x>0$ you fill the row with $-$,$-$,$+$ now to get the sign of the fraction you just multiply the signs of the column and you get,$+$,$-$,$+$
So the solution is $x\in (-\infty,-\frac{5}{3})\cup(0,+\infty)$
