We can say that a relationship between two variables is a function if, for every input, we get one and only one unique output. If we want to talk about y as a function of x in this problem, we want to make sure that every x has only one y associated with it. We can do this by solving our equation for x and seeing whether we can have more than one y that satisfies that equation. Here, I'll do the algebra and briefly explain my steps:
Multiply either side by x to cancel the denominator in 1/x:
![x((1)/(x)+3y)=x(-5)\\1+3xy=-5x](https://img.qammunity.org/2019/formulas/mathematics/middle-school/99yom2h8jkncj7fddns39rko2ls3dlobff.png)
Add 5x to either side to collect all of the x terms on one side, and subtract 1 to get all of the constants on one side:
![(1+3xy)+5x-1=-5x+5x-1\\3xy+5x=-1](https://img.qammunity.org/2019/formulas/mathematics/middle-school/67o2fx82xeafzt65t14vnakvzw6iqcrjc7.png)
Factor an x out of the left side of the equation:
![x(3y+5)=-1](https://img.qammunity.org/2019/formulas/mathematics/middle-school/5m9vena7zovwd70pc2t93iimgquof8wpln.png)
Divide either side by 3y + 5 to get the x by itself:
![(x(3y+5))/(3y+5)=(-1)/(3y+5) \\\\x=(-1)/(3y+5)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/c2fhmy9vfx8ra1js6pios2kqrj770w2vw7.png)
For every x we pick, only one y can satisfy the above equation, so we can say that yes,
is a function.