How do you graph #\frac { 3y - 5x } { 2} = \frac { y x } { 2} + 4#?

...there's more than one way to do this, here's how I did it.

Multiply both sides of the initial eq. by #2#:

#3y - 5x = yx + 8#

Subtract #yx# from both sides:

#3y - 5x - yx = 8#

Add #5x# to both sides:

#3y - yx = 8 + 5x#

Now, on the left side, factor out #y#:

#y(3 - x) = 8 + 5x#

Divide both sides by #(3 - x)#:

#y = (8+5x)/(3 - x)#

So now you have #y# expressed as a function of #x#.

As to graphing it, first find #y# when #x = 0#:

#y = 8/3#

and then, note that when #x = 3#, the denominator is zero. You can't divide by zero, so you can't graph this point. But imagine when #x# is just a TINY bit less than #3#. Then #3 - x# is very close to zero.

A number divided by a tiny, tiny number is very large. So you know that the graph runs off to positive infinity as #x# approaches #3# from the left.

Now, imagine #x# is just a tiny bit greater than #3#. Now #3 - x# is a tiny, tiny NEGATIVE number. So you then know that the graph climbs up from NEGATIVE infinity as #x# proceeds from points to the right of #x = 3#.

Now: this is algebra, not calculus. Calculus gives you some tools to better graph this function, but I can't use them here. So then calculate #y# for points where #x = 1, 2, 4, 5#, and maybe several more. and also #-1, -2, -3#, etc.

And then connect the dots!

You should get a form of a hyperbola. You can check the result:

https://www.desmos.com/calculator

(paste in (8+5x)/(3-x))