Question

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

Answer 1

First, separate the `x`'s and `y`'s. You want `y` as a function of `x`.

Explanation:

...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))