How do you graph \frac { 3y - 5x } { 2} = \frac { y x } { 2} + 43y5x2=yx2+4?

1 Answer
Aug 10, 2017

First, separate the xx's and yy's. You want yy as a function of xx.

Explanation:

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

Multiply both sides of the initial eq. by 22:

3y - 5x = yx + 83y5x=yx+8

Subtract yxyx from both sides:

3y - 5x - yx = 83y5xyx=8

Add 5x5x to both sides:

3y - yx = 8 + 5x3yyx=8+5x

Now, on the left side, factor out yy:

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

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

y = (8+5x)/(3 - x)y=8+5x3x

So now you have yy expressed as a function of xx.

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

y = 8/3y=83

and then, note that when x = 3x=3, the denominator is zero. You can't divide by zero, so you can't graph this point. But imagine when xx is just a TINY bit less than 33. Then 3 - x3x 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 xx approaches 33 from the left.

Now, imagine xx is just a tiny bit greater than 33. Now 3 - x3x is a tiny, tiny NEGATIVE number. So you then know that the graph climbs up from NEGATIVE infinity as xx proceeds from points to the right of x = 3x=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 yy for points where x = 1, 2, 4, 5x=1,2,4,5, and maybe several more. and also -1, -2, -31,2,3, 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))