Why Did My Cubic Equation Plot as a Curve Instead of a Line?

[Kyoma]

Okay, I was doing some questions concerning linear law when I came across a question that asked me to convert a cubic equation into a 'linear equation' so that I can draw it on a graph.

Cubic Equation:

p(x+y-q) = qx³

Then, I convert the above equation into:

y/x² = (q/p)x

However, when I tried plotting the graph. Letting y/x² to be the y-axis and x to be the x-axis, I got a curve, instead of a line!

What's wrong?

 

[nonequilibrium]

But then your y-axis is dependent on your x-axis, that's VERY odd.

The procedure you have to go through is probably more like the following:

Say y = e^x and we want to represent is linearly, what do we do? Well we know that to get x downstairs, we can use the ln function: ln(y) = x

Now we draw a regular y' = x and on the y'-axis we write y' = ln(y)

So it basically comes down to: changing your variables! I've just done an example where I replaced y, but I think I'd replace x in your example, so that you have to write on the x'-axis: x' = f(x); what is f?

 

[Kyoma]

?

The answer given to me is:

x+y = (q/p)x³ + q

Letting x+y to be the y-axis and x³ to be the x-axis, I indeed got a straight line graph...