1. ## Straight line graph

I have done i. With ii I cannot get an equation that does not include x as a part of the coefficient of x. The answer in the book is px=1-q(y/x). Compared to y=mx+c what are the equivalents of m and c?

2. $px^2 + qy = x$

Divide by x;

$px + \frac{qy}{x} = 1$

Move the qy/x around;

$px = 1 - \frac{qy}{x}$

Here, you can take the Y as being x and X as being y/x.

This gives:

$pY = 1 - qX$

You can put it in the form Y = mX + c now?

3. Yes, I suppose I would plot x against y/x.

I don't think the books answer is quite what is asked for. The left side needs to be divided by p and the right side multiplied by p.

4. Originally Posted by Stuck Man
Yes, I suppose I would plot x against y/x.

I don't think the books answer is quite what is asked for. The left side needs to be divided by p and the right side multiplied by p.
Not quite... when you divide a side by p, you need to divide both sides by p.

$\frac{pY}{p} = \frac{1-qX}{p}$

$Y= \frac{1}{p}-\frac{q}{p}X$

So, m becomes -q/p and c becomes 1/p

5. Thats what I was thinking of. Thanks.