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?

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- Aug 23rd 2010, 04:53 AMStuck ManStraight 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?

- Aug 23rd 2010, 05:17 AMUnknown008
$\displaystyle px^2 + qy = x$

Divide by x;

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

Move the qy/x around;

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

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

This gives:

$\displaystyle pY = 1 - qX$

You can put it in the form Y = mX + c now? (Happy) - Aug 23rd 2010, 05:29 AMStuck 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. - Aug 23rd 2010, 05:39 AMUnknown008
- Aug 23rd 2010, 05:43 AMStuck Man
Thats what I was thinking of. Thanks.