Originally Posted by

**preid** Hi

Can someone explain the thought process behind the rearrangement at the end for these two examples?

$\displaystyle

d(a + 9x) = y

$

$\displaystyle

da + 9dx = y

$

$\displaystyle

9dx = y - da

$

$\displaystyle

x = \frac{y-da}{9d}

$

I can't see how you get from that to:

Separate into difference of two fractions:

$\displaystyle {\color{red}x=\frac{y}{9d}-\frac{da}{9d}}$

$\displaystyle {\color{red}x=\frac{1}{9}\left(\frac{y}{d}\right)-\frac{1}{9}\left(\frac{da}{d}\right)}$

Factor out 1/9

$\displaystyle {\color{red}\frac{1}{9}\left(\frac{y}{d}-\frac{da}{d}\right)}$

Cancel out the d in the last fraction and you have:

$\displaystyle

x = \frac{1}{9}(\frac{y}{d} - a)

$

Same with this example:

$\displaystyle

f(a + x) = y

$

$\displaystyle

fa + fx = y

$

$\displaystyle

fx = y - fa

$

$\displaystyle

x = \frac{y - fa}{f}

$

To this:

$\displaystyle

x = \frac{y}{f} - a

$

Thanks