# Solve for X - Rearrangement question

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• May 7th 2010, 09:54 AM
preid
Solve for X - Rearrangement question
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:

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

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
• May 7th 2010, 10:05 AM
masters
Quote:

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

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