# Solve for X - Rearrangement question

• May 7th 2010, 10:54 AM
preid
Solve for X - Rearrangement question
Hi

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

$
d(a + 9x) = y
$

$
da + 9dx = y
$

$
9dx = y - da
$

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

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

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

Same with this example:

$
f(a + x) = y
$

$
fa + fx = y
$

$
fx = y - fa
$

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

To this:

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

Thanks
• May 7th 2010, 11:05 AM
masters
Quote:

Originally Posted by preid
Hi

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

$
d(a + 9x) = y
$

$
da + 9dx = y
$

$
9dx = y - da
$

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

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

Separate into difference of two fractions:

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

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

Factor out 1/9

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

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

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

Same with this example:

$
f(a + x) = y
$

$
fa + fx = y
$

$
fx = y - fa
$

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

To this:

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

Thanks

..