+0  
 
0
432
2
avatar

Um, hi. Small problem here. Make U the subject in:S= U*V / U + V (this being a fraction). Thanks.

 Mar 4, 2015

Best Answer 

 #2
avatar+21244 
+5

Make U the subject in:S= U*V / U + V (this being a fraction).

$$\small{\text{$
S= \dfrac{ U\cdot V } { U + V }
$}}\\\\
\small{\text{$
\dfrac{1}{S}= \dfrac{ U + V } { U\cdot V }
$}}\\\\
\small{\text{$
\dfrac{1}{S}= \dfrac{ U } { U\cdot V } + \dfrac{ V } { U\cdot V }
$}}\\\\
\boxed{\small{\text{$
\dfrac{1}{S}= \dfrac{ 1 } { V } + \dfrac{ 1 } { U }
$}}}\\\\
\small{\text{$
\dfrac{1}{U}= \dfrac{ 1 } { S } - \dfrac{ 1 } { V }
$}}\\\\
\small{\text{$
\dfrac{1}{U}= \dfrac{ V-S } { V\cdot S }
$}}\\\\
\boxed{
\small{\text{$
U= \dfrac{ V\cdot S } { V-S }
$}}}$$

.
 Mar 5, 2015
 #1
avatar+27475 
+5

$$S=\frac{U\times V}{U+V}$$

 

Multiply both sides by U + V:

 

$$S\times (U+V)=U\times V$$

 

Expand the left-hand side:

 

$$S\times U + S\times V = U\times V$$

 

Subtract U*V and S*V from both sides:

 

$$S\times U-U\times V=-S\times V$$

 

Factor out U on the left-hand side:

 

$$U\times (S-V)=-S\times V$$

 

Divide both sides by S - V:

 

$$U=-\frac{S\times V}{S-V}$$

.

 Mar 4, 2015
 #2
avatar+21244 
+5
Best Answer

Make U the subject in:S= U*V / U + V (this being a fraction).

$$\small{\text{$
S= \dfrac{ U\cdot V } { U + V }
$}}\\\\
\small{\text{$
\dfrac{1}{S}= \dfrac{ U + V } { U\cdot V }
$}}\\\\
\small{\text{$
\dfrac{1}{S}= \dfrac{ U } { U\cdot V } + \dfrac{ V } { U\cdot V }
$}}\\\\
\boxed{\small{\text{$
\dfrac{1}{S}= \dfrac{ 1 } { V } + \dfrac{ 1 } { U }
$}}}\\\\
\small{\text{$
\dfrac{1}{U}= \dfrac{ 1 } { S } - \dfrac{ 1 } { V }
$}}\\\\
\small{\text{$
\dfrac{1}{U}= \dfrac{ V-S } { V\cdot S }
$}}\\\\
\boxed{
\small{\text{$
U= \dfrac{ V\cdot S } { V-S }
$}}}$$

heureka Mar 5, 2015

39 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.