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how i can get the "r"

 May 31, 2017

Best Answer 

 #1
avatar+9465 
+3

\(\frac{E}{e}=\frac{R+r}{R-r}\)

Multiply both sides of the equation by  (R - r)  .

\((R-r)\,*\,\frac{E}{e}=R+r\)

Distribute the   \( \frac{E}{e} \)  .

\(R*\frac{E}{e}\,-\,r*\frac{E}{e}=R+r\)

Add   \( r*\frac{E}{e} \)   to both sides of the equation.

\(R*\frac{E}{e}=R+r+r*\frac{E}{e}\)

Subtract R from both sides of the equation.

\(R*\frac{E}{e}-R=r+r*\frac{E}{e}\)

Factor out an   r   on the right side.

\(R*\frac{E}{e}-R=r(1+\frac{E}{e})\)

Divide both sides of the equation by   \( (1+\frac{E}{e}) \)  .

\(\frac{R*\frac{E}{e}-R}{1+\frac{E}{e}}=r\)

 May 31, 2017
 #1
avatar+9465 
+3
Best Answer

\(\frac{E}{e}=\frac{R+r}{R-r}\)

Multiply both sides of the equation by  (R - r)  .

\((R-r)\,*\,\frac{E}{e}=R+r\)

Distribute the   \( \frac{E}{e} \)  .

\(R*\frac{E}{e}\,-\,r*\frac{E}{e}=R+r\)

Add   \( r*\frac{E}{e} \)   to both sides of the equation.

\(R*\frac{E}{e}=R+r+r*\frac{E}{e}\)

Subtract R from both sides of the equation.

\(R*\frac{E}{e}-R=r+r*\frac{E}{e}\)

Factor out an   r   on the right side.

\(R*\frac{E}{e}-R=r(1+\frac{E}{e})\)

Divide both sides of the equation by   \( (1+\frac{E}{e}) \)  .

\(\frac{R*\frac{E}{e}-R}{1+\frac{E}{e}}=r\)

hectictar May 31, 2017
 #2
avatar+118608 
0

Good work Hectictar :)

 

There is no way I could read this question, it was just too small.

 Jun 1, 2017
 #3
avatar
0

Solve for r:
E/e = (r + R)/(R - r)

E/e = (r + R)/(R - r) is equivalent to (r + R)/(R - r) = E/e:
(r + R)/(R - r) = E/e

Cross multiply:
e (r + R) = E (R - r)

Expand out terms of the left hand side:
e r + e R = E (R - r)

Expand out terms of the right hand side:
e r + e R = E R - E r

Subtract e R - E r from both sides:
r (E + e) = E R - e R

Divide both sides by E + e:
Answer: | r = (E R - e R)/(E + e)=R[E - e] / [E + e]

 Jun 1, 2017

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