Help with Fundamental Algebra Rule

[JvN]

For the question:
Make 'R' the subject of the formula:

1/R = 1/Ri + 1/Rii

I can find the answer through the 'long version' of multiplying in the common denominator to both sides as eventually isolating R. But I know that there is a shorter version to this answer based on some fundamental algebra rule.

I don't understand what that rule is or how it applies to shorten the length of solving the answer for this question.
Please help.

 

[Otis]

… Make 'R' the subject of the [equation]:

1/R = 1/Ri + 1/Rii …

Hello JvN. Do symbols R and i represent Real numbers?

Does the given equation look like one of the following?

\(\displaystyle \frac{1}{R} \;=\; \frac{1}{R \cdot i} \;+\; \frac{1}{R \cdot i^2}\)

\(\displaystyle \frac{1}{R} \;=\; \frac{1}{R} \cdot i \;+\; \frac{1}{R} \cdot i^2\)

I can find the answer through the 'long version' …

What answer did you get?

?

 

[JvN]

Hi Otis
Thanks for your reply

'i' and 'ii' are just used to differentiate each term from the other. The equation could just as easily be: 1/x = 1/y + 1/z

The 'long version' answer I got was: R = (Ri.Rii)/(Ri+Rii)
But this took several lines of work after multiplying in the LCD, rearranging the equation and cancelling like terms.

I did ask a math teacher and he showed me the shortened version, it takes two lines. Unfortunately, he only speaks Japanese, so I was unable to work out what the rule is for moving variables to the top of a fraction without the drawn out process of using the LCD.

 

[hoosie]

1/a = 1/b + 1/c
To find a in terms of b and c first try working with numbers:
Consider the numbers 2,3 and 6:
1/2 = 1/3 + 1/6 where a = 2, b = 3, c = 6
= 6/(3 × 6) + 3/(3 × 6)
= (6 + 3)/(3 × 6)

∴(3 × 6)/(3 + 6) = 2

2 = the product of 3 and 6 divided by the sum of 3 and 6.
∴a = the product of b and c divided by the sum of b and c.
a = bc/(b + c)

Try numbers 3,4 and 12:

3 = (4 × 12)/(4 + 12)

∴1/3 = 1/4 + 1/12

 

[Romsek]

assuming that the i's in question are not the i representing the square root of -1
there's really no magic trick that makes things easier.

you just have to do

\(\displaystyle \dfrac{1}{R} = \dfrac{R_{ii} + R_i}{R_{i}R_{ii}}\)

\(\displaystyle R = \dfrac{R_{i}R_{ii}}{R_{ii}+R_i}\)

it's only two steps

 

[lookagain]

Romsek,

your first line has a typo. The plus sign needs to be an equals sign (corrected) as I show it here:


\(\displaystyle \dfrac{1}{R} =\dfrac{R_{ii} + R_i}{R_{i}R_{ii}}\)

\(\displaystyle R = \dfrac{R_{i}R_{ii}}{R_{ii}+R_i}\)

 

[Romsek]

your first line has a typo. The plus sign needs to be an equals sign (corrected) as I show it here:

ok thank you