Question

Question #00584

Answer 1

`r=E/I-R` or `r=(E-IR)/I`

Explanation:

We want to isolate the `r` to be by itself.

First let's get rid of the coefficient affecting `r`. We do this by dividing each side by `I`.

`E/I = (I(R+r))/I`
`E/I = R+r`

Now we want to get `r` by itself, so we need to move the `R` somehow. We do this by subtracting `R` from both sides.

`E/I-R = R+r-R`
`E/I-R=r`

That's our answer, but in order to get it in the form of answer c we have to multiply the `R` by `I/I` which is technically 1, hence why it works.

`E/I-R(I/I)=r`
`(E-IR)/I=r`

Hope that helps!

Answer 2

See a solution process below:

Explanation:

First, divide each side of the equation by `color(red)(I)` to eliminate the need for parenthesis while keeping the equation balanced:

`E/color(red)(I) = (I(R + r))/color(red)(I)`

`E/I = (color(red)(cancel(color(black)(I)))(R + r))/cancel(color(red)(I))`

`E/I = R + r`

Next, subtract `color(red)(R)` from each side of the equation to solve for `r` while keeping the equation balanced:

`E/I - color(red)(R) = R + r - color(red)(R)`

`E/I - R = R - color(red)(R) + r`

`E/I - R = 0 + r`

`E/I - R = r`

`r = E/I - R`

Answer `b` is not an option. The other three answers have the solution for `r` as a fraction over `R`. Therefore, we need to get the `R` term over a common denominator:

`r = E/I - (I/I xx R)`

`r = E/I - (IR)/I`

`r = (E - IR)/I`