To solve for [tex]\( r \)[/tex] in terms of [tex]\( t \)[/tex] from the given equation:
[tex]\[
t = \frac{120r}{r + 120}
\][/tex]
we need to isolate [tex]\( r \)[/tex]. Here’s the step-by-step process to find the inverse:
1. Start with the given equation:
[tex]\[
t = \frac{120r}{r + 120}
\][/tex]
2. Multiply both sides by [tex]\((r + 120)\)[/tex] to eliminate the fraction:
[tex]\[
t(r + 120) = 120r
\][/tex]
3. Distribute [tex]\( t \)[/tex] on the left side:
[tex]\[
tr + 120t = 120r
\][/tex]
4. Move all terms involving [tex]\( r \)[/tex] to one side of the equation:
[tex]\[
tr - 120r = -120t
\][/tex]
5. Factor out [tex]\( r \)[/tex] on the left side:
[tex]\[
r(t - 120) = -120t
\][/tex]
6. Divide both sides by [tex]\((t - 120)\)[/tex] to solve for [tex]\( r \)[/tex]:
[tex]\[
r = \frac{-120t}{t - 120}
\][/tex]
Thus, the inverse of Janine’s equation, solved for [tex]\( r \)[/tex] in terms of [tex]\( t \)[/tex], is:
[tex]\[
r = \frac{-120t}{t - 120}
\][/tex]
This expression allows you to find the resistance of the second lightbulb when given the total resistance [tex]\( t \)[/tex] in the parallel circuit.
