How do you solve for #r# in #S=L(1-r)#?

Question

Hazel Anglin · Accepted Answer

See a solution process below:

First, divide each side of the equation by #color(red)(L)# to eliminate the need for parenthesis while keeping the equation balanced: #S/color(red)(L) = (L(1 - r))/color(red)(L)# #S/L = (color(red)(cancel(color(black)(L)))(1 - r))/cancel(color(red)(L))# #S/L = 1 - r# Next subtract #color(red)(S/L)# and add #color(blue)(r)# to each side of the equation to solve for #r# while keeping the equation balanced: #S/L - color(red)(S/L) + color(blue)(r) = 1 - color(red)(S/L) - r + color(blue)(r)# #0 + r = 1 - S/L - 0# #r = 1 - S/L# or #r = L/L - S/L# #r = (L - S)/L#

Hazel Anglin · Answer

undefined To solve for $ r $ in the equation $ S = L(1 - r) $, you can follow these steps:

1. Divide both sides of the equation by $ L $ to isolate the term $ 1 - r $.
2. Then, subtract $ 1 $ from both sides to isolate $ -r $.
3. Finally, multiply both sides by $ -1 $ to solve for $ r $.

The solution for $ r $ will be:

$$ r = 1 - \frac{S}{L} $$