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

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Answer 1 · 2018-03-13T12:54:44

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#