How do you solve for l in #S= (pi r^2 )+ (pi r l)#?

1 Answer
Feb 26, 2017

Answer:

See the entire solution process below:

Explanation:

First, factor out a #pir# from each term in the right side of the equation:

#S = pir(r + l)#

Next, divide each side of the equation by #color(red)(pir)# to eliminate the coefficient while keeping the equation balanced:

#S/color(red)(pir) = (pir(r + l))/color(red)(pir)#

#S/(pir) = (color(red)(cancel(color(black)(pir)))(r + l))/cancel(color(red)(pir))#

#S/(pir) = r + l#

Now, subtract #color(red)(r)# from each side of the equation to solve for #l#:

#-color(red)(r) + S/(pir) = -color(red)(r) + r + l#

#-r + S/(pir) = 0 + l#

#-r + S/(pir) = l#

#l = -r + S/(pir)#

If you want the right side of the equation to be over a common denominator you can multiply #-r# by #(pir)/(pir)# giving:

#l = ((pir)/(pir) xx -r) + S/(pir)#

#l = (-pir^2)/(pir) + S/(pir)#

#l = (-pir^2 + S)/(pir)#

#l = (S - pir^2)/(pir)#

The solution is: #l = -r + S/(pir)# or #l = (S - pir^2)/(pir)#