How do you solve for #r# in #r-s-12=P#?

1 Answer
Mar 18, 2017

See the entire solution process below:

Explanation:

Add #color(red)(s)# and #color(blue)(12)# to each side of the equation to solve for #r# while keeping the equation balanced:

#r - s - 12 + color(red)(s) + color(blue)(12) = P + color(red)(s) + color(blue)(12)#

#r - s + color(red)(s) - 12 + color(blue)(12) = P + s + 12#

#r - 0 - 0 = P + s + 12#

#r = P + s + 12#