How do you solve for #r# in the equation #9k + ar = r-2y#?

1 Answer
Ratnaker Mehta
Jun 13, 2016

#:. r=(9k+2y)/(1-a),#provided that #a!=1.#

Explanation:

Given that, #9k+ar=r-2y.# Transposing #9k# (from LHS to RHS) & #r# (from RHS to LHS),
#ar-r=-9k-2y.#
#:. r(a-1)=-(9k+2y).#
#:. r=-(9k+2y)/(a-1) = (9k+2y)/(1-a),#provided that #a!=1.#

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