How do you solve for R in # i= (N-P)/( Q-R)#?

3 Answers
May 13, 2016

Answer:

Use the rule #a/b = m/n -> b xx m = n xx a#

Explanation:

#i(Q - R) = N - P#

#Q - R = (N - P)/i#

#-R = (N - P)/i - Q#

#R = - (N - P)/i + Q#

Hopefully this helps!

May 13, 2016

Answer:

#R = (-N+P+iQ)/i#

Explanation:

The strategy is to get #R# out of the denominator using multiplication, then isolate all terms including #R#, then factor #R# out if necessary, and finally divide by the coefficient of #R#.

#i = (N-P)/(Q-R)#

Multiply each side by #Q-R#.

#=> i(Q-R) = (N-P)/(Q-R)(Q-R) = N-P#

Apply the distributive property to the left hand side.

#=> iQ - iR = N-P#

Subtract #iQ# from each side.

#=> iQ - iR - iQ = N-P-iQ#

#=> -iR = N-P-iQ#

Divide each side by #-i#

#=> (-iR)/(-i) = (N-P-iQ)/(-i)#

#=> R = -(N-P-iQ)/i = (-N+P+iQ)/i#

May 14, 2016

Answer:

#R = (P-N)/i + Q# or #R = Q - (N-P)/i#

Explanation:

Here is another method. The biggest problem is that #R# is in the denominator. However, there is only one term on each side of the equal sign., so we can simply invert the entire equation

#1/i = (Q - R)/(N - P)#

Multiply by #(N - P) rArr (N-P)/i = Q - R#

Now : EITHER..... Move #R# to the left and the whole of the term on the left to the right, remembering to change the signs.

#R = Q - (N-P)/i#

OR: Move the Q to the left and then multiply through by -1 to make
-R into +R

#(N-P)/i = Q - R#
#(N-P)/i - Q = - R#

#(P-N)/i + Q = R#