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\[Solve P = 1/6(r + 5t) for r.
You need to isolate for r be applying inverse operations (reverse PEDMAS) to both sides -multiply by 6 - subtract 5t
but is 1/6 not just 6
multiply by 6 can get rid of the 1/6 1/6 x 6 = 1
Yep. So think of that as the stuff inside the brackets being divided by 6. This is because dividing by 6 and multiplying by 1/6 is the same thing. In fact this is probably how you learned to do fraction division. Either way, when you multiply 1/6 by 6/1 you get 6/6 which is just 1.
Should P = 1/6(r + 5t) be P = 1/ [6(r + 5t)]? Grouping symbols are helpful and sometimes critical to a problem solution.
if it had of been multiplied by 5/7, we would have multiplied it by 7/5 (its inverse), which results in 1
ok so im confused on what to do. can someone show me step by step ?
\[P=1/6(r+5t)\]multiply by 6 \[6P=(r+5t)\] Subtract 5t \[6P-5t=r+5t-5t\] \[6P-5t=r\]
iif you multiply by 6 wouldnt there be a 35t?
P = (1/6) (r + 5t) 6 times P = 6 times (1/6) times (r + 5t) 6P = [6 times (1/6]) times (r + 5t) 6P = one times (r + 5t) 6P = r + 5t You are thinking about the Distributive Property which does not apply in this solution. Multiply one term of the right side by 6; otherwise, you are multiplying it by 36.