How do you solve #6/2=4/p#?

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Answer 1 · 2017-10-08T10:33:11

# p = 1 1/3#

#6/2=4/p#

multiply all terms by #20p#

#6/2\times(20p)=4/p\times(20p)#

#6/cancel2^1\times(cancel20^10p)=4/cancelp\times(20cancelp)#

# 60p = 80#

Divide both sides by 60:

# (60p)/60 = 80/60#

# cancel((60) \ timesp)/cancel60 = cancel80^4/cancel60^3#

#p = 4/3 = 1 1/3#

# p = 1 1/3#