How do you solve the equation and identify any extraneous solutions for #5/(3c)=2/3#?

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Answer 1 · 2015-07-12T03:19:26

#c=5/2#
#c!=0#

#5/(3c)=2/3#

#c!=0#

Multiply both sides by #3c#.

#5=2/3*3c#

Cancel the #3#.

#5=2/cancel 3*cancel3c#

#5=2c#

Divide both sides by #2#.

#5/2=c#

Switch sides.

#c=5/2#