How do you solve #5/(3c)=2/3# and find any extraneous solutions?

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Answer 1 · 2017-01-29T15:07:09

#c=5/2#

#c=0# is an extraneous solution

Extraneous solution is such that we are not able to divide by 0.

So #3c !=0=> c=0# is the extraneous solution.

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#color(blue)("First principle method")#

#color(green)(5/(3c)=2/3color(red)(xx1))#

#color(green)(5/(3c)=2/3color(red)(xxc/c))#

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

As the denominators (bottom values) are both the same we only need compare the numerators (top values).

Thus #5=2c#

#=>c=5/2#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Short cut method")#

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

Cross multiply

#5xx3=2xx3c#

#15=6c#

#c=15/6 = (15-:3)/(6-:3)=5/2#