How do you solve #\frac { 6} { ( c - 9) } + \frac { 6} { c } = 1#?

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Answer 1 · 2018-03-31T03:38:36

Solution: #c=3, c=18#

#6/(c-9)+6/c=1# Multiplying by #c(c-9)# on both sides we get,

#6c+6(c-9)=c(c-9) or 6c+6c-54=c^2-9c# or

#c^2-9c-12c+54=0 or c^2-21c+54=0 # or

#c^2-18c-3c+54=0 or c(c-18)-3(c-18)=0 # or

#(c-18)(c-3)=0 :.# either #c-18=0 :. c=18 # or

#c-3=0 :. c=3 #

Solution: #c=3, c=18# [Ans]