How do you solve #n/(n-12)=9/5#?

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Answer 1 · 2017-02-16T12:37:12

#n = 27#

#n/(n-12) = 9/5#

In a case like this where there is #"one fraction" = "one fraction"#,

the quickest method is to cross-multiply which will get rid of the fractions.

#9xx(n-12) = 5xx n#

#9n -108 = 5n#

#9n -5n = 108#

#4n = 108#

#n = 27#