How do you solve #(e-2)/2=(e+2)/3#?

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Answer 1 · 2016-08-19T11:40:36

#e=10#

#(e-2)/2=(e+2)/3#

or

#3(e-2)=2(e+2)#

or

#3e-6=2e+4#

or

#3e-2e=4+6#

or

#e=10#

Answer 2 · 2016-08-19T14:24:51

#e=10#.

We rewrite the eqn. as #(e+2)/(e-2)=3/2#, and, use a Result called

Compodendo Dividendo # : a/b=c/d iff (a+b)/(a-b)=(c+d)/(c-d)#

utilising this in our eqn. we get,

#((e+2)+(e-2))/((e+2)-(e-2))=(3+2)/(3-2)#

#(2e)/4=e/2=5 rArr e=10#, as obtained by Deepak G. !