If #(2x-3y)/(3x-2y)=2/5#, what is #x/y#?

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Answer 1 · 2016-08-12T04:52:49

#11/4#.

#(2x-3y)/(3x-2y)=2/5#

Cross-multiplication gives, #10x-15y=6x-4y#

#rArr 4x=11y#

#rArrx/y=11/4#.

Another way to solve the problem, is, that, to suppose #m# for

the reqd. value, i.e., #m=x/y#. Then, we have, #x=my#.

We submit this #x# in the given eqn., to get,

# {2(my)-3y}/{3(my)-2y}=2/5#

#:. (cancely(2m-3))/(cancely(3m-2))=2/5#

#:. 5(2m-3)=2(3m-2)#.

#:. 10m-15=6m-4#

#:. 10m-6m= **4m=** 15-4= **11**#.

#:.# Reqd. Value #m=11/4#, as before!

Enjoy maths.!