Question 1008688

let numbers be {{{x}}} and {{{y}}}
two number have the ratio {{{x:y=3:4}}} or 

{{{x/y=3/4}}}....solve for {{{x}}}

{{{x=3y/4}}}.........eq.1

if {{{5}}} is subtracted from each of the number the difference have ratio of {{{1:3}}}, we have

{{{(x-5)/(y-5)=1/3}}}.........substitute {{{x}}} from eq.1

{{{(3y/4-5)/(y-5)=1/3}}}...solve for {{{y}}}, cross multiply first

{{{3(3y/4-5)=1*(y-5)}}}

{{{9y/4-15=y-5}}}

{{{9y/4-y=15-5}}}

{{{9y/4-4y/4=10}}}

{{{5y/4=10}}}

{{{5y=40}}}

{{{y=40/5}}}

{{{highlight(y=8)}}}

go back to {{{x=3y/4}}}.........eq.1 substitute in {{{8}}} for {{{y}}}

{{{x=3*8/4}}}

{{{x=3*cross(8)2/cross(4)}}}

{{{x=3*2}}}

{{{highlight(x=6)}}}