Question 1122916

let's numbers be {{{x}}} and {{{y}}}

given: 
the quotient of two numbers is {{{4}}} which meas

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


 and there difference is {{{39}}} which meas

{{{x-y=39}}}...eq.2

start with eq.1 and solve it for {{{x}}}:

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

{{{x=4y}}}...eq.1a

go to eq.2, substitute {{{x}}} from eq.1a

{{{4y-y=39}}}

{{{3y=39}}}

{{{y=39/3}}}

{{{y=13}}}

go back to eq.1a, find {{{x}}}

{{{x=4*13}}}

{{{x=52}}}

so, your numbers are: {{{13}}} (the smaller number) and {{{52}}}