Question 1152558

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

if one third the sum of two numbers is {{{50}}}, we have

{{{(1/3)(x+y)=50}}}......eq.1

and, if one fifth of their difference is {{{2}}}, we have

{{{(1/5)(x-y)=2}}}......eq.2

solve the system:

{{{(1/3)(x+y)=50}}}......eq.1

{{{(1/5)(x-y)=2}}}......eq.2
----------------------------

start with

{{{(1/3)(x+y)=50}}}......eq.1, both sides multiply by {{{3}}}

{{{x+y=150}}}......solve for {{{x}}}


{{{x=150-y}}}......eq.1a

go to

{{{(1/5)(x-y)=2}}}......eq.2, substitute {{{x}}}

{{{(1/5)(150-y-y)=2}}}.......solve for {{{y}}}

{{{(1/5)(150-2y)=2}}}...........both sides multiply by {{{5}}}

{{{150-2y=10}}}

{{{150-10=2y}}}

{{{140=2y}}}

{{{y=70}}}

go back to eq.1a

{{{x=150-y}}}

{{{x=150-70}}}

{{{x=80}}}


answer: the numbers are {{{70}}} and {{{80}}}