Question 826767
let numbers be {{{x}}} (larger) and {{{y}}} (smaller)

if the difference between two numbers is {{{28}}}, we have

{{{x-y=28}}}....eq.1

if the larger number is {{{8}}} less than twice the smaller number, we have

{{{x=2y-8}}}....eq.2

now solve this system:


{{{x-y=28}}}....eq.1
{{{x=2y-8}}}....eq.2
_______________________

{{{x-y=28}}}....eq.1...substitute {{{x}}} from eq.2

{{{2y-8-y=28}}}....solve for {{{y}}}

{{{2y-y=28+8}}}

{{{highlight(y=36)}}}

go back to eq.2 and find {{{x}}}

{{{x=2y-8}}}....eq.2..............plug in {{{y}}}

{{{x=2*36-8}}}

{{{x=72-8}}}

{{{highlight(x=64)}}}

so, your numbers are {{{64}}} and {{{36}}}