Question 1123787


let's the larger number be {{{x}}} and the smaller number {{{y}}}

if the difference of two positive numbers is {{{8}}}, we have

{{{x-y=8}}}...eq.1

if the larger number is {{{twenty-one}}} {{{less}}} than {{{twice}}} the smaller number, we have

{{{x+21=2y}}}........eq.2


 to find the numbers, solve this system:

{{{x-y=8}}}...eq.1
{{{x+21=2y}}}........eq.2
----------------------------substitute eq.2 from eq.1

{{{x-y-(x+21)=8-2y}}}
{{{x-y-x-21=8-2y}}}
{{{-y-21=8-2y}}}
{{{2y-y=8+21}}}
{{{y=29}}}

substitute in eq.2, and solve for {{{x}}}

{{{x+21=2y}}}........eq.2

{{{x+21=2*29}}}

{{{x=58-21}}}

{{{x=37}}}

so, the larger number is {{{37}}} and the smaller number is {{{29}}}