Question 1126805

let one number be {{{x}}}, and other number be {{{y}}}

if {{{x}}} is {{{seven}}}{{{ more}}} than {{{twice}}} {{{y}}}, we have

{{{x=2y+7}}}.......eq.1


 if their {{{sum}}} is {{{decrease}}} by {{{ten}}} the {{{result}}} is {{{twelve}}}; so, we have 

{{{x+y-10=12}}}....eq.2 

substitute {{{x}}} from eq.1

{{{2y+7+y-10=12}}}....eq.2 ...solve or {{{y}}}

{{{3y-3=12}}}

{{{3y=12+3}}}

{{{3y=15}}}

{{{y=15/3}}}

{{{y=5}}}

now find {{{x}}}

{{{x=2y+7}}}.......eq.1

{{{x=2*5+7}}}

{{{x=10+7}}}

{{{x=17}}}


so, one number is {{{17}}}, and the other number is {{{5}}}