Question 1011503
let the numbers be {{{x}}} and {{{y}}}
if the sum of two numbers is {{{26}}}, we have

{{{x+y=26}}}...solve for {{{x}}}

{{{x=26-y}}}.......eq.1

if twice the first ({{{2x}}}) plus half the second ({{{y/2}}}) is {{{10}}}, we have 

{{{2x+y/2=10}}}...............substitute {{{x}}} from eq.1

{{{2(26-y)+y/2=10}}}......solve for {{{y}}}

{{{52-2y+y/2=10}}}

{{{52-4y/2+y/2-10=0}}}

{{{42-3y/2=0}}}

{{{42=3y/2}}}

{{{84=3y}}}

{{{84/3=y}}}

{{{highlight(y=28)}}}

go to {{{x=26-y}}}.......eq.1 and plug in {{{y}}}

{{{x=26-28}}} 

{{{highlight(x=-2)}}}

so, your numbers are {{{highlight(-2)}}} and {{{highlight(28)}}}