Question 1020887

let numbers be {{{x}}} and {{{y}}}
if numbers is differ by {{{8}}}, we have
{{{x-y=8}}}...........eq.1 (assuming {{{x}}} is greater number)

if one-half the large, which is {{{(1/2)x}}} is added to one-fourth the smaller, which is {{{(1/4)y}}}, the resulting some is {{{22}}}, we have

{{{(1/2)x +(1/4)y=22}}}........eq.2

go to {{{x-y=8}}}...........eq.1 and solve for {{{x}}}

{{{x=8+y}}}...substitute in eq.2

{{{(1/2)(8+y) +(1/4)y=22}}}........eq.2 and solve for {{{y}}}

{{{(1/2)(8+y) +(1/4)y=22}}}.....both sides multiply by {{{4}}}

{{{4(1/2)(8+y) +4(1/4)y=4*22}}}

{{{2(8+y) +y=88}}}

{{{16+2y +y=88}}}

{{{3y=88-16}}}

{{{3y=72}}}

{{{y=72/3}}}

{{{highlight(y=24)}}}

now find {{{x}}}

{{{x=8+y}}}

{{{x=8+24}}}

{{{highlight(x=32)}}}


so, your numbers are: {{{highlight(24)}}} and {{{highlight(32)}}}