Question 675676

let numbers be {{{x}}} and {{{y}}}

given:
two numbers are in the ratio {{{3}}} to {{{5}}}

so, we have {{{x/y=3/5}}}......eq. 1


 if {{{9}}} is added to their {{{sum}}} the result is {{{41 }}}


{{{x+y+9=41}}}......eq. 2


solve this system:

so, we have {{{x/y=3/5}}}......eq. 1

{{{x+y+9=41}}}......eq. 2
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{{{x/y=3/5}}}......solve for {{{x}}}

{{{x=(3/5)y}}}......substitute in eq. 2

{{{(3/5)y+y+9=41}}}......eq. 2......solve for {{{y}}}

{{{(3/5)y+y=41-9}}}

{{{(3y/5)+y=32}}}...both sides multiply by {{{5}}}

{{{5(3y/5)+5y=32*5}}}

{{{3y+5y=160}}}

{{{8y=160}}}

{{{y=160/8}}}

{{{highlight(y=20)}}}


now find {{{x}}}


{{{x=(3/5)y}}}

{{{x=(3/5)20}}}

{{{x=(3/cross(5))cross(20)4}}}

{{{x=3*4}}}

{{{highlight(x=12)}}}