Question 803861

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

if one number is {{{15}}} greater than the other number, you have {{{x=y+15}}}....eq.1

if the product of the two numbers is {{{286}}}, you have {{{x*y=286}}}.....eq.2

solve the system:

{{{x=y+15}}}....eq.1....substitute in eq.2
{{{x*y=286}}}.....eq.2
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{{{(y+15)*y=286}}}.....eq.2

{{{y^2+15y=286}}}

{{{y^2+15y-286=0}}}.......factor, write {{{15y}}} as {{{26y-11y}}}

{{{y^2+26y-11y-286 = 0}}}......group

{{{(y^2+26y)-(11y+286) = 0}}}

{{{y(y+26)-11(y+26) = 0}}}

{{{(y-11)(y+26) = 0}}}

solutions:

if {{{y-11  = 0}}} => {{{y=11}}}

if {{{ y+26 = 0}}} => {{{y=-26}}}


Assuming both numbers are positive, we will use only {{{highlight(y=11)}}} and calculate {{{x}}}

{{{x=y+15}}}....eq.1

{{{x=11+15}}} 

{{{highlight(x=26)}}} 

so, your numbers are {{{highlight( 26)}}} and {{{highlight(11)}}}