Question 1198032

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

if two numbers differ by {{{8}}} we have
 
{{{x-y=8}}}............solve for {{{x}}}

{{{x=8+y}}}..........eq.1


  if  their product is {{{65}}} we have

{{{xy=65}}}........sustitute  {{{x}}}


{{{(8+y)y=65}}}

{{{8y+y^2=65}}}

{{{y^2+8y-65=0}}}......factor

{{{(y - 5)(y + 13) = 0}}}

solutions:

{{{(y - 5)= 0}}}->{{{y=5}}}

{{{(y + 13) = 0}}}->{{{y=-13}}}


then

{{{x=8+y}}}->{{{x=8+5}}}->{{{x=13}}}

{{{x=8+y}}}->{{{x=8-13}}}->{{{x=-5}}}



 the numbers are:

{{{x=13}}},{{{y=5}}}

or

{{{x=-5}}},{{{y=-13}}}