Question 490009
let two numbers be {{{x}}} and {{{y}}}

the difference of two numbers is {{{3}}}

 {{{x-y=3}}}.....1

and the product is {{{10}}} 

{{{xy=10}}}......2


to find out what are the two numbers, solve this system:


 {{{x-y=3}}}.....1

{{{xy=10}}}......2
--------------------------

 {{{x-y=3}}}.....1...solve for {{{y}}}

{{{x-3=y}}}......plug it in 2


{{{x(x-3)=10}}}......2

{{{x^2-3x=10}}}

{{{x^2-3x-10=0}}}...use quadratic formula to solve for {{{x}}}


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}....note that {{{a=1}}}, {{{b=-3}}}, and {{{c=-10}}}



{{{x = (-(-3) +- sqrt( (-3) ^2-4*1*(-10)  ))/(2*1) }}}


{{{x = (3 +- sqrt(9+40 ))/2 }}}

{{{x = (3 +- sqrt(49))/2 }}}

{{{x = (3 +- 7)/2 }}}

solutions:

{{{x = (3 +7)/2 }}}

{{{x = 10/2 }}}

{{{x = 5 }}}

or


{{{x = (3 -7)/2 }}}

{{{x = (-4)/2 }}}

{{{x = -2 }}}

now find {{{y}}}



{{{x-3=y}}}...plug in {{{x = 5 }}}

{{{5-3=y}}}

{{{2=y}}}


or

{{{x-3=y}}}...plug in {{{x = -2 }}}

{{{-2-3=y}}}

{{{-5=y}}}

so, you have two sets of solutions, and they are:

1.

{{{x = 5 }}}
{{{2=y}}}

2.
{{{x = -2 }}}
{{{-5=y}}}


check first set:

{{{x = 5 }}}
{{{2=y}}}


{{{x-y=3}}}.....1

{{{ 5-2=3}}}

{{{ 3=3}}}

and the product is {{{10}}} 

{{{5*2=10}}}......2

{{{10=10}}}