Question 1203982

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

if {{{2}}} numbers add to get {{{-4}}}, we have

 {{{x+y=-4}}}...eq.1


and if multiply to get {{{3}}}, we have


{{{x*y=3}}}...eq.2... solve for {{{y}}}

{{{y=3/x}}}..eq.2a


substitute in eq.1


{{{x+3/x=-4}}}...eq.1.. solve for {{{x}}}, both sides multiply by {{{x}}} 

{{{x^2+3=-4x}}}

{{{x^2+4x+3=0}}}...factor

{{{(x + 1) (x + 3) = 0}}}


solutions:

{{{x=-1}}}
or
{{{x=-3}}}


go to

{{{y=3/x}}}..eq.2a, substitute {{{x}}}

{{{y=3/-1=-3}}}
or
{{{y=3/-3=-1}}}



so, solutions are

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

or
{{{x=-3}}}, {{{y=-1}}}