Question 698830

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

if two numbers add to {{{-3/2}}}, we have

{{{x+y=-3/2}}}....eq. 1

 and if they multiply to {{{-1}}}, we have

{{{xy=-1}}}...eq. 2

solve the system:

{{{x+y=-3/2}}}....eq. 1
{{{xy=-1}}}...eq. 2
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{{{xy=-1}}}...eq. 2....solve for {{{x}}}

{{{x=-1/y}}}....substitute in eq. 1


{{{-1/y+y=-3/2}}}....eq. 1...solve for {{{y}}}


{{{-1/y+y*y/y=-3/2}}}

{{{(-1+y*y)/y=-3/2}}}

{{{(-1+y^2)/y=-3/2}}}...cross multiply

{{{(-1+y^2)*2=-3*y}}}

{{{-2+2y^2=-3y}}}

{{{2y^2+3y-2=0}}}...use quadratic formula

{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{y = (-3 +- sqrt( 3^2-4*2*(-2) ))/(2*2) }}}

{{{y = (-3 +- sqrt( 9+16 ))/4 }}}

{{{y = (-3 +- sqrt( 25 ))/4 }}}

{{{y = (-3 +- 5 )/4 }}}

solutions:

{{{y = (-3 + 5 )/4 }}}

{{{y = 2/4 }}}

{{{y = 1/2 }}}

or

{{{y = (-3 - 5 )/4 }}}

{{{y = -8/4 }}}

{{{y = -2 }}}


now find {{{x}}}

{{{x=-1/y}}}....plug in {{{y = 1/2 }}}

{{{x=-1/(1/2)}}}

{{{x=-(1*2)/1}}}

{{{x=-2}}}

and

{{{x=-1/y}}}....plug in {{{y = -2 }}}

{{{x=-1/(-2)}}}

{{{x=1/2}}}


so, the numbers are:

{{{x=-2}}} and {{{y = 1/2 }}}

or

{{{x=1/2}}} and {{{y = -2 }}}