Question 123198

The difference between two positive nubers is 5

if numbers are {{{x}}} and {{{y}}} where {{{x>y}}}, then:

{{{x-y=5}}}

 and the sum of their squares is 233

{{{x^2 + y^2 = 233}}}



What are the numbers

{{{x-y=5}}}......=>.....{{{x=y+5}}}}

substitute {{{x}}} in {{{x^2 + y^2 = 233}}}

{{{(y+5)^2 + y^2 = 233}}}

{{{y^2 + 10y + 25 + y^2 = 233}}}

{{{2y^2 + 10y + 25 - 233 = 0}}}


{{{2y^2 + 10y - 208= 0}}}

{{{y^2 + 5y - 104 = 0}}}


{{{y[1,2] = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{y[1,2] = (-5 +- sqrt( 5^2 - 4*1*(-104) ))/(2*1) }}}

{{{y[1,2] = (-5 +- sqrt( 25 + 416 ))/2 }}}

{{{y[1,2] = (-5 +- sqrt( 441))/2 }}}

{{{y[1,2] = (-5 +- 21)/2 }}}

{{{y[1] = (-5 + 21)/2 }}}

{{{y[1] = 16/2 }}}

{{{y[1] = 8 }}}


{{{y[2] = (-5 - 21)/2 }}}

{{{y[2] = ( -26)/2 }}}

{{{y[2] =  -13}}}



......=>.....{{{x1=y1+5= 8+5=13}}}}


......=>.....{{{x2=y2+5= -13+5=-8}}}}


check:
{{{x-y=5}}}......=>.....{{{13-8=5}}}

and {{{-8-(-13)=-8+13=5}}}


{{{13^2 + 8^2 = 233}}}
{{{169 + 64 = 233}}}
{{{233= 233}}}

and

{{{(-8)^2 + (-13)^2 = 233}}}
{{{64 + 169  = 233}}}
{{{233= 233}}}