Hi,

the sum of two numbers&  is 19 and the sum of their squares is 221

Question states*** 

  x + y = 19  0r  

x^2 + y^2 = 221

 x^2 + (19-x)^2 = 221

 x^2 + 361-38x + x^2 = 221

  2x^2 - 38x  + 140 = 0

   x^2 - 19x + 70 = 0

factoring

(x - 14)(x-5) = 0

(x - 14)= 0,  x = 14   and y = 5

0r(x-5) = 0,  x =  5   and y = 14

numbers are 5 and 14 

Answer by aaronwiz(69)   (Show Source): You can put this solution on YOUR website!
 Hi, my name is Aaron I am in 10th grade and im in honors trig/pre-calc. 

let

x=first number

y=second number



x+y=19

x^2+Y^2=221



x=19-y

x^2+Y^2=221



Now simply substitute



(19-y)^2+y^2=221



foil



(361-38y+y^2)+y^2=221



put into standard form



2y^2-38y+140=0



simplify



y^2-19y+70=0



factor



(y-14)(y-5)=0



y=14 or 5



plug y=5 back in to x^2+Y^2=221



25+x^2=221

x^2=196 

thus x= plus or minus 14 when y=5



plug y=14 back in to x^2+Y^2=221

x^2+196=221

x^2=25

thus x= plus or minus 5 when y=14



However when negative numbers are plugged into x+y=19 it is never true. That means only positive solutions work. So one of the numbers is 5 and the other one is 14.



Hope I helped. I would really appreciate your recommendation, their are plenty of other things I could be doing rather then spending time on this site helping people. Good luck! :)  

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