Respostas

2014-03-20T19:06:59-03:00
 \left \{ {{y+y=14} \atop {x^{2}+y^{2}=100}} \right.
(x+y)^{2} =  x^{2} +y^{2} +2xy
(14)^{2}=100+2xy
196=100+2xy
xy= \frac{196-100}{2} = \frac{96}{2} =48

Então y= \frac{48}{x} , substituindo na primeira equação

x+y =x+ \frac{48}{x} =14

 x^{2} +48=14x
 x^{2} -14x+48=0
 x_{1} =8
 x_{2} =6

 \left \{ {{para~~ x=6 ~,~entao ~:~y= \frac{48}{6}=8  } \atop {{para~~ x=8 ~,~entao ~:~y= \frac{48}{8}=6}} \right.