Question

Find all pairs of real numbers $(x,y)$ such that $x + y = 6$ and $x^2 + y^2 = 28$. if you find more than one pair, then list your pairs in order by increasing $x$ value, separated by commas. for example, to enter the solutions $(2,4)$ and $(-3,9)$, you would enter "(-3,9),(2,4)" (without the quotation marks).

Answer 1

1. $x^2+y^2=28$ and x+y=6

2.  x^2+y^2 and x+y remind us of the formula: $(x+y)^{2}= x^2+y^2+2xy$

substituting we get: 

 $6^{2}= 28+2xy$

2xy=36-28
xy=8/2=4

3 Let's solve the system of equations i)x+y=6 and ii)xy=4

substitute y=4/x in (i): 

$x+ \frac{4}{x} =6$

multiply all sides by x:

$x^{2} +4=6x$

$x^{2} -6x+4=0$

complete the square:

$x^{2} -6x+4=x^{2} -2*3x+9-9+4= (x-3)^{2}-5=0$

$x-3=\sqrt{5}$ or $x-3=-\sqrt{5}$ 

$x=3+\sqrt{5}$ or $x=3-\sqrt{5}$

so from x+y=6, y=6-x

case 1,  $x=3+\sqrt{5}$ gives $y=3-\sqrt{5}$

case 2, $x=3-\sqrt{5}$ gives $y=3+\sqrt{5}$ 


Answer: 

$(3-\sqrt{5}, 3+\sqrt{5})$, $(3+\sqrt{5}, 3-\sqrt{5})$