Find all pairs (x,y) of real numbers such that x + y = 10 and x^2 + y^2 = 56 + 3xy. For example, to enter the solutions (2,4) and (-3,9), you would enter "(2,4),(-3,9)" (without the quotation marks).
+129850 x + y = 10
y =10 - x
So
x^2 + y^2 = 56 +3xy
x^2 + (10 - x)^2 = 56 + 3x ( 10 - x)
x^2 + x^2 - 20x + 100 = 56 + 30x - 3x^2
5x^2 - 50x + 44 = 0
x = [ 50 ± sqrt ( 50^2 - 4*5 *22) ] / (2*5) =
[ 50 ± sqrt ( 1620) ] /10 =
[ 50 ± 18sqrt (5) ] 10 = 5 ± (9/5)sqrt (5)
x = 5 + ( 9/5)sqrt (5) and 5 - (9/5)sqrt (5)
y = 10 - x ....so....
y = 5 - (9/5) sqrt (5) and 5 + (9/5)sqrt (5)
