SOLUTION: 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 comma

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: 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 comma      Log On


   

Question 1086028: 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).
*That's a quadratic in x
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
x=6-y
Substituting,
%286-y%29%5E2%2By%5E2=28
y%5E2-12y%2B36%2By%5E2=28
2y%5E2-12y%2B8=0
y%5E2-6y%2B4=0
y%5E2-6y%2B9%2B4=9
%28y-3%29%5E2=5
y-3=0+%2B-+sqrt%285%29
y=3+%2B-+sqrt%285%29
So then,
x=6-y
Solve for x.
.
.
.
.