Find all pairs \( $(x,y)$\) of real numbers \( $(x,y)$\) such that \( $x + y = 10$ \) and \($x^2 + y^2 = 56$.\)
+23246 x + y = 10 ---> y = 10 - x
Substitute this value into: x2 + y2 = 56
x2 + (10 - x)2 = 56
x2 + (100 - 20x + x2) = 56
2x2 - 20x + 100 = 56
2x2 - 20x + 44 = 0
x2 - 10x + 22 = 0
Using the quadratic formula with a = 1, b = -10, and c = 22:
x = [ 10 +/- sqrt( 102 - 4(1)(22) ) ] / ( 2·1 )
x = [ 10 +/- sqrt(12) ] / 2
x = 5 +/- sqrt(3)
Substituting these values back into y = 10 - x ---> y = 5 -/+ sqrt(3)
Answers: ( 5 + sqrt(3), 5 - sqrt(3) ) and ( 5 - sqrt(3), + sqrt(3) )
