SOLUTION: x+y=3
x^2+y^2=369
Please show me the steps to solve this in any way.
Algebra.Com
Question 447410: x+y=3
x^2+y^2=369
Please show me the steps to solve this in any way.
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
x+y=3
x^2+y^2=369
x=3-y
substitute the x in eq 2
(3-y)^2+y^2=369
9-6y+y^2+y^2=369
2y^2-6y-360=0
/2
y^2-3y-180=0
Find the roots of the equation by quadratic formula
a=1, b=-3, c=-180
b^2-4ac= 9-(-720)
b^2-4ac= 729
=27
x1=(3+27 )/2
x1=15
x2=(3-27 )/2
x2= -12
(15 & -12)
m.ananth@hotmail.ca
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