SOLUTION: solve algebraically for each system given below x^2+y^2=100 x+y=2

Algebra ->  Test -> SOLUTION: solve algebraically for each system given below x^2+y^2=100 x+y=2       Log On


   

Question 931584: solve algebraically for each system given below
x^2+y^2=100
x+y=2
Found 2 solutions by Fombitz, ewatrrr:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
From eq. 2,
x=2-y
Substitute into eq. 1,
%282-y%29%5E2%2By%5E2=100
4-4y%2By%5E2%2By%5E2=100
2y%5E2-4y%2B4=100
2y%5E2-4y-96=0
y%5E2-2y-48=0
%28y-8%29%28y%2B6%29=0
Two solutions:
y-8=0
y=8%7D%7D%0D%0AThen%2C%0D%0A%7B%7B%7Bx=2-8
x=-6
and
y%2B6=0
y=-6
Then,
x=2-%28-6%29
x=8
.
.
.
(-6,8) and (8,-6)

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
x+y=2 0r y = 2-x
......
x^2+y^2=100
x^2 + (2-x)^2 = 100
x^2 + 4 - 4x + x^2 = 100
2x^2 - 4x - 96 = 0
x^2 -2x - 48 = 0
(x+6)(x-8)= 0
x = -6,8
y = 8, -6
(-6,8) and (8, -6) solutions for this System