solving systems for points of intersection (Alg 2)?

Question

Help!!! I'm having trouble solving this system:

x^2 + y^2 = 4
x = y

Answer 1

x^2+y^2=4
x=y
If x=y then substitute y for x
then
y^2+y^2=4
simplify
2y^2=4
Divide both sides of the equation by 2.
then
y^2=2
take the square root of both sides of the equation
then
y= 2^(0.5) or square root of 2
and because x and y are equal
x= 2^(0.5) or square root of 2

Answer 2

x^2 + y^2 = 4 is a circle, center at (0,0), with a radius of 2.
x=y is a line; it goes through (1,1) and (-1,-1).
There are two points the line intersects the circle: (sqrt(2),sqrt(2)) and (-sqrt(2),-sqrt(2)).

Algebraically, square the second equation, which gives you x^2=y^2.
Substitute y^2 into the first equation for x^2 and you get 2(y^2) = 4
Then divide by 2: y^2 = 2, so y=+/-sqrt(2).
Plug this into the second equation and you get x=+/-sqrt(2)

Answer 3

plug in x for y and you get x^2+x^2=4

2x^2=4 divide by 2

x^2=2 take the square root of both sides

x= positive or negative the square root of 2

meaning the points of intersection for the system is the positive and negative square root of 2

Answer 4

Sub. x=y into x^2 + y^2 = 4
y^2 + y^2 = 4

2 y^2 = 4

y^2 = 2

y = +/-sqrt(2)

x = y = +/-sqrt(2)

The answer is not a whole number, is that why you can't solve it?

Answer 5

(+-√2,+-√2)

either replace x with y or y with x and solve.

or graph

y=x

and

y = 2-x

Answer 6

x and y = sq, rt 2