graph the system of equations. then solve: x^2+ y^2=10 and xy=3?

Answer 1

x^2+ y^2=10 xy=3

x^2+ y^2= (x+y)^2 - 2xy

so

x^2+ y^2= (x+y)^2 - 2xy=10

(x+y)^2 - 2xy=10 -> (x+y)^2 - 2*3=10

so (x+y)^2 =16

so (x+y)= 4 or - 4

now (x+y) = 4 and xy=3 so x=1 y=3

or

(x+y)= - 4 and xy=3

so x= - 1 y= - 3

Answer 2

The first is a circle with radius sqrt(10), the second is a hyperbola with the coordinate axes as asymptotes. You can get some idea of what they look like by using the calculator. Graph y1 = sqrt(10 - x^2), y2 = -sqrt(10 - x^2), and y3 = 3/x. The circle will have gaps, and it might look like an oval. Anyway, you can see that there are 4 intersection points.

To find them:
xy = 3 , so y = 3/x
x^2 + y^2 = 10
substitute: x^2 + (3/x)^2 = 10
x^2 + 9/x^2 = 10
multiply through by x^2:
x^4 + 9 = 10x^2
x^4 - 10x^2 + 9 = 0
(x^2 - 9)(x^2 - 1) = 0
From this, you get x = +/- 1 or x = +/- 3. Substitute into y = 3/x to get the y-coordinates.

Answer 3

Circle C(0,0) r=√10
Hiperbola: y=3/x [asimptotes x=0; y=0]


Solutions:

x^2+ y^2 + 2xy=10+6 --> (x+y)^2=16 --> x+y= ±4
x^2+ y^2 - 2xy=10-6 --> (x+y)^2=4 --> x-y= ±2

2x=6 --> x=3 y=1 --> (3,1)
2x=2 --> x=1 y=3 --> (1,3)

2x=-2 --> x=-1 y=-3 -->(-1,-3)
2x=-6 --> x=-3 y=-1 --> (-3,-1)

Saludos.

Answer 4

I can't graph it on here, but it is true in two cases. If x=-1 and y=-3 or if x=1 and y=3.

Answer 5

x = 7y