Question

solve the system algebraically
x^2+y^2=36
x^2-y^2=9

Answer 1

add those two equations to get:
2x^2 = 45

these are your x-coordinates of where the intersections occur. then plug them back in to get your y-coordinates.

Answer 2

Add the equations:


x^2+y^2=36
x^2-y^2=9
------------
2x^2 + 0y^2 = 45


So we have 2x^2 = 45, solve this for x


2x^2 = 45

x^2 = 45/2

x = sqrt(45/2) or x = -sqrt(45/2)

x = sqrt(45)/sqrt(2) or x = -sqrt(45)/sqrt(2)

x = (sqrt(45)*sqrt(2))/sqrt(2)*sqrt(2)) or x = -(sqrt(45)*sqrt(2))/sqrt(2)*sqrt(2))

x = sqrt(45*2)/sqrt(2*2) or x = -sqrt(45*2)/sqrt(2*2)

x = sqrt(90)/sqrt(4) or x = -sqrt(90)/sqrt(4)

x = sqrt(3^2*10)/sqrt(2^2) or x = -sqrt(3^2*10)/sqrt(2^2)

x = sqrt(3^2)*sqrt(10)/sqrt(2^2) or x = -sqrt(3^2)*sqrt(10)/sqrt(2^2)

x = 3*sqrt(10)/2 or x = -3*sqrt(10)/2


Now that you know the value of x, you can use it to find the value of y.


So plug in x = 3*sqrt(10)/2 into either equation to find the corresponding value for y


Also, plug in x = -3*sqrt(10)/2 to find the other solution for y.

Answer 3

i think elimination is meant to be used, thats more graphically

Answer 4

side note:
the system is that of a circle and hyperbola both centered at the origin. they intersect at 4 points so that should coincide with the number of solutions you get.

Answer 5

what u talking about? when you add those equations, you eliminated the y^2... i did use elimination...

Answer 6

x=22.5 , y= 21.68