Question

Are there 2 solutions for the system shown below?

x^2+y^2=25
x-y^2=-5

Answer 1

2nd equation :
x-y^2=-5
x = y^2 - 5
subtitute the equation above into 1st equation :
(y^2 - 5)^2+y^2=25
y^4 -9y^2 = 0
y^2(y^2 - 9) = 0
y^2 (y+3)(y-3) = 0

now how many solution make y be equals zero ?

Answer 2

just one equation makes y equal to zero, right?

Answer 3

Be sure and test possible answers. Some of these may not work for the system:
(4, -3) (4, 3) (-5,0) (5,0)

Answer 4

setting each factors bellow to zero, then solve for y
y^2 (y+3)(y-3) = 0

y^2 = 0
solve for y ?

y + 3 = 0
solve for y ?

y - 3 = 0
solve for y ?

how many different of y's ?

Answer 5

only y^2=0 gives me 0. y+3 gives me -3 and y-3 gives me positive 3

Answer 6

btw, (5,0) cant be satisfies for the 2nd equation :). @radar
x-y^2=-5

5 - 0^2 = -5
5 = -5 (incorrect)

Answer 7

yes, you are right, @ali1029

Answer 8

That was my point,.....

Answer 9

so then there is only 1 solution possible?

Answer 10

actually, if you have 3 y's as solutions then you have 3 x's too.
therefore, we have 3 solutions for (x,y)

Answer 11

oh I see, that completely makes sense! thank you :)

Answer 12

welcome