Question

How many solutions are there for the system x^2+4y^2=100 and 4y-x^2=-20

Answer 1

There are 5 solutions for this system.

x^2 + 4y^2 = 100  ____1
4y - x^2 = -20  ____2
Add both 1 & 2 together. x^2 gets cancelled
4y^2 + 4y = 80   (send 80 to the other side and divide by 4)
Then equation the becomes : y^2 + y -20 =0
Now factorise the equation: (y+5) (y-4) = 0
Solve for y :  y = -5 and y = 4
Using the values of y to find the values of x. From equation 1:
x^2 = 100 - 4y^2    x = /100 - 4y^2  (/ means square root) Replace values of y
y = -5, x = /100 - 4(-5)^2 = /100 - 100 = 0
y = 4, x = /100 - 4(4)^2 = / 100 - 64 = /36 = -6 or 6
Thus we have 6 solutions y = -5, 4 and x = -6, 0, 6