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Anastasy [175]
3 years ago
15

PLEASE HELP MEEEEEEEEEEEEEEEEEEEE

Mathematics
1 answer:
Vesnalui [34]3 years ago
6 0

Answer:

A, the correct answer is A

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How many solutions are there for the system x^2+4y^2=100 and 4y-x^2=-20
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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
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Thus we have 6 solutions y = -5, 4 and x = -6, 0, 6
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