Question

Please help! Solve the following system using any method:
x+y=2
x^2-y=0
Another solution is:
a. (1,1)
b. (-1,1)
c. (-1,-1)
d. (1, -1)

Answer 1

from 2nd equation
y=x^2
put this value of y in 1st equation
x+x^2=2
x^2+x=2
x^2+x-2=0
x^2+2x-x-2=0
x(x+2)-1(x+2)=0
(x-1)(x+2)=0
x=1,x=-2

Answer 2

when x=1,y=1
and when x=-2,y=4

Answer 3

so A is the answer

Answer 4

Thanks!

Answer 5

From first equation: x = 2 - y
Put this in second equation:

(2 - y)^2 - y = 0

\[y^2 - 4y + 4 -y = 0\]

\[y^2 - 5y +4 = 0\]

\[(y-4)(y-1) = 0\]

y = 4 or y = 1

So, from this x = 1 when y = 1
and x = -2 when y = 4

So, the required answer is : (1, 1) i.e, A..

Answer 6

Thank you!

Answer 7

Welcome dear...