find exact solutions of system of equations....?

Question

can someone take me step by step....for this : y=x+2
y=x^2

Answer 1

y=x+2

y =x^2
So
x+2 =x^2
so x^2 -x-2=0
x^2 -2x+x-2=0
x(x-2) +1(x-2)=0
(x-2)(x+1)=0

so x = 2 or -1

so if x=2 y = 4
if x = -1 then y = 1

Answer 2

y = x +2
y = x^2

therefore equate the two

x^2 = x + 2

then bring it over to one side by subtraction

x^2 - x - 2 = 0

then factorize
in the format of

(x + )(x - )

to which i get

(x+1)(x-2) (plese check this out to be sure this is correct)

and this will give you 2 possible solutions to the above

Answer 3

substitute one solution for another i.e
x^2=x+2
x^2-x-2=0
(x-2)(x+1)=0
x = 2 or x=-1
therefore y=4 0r y=1

Answer 4

x² = x + 2
x² - x - 2 = 0
(x - 2).(x + 1) = 0
x = 2 , x = - 1
Possible values for y are then:-
y = 4 , y = 1
Ordered pairs are:-
(2,4) , (- 1, 1)

Answer 5

=>x+2=x^2
x^2-x-2=0
x^2+x-2x-2=0
x(x+1)-2(x+1)=0
(x-2)(x+1)=0
x=2,x=-1

Answer 6

y=x+2
y=x^2
So x+2=x^2

So x^2-x-2=0

So (x-2)(x+1)=0

So since one of the terms must be zero,
x=2 or x=-1

If x=2, y=4
If x=-1, y=1

Hope this helps?
.

Answer 7

i told ya solve it on your own, dude