find the point of intersection. x^2-x-y=1 and -x+y=-1?

Question

hey guys can u please help me with this question as i have no idea about what to do. i have the answer which is (0,-1)and(2,1)

thank you guys in advance

Answer 1

x^2 -x -y =1
-x +y = -1

from eq 2, y = x-1
substitute in eq 1:
x^2 -x -(x-1) =1
x^2 -2x +1 =1
x^2 -2x =0
x(x-2) =0 => x=0 or x=2,
if x =0, then y= 0-1 = -1, and the point is ( 0, -1)
if x=2, then y= 2-1, and the point is ( 2, 1) .

Answer 2

x^2 - x - y = 1
-x + y = -1

All you have to do is solve for y in the second equation, and then plug it into the first equation.

-x + y = -1, means y = x - 1. Plugging this into the first equation, we have

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

Now, we factor
x(x - 2) = 0. This gives us TWO solutions for x:
x = {0, 2}

Now that we have TWO values for x, we'll have two values for y. Using our earlier substitution of y = x - 1:

When x = 0, y = 0 - 1 = -1
When x = 2, y = 2 - 1 = 1

Therefore, y = {-1, 1}

So our two solutions are:
x = 0, y = -1
x = 2, y = 1

OR

(0, -1) and (2, 1)

Answer 3

x^2-x-y=1 <=> y = x^2-x-1
-x+y=-1 <=> y = x - 1
=> x^2 - x -1 = x -1 (=y)
=> x^2 - 2x = 0
=> x = 0 or x = 2
the points found are : (0,-1) (2,1)