Find the point(s) of intersection of the graphs of the equat

[jake_k2011]

Find the point(s) of intersection of the graphs of the equations.

X=1-y^2
y=x-1

For one of the points I get is (1,0) and (2,1). i'm taking an online class and (1.0) seems to be correct but (2,1) is wrong. Did I make a mistake?? I keep getting (2,1) over and over no matter what method I use. Help??

 

[Loren]

If we can't see your work we can't tell if you made a mistake. Please show your work or tell us what steps you took..

 

[jake_k2011]

here's what I've done

X=1-(x-1)^2
X=1(-x+1)^2
X=1(-x+1) (-x+1)
X=1 X^2-2X+1
0= x^2-3x+2
0= (x-2)(x-1)

soo x=2 and x=1

plug into the equation y=x-1 for each to get yet

y=x-1
y=2-1
y=1

y=x-1
y=1-1
y=0

 

[Loren]

I'm wondering if X and x are the same or different?

here's what I've done

X=1-(x-1)^2
X=1(-x+1)^2 <<< should be X = 1 - (x^2 - 2x + 1) = 1 - x^2 + 2x - 1 = -x^2 + 2x.
X=1(-x+1) (-x+1)
X=1 X^2-2X+1
0= x^2-3x+2
0= (x-2)(x-1)

soo x=2 and x=1

plug into the equation y=x-1 for each to get yet

y=x-1
y=2-1
y=1

y=x-1
y=1-1
y=0

 

[Mrspi]

here's what I've done

X=1-(x-1)^2
X=1(-x+1)^2<----here's your error. You MUST follow the order of operations. Do the squaring FIRST....THEN the subtraction
X=1(-x+1) (-x+1)
X=1 X^2-2X+1
0= x^2-3x+2
0= (x-2)(x-1)

soo x=2 and x=1

plug into the equation y=x-1 for each to get yet

y=x-1
y=2-1
y=1

y=x-1
y=1-1
y=0

y = 1 - (x - 1)[sup:346ctyiz]2[/sup:346ctyiz]
y = 1 - (x[sup:346ctyiz]2[/sup:346ctyiz] - 2x + 1)

NOW do the subtraction:

x = 1 - x[sup:346ctyiz]2[/sup:346ctyiz] + 2x - 1
x = -x[sup:346ctyiz]2[/sup:346ctyiz] + 2x
x[sup:346ctyiz]2[/sup:346ctyiz] - x = 0
x(x - 1) = 0

so, x = 0 or x - 1 = 0, which gives x = 1.

Use these two values to get y.

y = x - 1

If x = 0, y = 0 - 1 or y = -1. (0, -1) is one point of intersection.
If x = 1, y = 1 - 1, or y = 0. (1, 0) is the other point of intersection.