Answer:
Option C (1, 0)
Step-by-step explanation:
We have a system with the following equations:
[tex]y = x ^ 2-1\\\\y = 2x-2[/tex]
The first equation is a parabola.
The second equation is a straight line
To solve the system, substitute the second equation in the first and solve for x.
[tex]2x-2 = x ^ 2 -1[/tex]
Simplify
[tex]x ^ 2-2x + 1 = 0[/tex]
You must search for two numbers that when you add them, obtain as a result -2 and multiplying both results in 1.
These numbers are -1 and -1
Therefore
[tex]x ^ 2-2x + 1= (x-1)(x-1)\\\\x ^ 2-2x + 1= (x-1)^2=0[/tex]
Finally the solutions are
[tex]x = 1\\y =0[/tex]
Answer:
The correct answer option is C. (1,0).
Step-by-step explanation:
We are given the following equations and we are to find its solution:
[tex]y=x^2-1[/tex]
[tex]y=2x-2[/tex]
So we will check each point if its the solution.
A. (-1, 0):
[tex]y=x^2-1[/tex] ---> [tex]0=(-1)^2-1[/tex] ---> [tex]0=0[/tex]
[tex]y=2x-2[/tex] ---> [tex]0=2(-1)-2[/tex] ---> [tex]0\neq -4[/tex]
B. (-2, 0):
[tex]y=x^2-1[/tex] ---> [tex]0=(-2)^2-1[/tex] ---> [tex]0\neq 3[/tex]
[tex]y=2x-2[/tex] ---> [tex]0=2(-2)-2[/tex] ---> [tex]0\neq -6[/tex]
C. (1, 0):
[tex]y=x^2-1[/tex] ---> [tex]0=(1)^2-1[/tex] ---> [tex]0=0[/tex]
[tex]y=2x-2[/tex] ---> [tex]0=2(1)-2[/tex] ---> [tex]0=0[/tex]
D. (0, 1):
[tex]y=x^2-1[/tex] ---> [tex]1=(0)^2-1[/tex] ---> [tex]1\neq -1[/tex]
[tex]y=2x-2[/tex] ---> [tex]1=2(0)-2[/tex] ---> [tex]1\neq -2[/tex]
