Question

Find the solutions to the system of equations. Select all that apply. Y=x^2-1 y=2x-2

A.(-1,0)
B.(-2,0)
C.(1,0)
D.(0,1)

Answer 1

The correct answer option is C. (1,0).

Explanation:

We are given the following equations and we are to find its solution:

`y=x^2-1`

`y=2x-2`

So we will check each point if its the solution.

A. (-1, 0):

`y=x^2-1` --->
`0=(-1)^2-1` --->
`0=0`

`y=2x-2` --->
`0=2(-1)-2` --->
`0\\eq -4`

B. (-2, 0):

`y=x^2-1` --->
`0=(-2)^2-1` --->
`0\\eq 3`

`y=2x-2` --->
`0=2(-2)-2` --->
`0\\eq -6`

C. (1, 0):

`y=x^2-1` --->
`0=(1)^2-1` --->
`0=0`

`y=2x-2` --->
`0=2(1)-2` --->
`0=0`

D. (0, 1):

`y=x^2-1` --->
`1=(0)^2-1` --->
`1\\eq -1`

`y=2x-2` --->
`1=2(0)-2` --->
`1\\eq -2`

Answer 2

Option C (1, 0)

Explanation:

We have a system with the following equations:

`y = x ^ 2-1\\\\y = 2x-2`

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.

`2x-2 = x ^ 2 -1`

Simplify

`x ^ 2-2x + 1 = 0`

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

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

Finally the solutions are

`x = 1\\y =0`