Based on the graphs of the equations [tex]y = -2x + 3[/tex] and [tex]y = x^2 - x + 1[/tex], the solutions are located at points:

A. [tex]\((0.5, 0.75)\)[/tex] and [tex]\((1, 1)\)[/tex]

B. [tex]\((-2, 7)\)[/tex] and [tex]\((0, 1)\)[/tex]

C. [tex]\((0, 1)\)[/tex] and [tex]\((0.5, 0.75)\)[/tex]

D. [tex]\((-2, 7)\)[/tex] and [tex]\((1, 1)\)[/tex]



Answer :

Alright, let's analyze the problem step-by-step.

We are working with the following pair of equations:
1. [tex]\( y = -2x + 3 \)[/tex]
2. [tex]\( y = x^2 - x + 1 \)[/tex]

We need to find the points where these two equations intersect. Let's check each of the given points to see if they satisfy both equations:

1. Point [tex]\( (0.5, 0.75) \)[/tex]

For [tex]\( y = -2x + 3 \)[/tex]:
[tex]\[ y = -2(0.5) + 3 = -1 + 3 = 2 \][/tex]

For [tex]\( y = x^2 - x + 1 \)[/tex]:
[tex]\[ y = (0.5)^2 - 0.5 + 1 = 0.25 - 0.5 + 1 = 0.75 \][/tex]

As we see, [tex]\( (0.5, 0.75) \)[/tex] does not satisfy the first equation because the calculated [tex]\( y \)[/tex] value is [tex]\( 2 \)[/tex], not [tex]\( 0.75 \)[/tex].

2. Point [tex]\( (1, 1) \)[/tex]

For [tex]\( y = -2x + 3 \)[/tex]:
[tex]\[ y = -2(1) + 3 = -2 + 3 = 1 \][/tex]

For [tex]\( y = x^2 - x + 1 \)[/tex]:
[tex]\[ y = (1)^2 - 1 + 1 = 1 - 1 + 1 = 1 \][/tex]

The point [tex]\( (1, 1) \)[/tex] satisfies both equations.

3. Point [tex]\( (-2, 7) \)[/tex]

For [tex]\( y = -2x + 3 \)[/tex]:
[tex]\[ y = -2(-2) + 3 = 4 + 3 = 7 \][/tex]

For [tex]\( y = x^2 - x + 1 \)[/tex]:
[tex]\[ y = (-2)^2 - (-2) + 1 = 4 + 2 + 1 = 7 \][/tex]

The point [tex]\( (-2, 7) \)[/tex] satisfies both equations.

4. Point [tex]\( (0, 1) \)[/tex]

For [tex]\( y = -2x + 3 \)[/tex]:
[tex]\[ y = -2(0) + 3 = 3 \][/tex]

For [tex]\( y = x^2 - x + 1 \)[/tex]:
[tex]\[ y = (0)^2 - 0 + 1 = 1 \][/tex]

The point [tex]\( (0, 1) \)[/tex] does not satisfy the first equation because the calculated [tex]\( y \)[/tex] value is [tex]\( 3 \)[/tex], not [tex]\( 1 \)[/tex].

So summarizing the results:
- The point [tex]\((0.5,0.75)\)[/tex] does not satisfy both equations.
- The point [tex]\((1,1)\)[/tex] satisfies both equations.
- The point [tex]\((-2,7)\)[/tex] satisfies both equations.
- The point [tex]\((0,1)\)[/tex] does not satisfy both equations.

Therefore, the intersection points from the given list of options where both of the given equations hold true are:

- Points [tex]\( (-2, 7) \)[/tex] and [tex]\( (1, 1) \)[/tex].