Answer:
(1, -2), (4, 1)
Step-by-step explanation:
Here we want to solve the system of equations:
y = x^2 - 4*x + 1
y = x - 3
First, we can see that in both parts we have isolated the variable "y", so we can just write:
x - 3 = y = x^2 - 4*x + 1 = y
removing the "y"s, we get:
x - 3 = x^2 - 4*x + 1
Now we can solve this for x
0 = x^2 - 4*x + 1 - x + 3
0 = x^2 - 5*x + 4
This is a quadratic equation, the solutions are given by the Bhaskara's formula, that says that for a general quadratic equation:
0 = a*x^2 + b*x + c
The solutions are given by:
[tex]x = \frac{-b \pm \sqrt{b^2 - 4*a*c} }{2*a}[/tex]
So for the case of our equation:
0 =x^2 - 5*x + 4
The solutions are given by:
[tex]x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4*1*4} }{2*1} = \frac{5 \pm 3}{2}[/tex]
So the two solutions are:
x₁ = (5 + 3)/2 = 4
x₂ = (5 - 3)/2 = 1
To find the ordered pair, we need to replace these values in one of the equations of the system, let's use the second:
y = x₁ - 3 = 4 - 3 = 1
Then we have one solution at (4, 1)
And for the other:
y = x₂ - 3 = 1 - 3 = -2
then the ordered pair is (1, -2)
Now we want to write from least to greatest with respect to the x-coordinate, then the correct order is:
(1, -2), (4, 1)
