For this case we have the following system of equations:
[tex]y = x + 3\\y = x ^ 2-x[/tex]
Equating the equations we have:
[tex]x ^ 2-x = x + 3\\x ^ 2-x-x-3 = 0\\x ^ 2-2x-3 = 0[/tex]
To find the solutions we factor the equation. To do this, we look for two numbers that, when multiplied, result in -3 and when added, result in -2.
These numbers are: -3 and +1
[tex]-3 + 1 = -2\\-3 * (+ 1) = - 3[/tex]
Thus, the factored equation is:
[tex](x-3) (x + 1) = 0[/tex]
Then, the solutions for the variable "x" are:
[tex]x_ {1} = 3\\x_ {2} = - 1[/tex]
We find the solutions for the variable "y":
[tex]y_ {1} = x_ {1} + 3 = 3 + 3 = 6\\y_ {2} = x_ {2} + 3 = -1 + 3 = 2[/tex]
Thus, the solutions are:
[tex](x_ {1}, y_ {1}): (3,6)\\(x_ {2}, y_ {2}): (-1,2)[/tex]
ANswer:
[tex](x_ {1}, y_ {1}): (3,6)\\(x_ {2}, y_ {2}): (-1,2)[/tex]
