Question

Which ordered pair is a solution to the system of equations y=x+3 and y=x^2-x

Answer 1

For this case we have the following system of equations:

$y = x + 3\\y = x ^ 2-x$

Equating the equations we have:

$x ^ 2-x = x + 3\\x ^ 2-x-x-3 = 0\\x ^ 2-2x-3 = 0$

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

$-3 + 1 = -2\\-3 * (+ 1) = - 3$

Thus, the factored equation is:

$(x-3) (x + 1) = 0$

Then, the solutions for the variable "x" are:

$x_ {1} = 3\\x_ {2} = - 1$

We find the solutions for the variable "y":

$y_ {1} = x_ {1} + 3 = 3 + 3 = 6\\y_ {2} = x_ {2} + 3 = -1 + 3 = 2$

Thus, the solutions are:

$(x_ {1}, y_ {1}): (3,6)\\(x_ {2}, y_ {2}): (-1,2)$

ANswer:

$(x_ {1}, y_ {1}): (3,6)\\(x_ {2}, y_ {2}): (-1,2)$