Answer
$$(x_{1},y_{1})=(2,2), and,(x_{2},y_{2})=(-1,5)$$
Work Step by Step
$ Step1: $ let y =4-x
$Step2:$ substitute above equation into: $x^{2}+y=6$
$Step3:$ $x^{2}+4+-x=6$
$Step4:$ After simplifying we have:
$$x^{2}-x-2=0$$
$Step5:$ Solving the derive quadratic equation by factoring method we have two values of x:
$$ (x-2)(x+1)=0$$
$$x=2, or x=-1$$
$Step6:$ we ready to find the values of y values by substituting the above x values into the following equation:
$$y=4-x$$
$for$ x=2, y-2=2, and for x=-1, y=5
$Step7:$ Our system of equations
$$x^{2}+y=6$$ and
$$x+y=4$$
They intersect at the following two points:
$$(x_{1},y_{1})=(2,2), and,(x_{2},y_{2})=(-1,5)$$
