Calculus 10th Edition

$$(x_{1},y_{1})=(2,2), and,(x_{2},y_{2})=(-1,5)$$
$Step1:$ 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: $$(x_{1},y_{1})=(2,2), and,(x_{2},y_{2})=(-1,5)$$