Graphically, these are two parabolas (an upright and a sideways parabola) intersecting at 4 intersection points, so there are 4 (x,y) values each corresponding to a solution.
y=7-x2, so x+(7-x2)2=11,
x+49-14x2+x4=11, or x4-14x2+x+38=0 (quartic equation).
Let's try some possible rational factors (38=2×19). When x=2, we get 16-56+2+38=0, so x-2 is a factor.
Therefore, y=7-22=3 and x+y2=2+9=11, so (2,3) is a solution, but not the only solution.
Divide by x-2 (synthetic division):
2 | 1 0 -14 1 | 38
1 2 4 -20 | -38
1 2 -10 -19 | 0 = x3+2x2-10x-19.
Let f(x)=x3+2x2-10x-19 then apply Newton's Iteration Method:
f'(x)=df/dx=3x2+4x-10;
xn+1=xn-f(xn)/f'(xn). Let x0=0, so:
x1=0+19/(-10)=-1.9,
x2=-1.846677 approx
x3=-1.848125,
x4=-1.848127,
x5=-1.848127, so this is the solution for x. y=√(11+1.848127)=3.584428.
Now we have two solutions, so the cubic can be reduced to a quadratic.
Let this solution for x be represented by X. Using synthetic algebraic division:
X | 1 2 -10 | -19
1 X X2+2X | X3+2X2-10X
1 X+2 X2+2X-10 | 0 = x2+(X+2)x+(X2+2X-10), which can be solved using the quadratic formula.
In this formula, a=1, b=X+2, c=X2+2X-10.
x=(-b±√(b2-4ac)/2a. This gives us x=3.131313 and -3.283186.
The corresponding y values are -2.805118 and -3.779310.
The solutions are (2,3), (-1.848127,3.584428), (3.131313,-2.805118), (-3.283186,-3.779310).