For the first one, I will do it algebraically first. You have: y = x^2 - 4 and y + x^2 = 4. If you like, rearrange the second one so the terms match:
y = x^2 - 4
y = -x^2 + 4
You are looking for x and y values that satisfy both of these equations. We know that y is explicitly equal to both x^2 - 4 and -x^2 + 4 so we can set both of these equal and then solve for x:
x^2 - 4 = -x^2 + 4 (and then we can use our algebra skills)
2x^2 = 8
x^2 = 4
x = -2 and 2
Now we plug -2 in for x in one of the original equations to check which y-value corresponds to the solution with this x-value, so:
y = (-2)^2 - 4 = 0, so the solution is x=-2 and y=0, or (-2,0).
But wait, we have another x-value: 2. We plug this into one of the above equations (it doesn't matter which one) and still get y=0, so our second solution is (2,0).
We have solutions: (-2,0) and (2,0).
If we want to do this geometrically, the values of x and y that satisfy both of these equations are the same as the x and y values that the graphs of these functions share (where they overlap).
If we graph y = x^2 - 4, we have the graph of parabola y=x^2, but moved downward four values. If you don't know how to graph this function, then use an x-y chart and pick x-values, such as 0,1,-1,2,etc... and find the y-values, then graph these points and connect the dots.
If we graph y = -x^2 + 4 we have an upside down parabola y=x^2 but moved upward four values. And we see that these graphs overlap at (2,0) and (-2,0), which is the same as our algebraic solutions.
For the second equation, let us do it algebraically first.
We know: y = x^2 - 5x + 8 and y = -x^2 + 7x - 10, and we are looking for the values of x and y that satisfy both of these equations. Like before, we know that y equals both of the right sides, so we can plug one of the right sides in for y in the other equation (or set both expressions equal to each other), and we get:
x^2 - 5x + 8 = -x^2 + 7x - 10 (move everything to one side)
2x^2 - 12x + 18 = 0 (we can divide by 2 and not change the equality
)
x^2 - 6x + 9 = 0 (now we have to experiment and factor)
(x - 3)(x - 3) = 0
(x - 3)^2 = 0
If we set x = 3 in that last line, we will get 0 = 0 so we know that x=3 is a solution and the only solution since no other x-values turn the left side into 0. (The squared part does say something about the originally functions' behavior but it's not necessary to get into that I don't think.)
Now we need to find the y-value that corresponds to this x-value in solving both of the equations. (It doesn't matter which originally equation we use.) So let's plug x=3 into one of them:
y = 3^3 - 5(3) + 8 = 2
So our solution is (3,2).
To find this solution geometrically:
Graphing these equations without a graphing calculator is pretty easy but requires "completing the square" to change the equations into forms which tell you the exact x- and y-intercepts and such. After all of this typing it seems like it would just be too much heh.
But if we graph both of these equations on the graphing calculator we see that there is one parabola suspended where its vertex is in the 1st quadrant and another upside down parabola with vertex in the 1st quadrant, and they both touch at (3,2) and only at that point.
If any of this is confusing let me know, haven't had much sleep in the past two days..