What is the solution set of this system of equations y = x^2 – 3x + 4 and x + y = 8?

To find the solution set of the system of equations given by y = x² – 3x + 4 and x + y = 8, we need to substitute the expression for y from the first equation into the second equation.

Starting with the second equation:

x + y = 8

We can substitute y:

x + (x2 - 3x + 4) = 8

This simplifies to:

x2 - 2x + 4 = 8

Next, we move everything to one side of the equation:

x2 - 2x + 4 - 8 = 0

Which simplifies to:

x2 - 2x - 4 = 0

Now, we can use the quadratic formula to solve for x:

x = rac{-b pm sqrt{b^2 - 4ac}}{2a}

Here, a = 1, b = -2, and c = -4. Plugging in these values:

x = rac{2 pm sqrt{(-2)^2 - 4*1*(-4)}}{2*1}

This becomes:

x = rac{2 pm sqrt{4 + 16}}{2} = rac{2 pm sqrt{20}}{2} = 1 pm sqrt{5}

So we have two possible values for x:

x = 1 + sqrt{5} 
ewline x = 1 - sqrt{5}

Next, we need to find the corresponding y-values for each x by substituting back into the first equation:

y = (1 + sqrt{5})2 - 3(1 + sqrt{5}) + 4

After calculating:

y = 1 + 2sqrt{5} + 5 - 3 - 3sqrt{5} + 4 = 7 - sqrt{5}

For the second x value:

y = (1 - sqrt{5})2 - 3(1 - sqrt{5}) + 4

Doing the math gives:

y = 1 - 2sqrt{5} + 5 - 3 + 3sqrt{5} + 4 = 7 + sqrt{5}

Thus, the complete solution set is:

(1 + sqrt{5}, 7 - sqrt{5}) 
ewline (1 - sqrt{5}, 7 + sqrt{5})

These pairs represent the points at which both equations intersect, forming the solution set for the system.