To solve this system of equations, we can use substitution or elimination method.

First, we rearrange the first equation to solve for y:

3x + y = 10
y = 10 - 3x

Next, substitute the expression for y into the second equation:

x^2 - $10-3x$ = 8
x^2 - 10 + 3x = 8
x^2 + 3x - 18 = 0

Now, we can factor the quadratic equation:

$x+6$$x-3$ = 0

Setting each factor to zero:

x + 6 = 0
x = -6

x - 3 = 0
x = 3

Now that we have the values for x, we can substitute them back into either equation to solve for y:

For x = -6:
y = 10 - 3$-6$ y = 10 + 18
y = 28

For x = 3:
y = 10 - 3$3$ y = 10 - 9
y = 1

Therefore, the solutions to the system of equations are x = -6, y = 28 and x = 3, y = 1.