The set of simultaneous equations to be solved is:

y=x^2-3x+1 ...(1)

y=2x^2+x+4 ...(2)

Equate (1) and (2)

=> x^2-3x+1 = 2x^2+x+4

=> x^2 + 4x + 3 = 0

=> x^2 + 3x + x + 3 = 0

=> x(x + 3) + 1(x + 3) = 0

=> (x + 1)(x + 3) = 0

We get x = -1 and x = -3

Substitute in (1)

For x = -1, y = 1 + 3 + 1 = 5

For x = -3, y = 2*9 - 3 + 4 = 18 + 1 = 19

**The solution of the equations are (-1, 5) and (-3 , 19)**

We'll equate the given equations and we'll get:

x^2-3x+1 = 2x^2+x+4

We'll move all terms to the left:

x^2 - 2x^2 - 3x - x + 1 - 4 = 0

-x^2 - 4x - 3 = 0

We'll multiply by -1:

x^2 + 4x + 3 = 0

We'll apply quadratic formula:

x1 = [-4+sqrt(16 - 12)]/2

x1 = (-4+2)/2

x1 = -1

x2 = -3

Therefore y1 = 1 + 3 + 1 = 5 and y2 = 9 + 9 + 1 = 19

**The solutions of the equation are represented by the pairs: (-1 ; 5) and (-3 ; 19).**