Solve the system of equations y+3x-x^2=1 and y-x=2x^2+4.

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

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

let us rewrite (1) and (2) as function os
x.

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

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

Then (1) =
(2).

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

==> Combine like
terms.

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

Now we
will find the roots.

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

==> x1= -1 ==> y1 = x^2 -3x + 1 = 1+3+1=
5

==> x2= -3 ==> y2= 9 + 9 +1 =
19

Then we have two sets of solutions to the
equation.

==> The solution is the pair ( -1, 5)
and the pair ( -3, 19).