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

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### 2 Answers

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).**

We'll isolate y to the left side, moving all the rest of terms to the right side:

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

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

We'll equate (1) = (2).

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

We'll subtract boths sides x^2-3x+1 and we'll use symmetric property:

x^2 + 4x + 3 = 0

We'll apply quadratic formula:

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

x1 = (-4+2)/2

x1 = -1

x2 = (-4-2)/2

x2 = -3

For x1 = -1, y1 = 1+3+1 = 5

For x2 = -3, y = 9+9+1 = 19

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