simultaneous equations! please help!?

Question

y+x=4
y=3x^2+2

help!

Answer 1

y+x=4
y=3x^2+2


You know what y equals (y=3x^2+2) so put that in place of the y in the first equation

3x^2+2+x=4

subtract 2 from each side

3x^2+x = 2

subtract x from each side (and for purposes here, I'm writing out x^2)

3(x)(x) = 2 - x

divide each side by 3

(x)(x) = 2/3 - x/3

divide each side by x

x = 2/3x - x/3x
becomes
x = 2/3x - 1/3

subtract 2/3x

1/3(x) = - 1/3

multiply each side by 3

x = -1

Check it by plugging it into the equation

3x^2+2+x=4

3(-1)(-1) + 2 + (-1) = 4
3 + 2 -1 = 4
5 - 1 = 4
4=4

So if x = -1, you can then find y

y=3x^2+2

y = 3(-1)(-1) + 2
y = 3 +2
y = 5

End result, x = -1, y = 5

Answer 2

It's probably easiest to solve this by substitution.

y + x = 4
y = 4 - x

Plug this into the other equation:

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

Rearrange

3x^2 + x - 2 = 0

Use the quadratic formula. a = 3, b = 1, c = -2
Solving the quadratic formula gives x = 2/3 or -1
Plug these into the first equation, y + x = 4

If x = 2/3, then y = 10/3
If x = -1, then y = 5

Answer 3

y + x = 4

so,

y = 4-x

Therefore

y = 3x^2 + 2

so,

4-x = 3x^2 + 2

2 = 3x^2 + x
0 = 3x^2 +x -2
0 = (x+1)(3x-2)

x = -1
x = 2/3

So, back to the original equation
y+x = 4
y+(-1) = 4
y-1 = 4
y = 5

And,

y + 2/3 = 4
y = 4 - 2/3
y = 3 1/3, or (10/3)

Therefore the two points are (-1, 5) and (2/3, 3 1/3)

Answer 4

4 - x = 3x² + 2
3x² + x - 2 = 0
(3x - 2) (x + 1) = 0
x = 2 / 3 , x = - 1
y = 10/3 , y = 5

(2/3,10/3) , (-1,5)