Simultaneous Equations help please?

y=x+2

y=3x^2

So all you have to do here, is substitue 3x^2 in to the first equation's 'y':

3x^2=x+2

3x^2-x-2=0

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

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

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

x=-2/3 or 1

Now sub 1 in to the first equatoin for x:

y=3

So one solution could be: (1,3)

Now let's sub -2/3 in to the first equation for x:

y=-2/3+2

y=4/3

And another solution is:

(-2/3, 4/3)

Have a nice day :)

if y=x+2 and y=3x^2, that means that x+2=3x^2, move them all to one side and you get 3x^2-x-2=0.

This factors to (3x+2)(x-1)=0. In order for that expression to equal zero, one of the two factors must equal zero. So set 3x+2=0 and you solve that x=-2/3 also set x-1=0 and you get x=1.

So your answers are x=-2/3 and x=1

substitute 3x^2 for y in eq # 1 to get 3x^2 = x + 2 so 3x^2 - x - 2 = 0

using then quadratic formula x = (1 +/- sqrt(1 + 24)/6 = (1 +/- 5)/6 = 1, -2/3

y=x+2

y=3x^2

=> x+2 = 3x^2

0 = 3x^2 - x - 2

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

x = -2/3 and 1, => y = 4/3 and 3

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