1. ## solving quadratic system

Hi;
I have the equations y = 2x^2 - 2x +3 and y = x^2 + 5x -7 therefor they are both equal so set them equal and solve I'm ok with that but...

I thought I could set them both to zero and solve for x by elimination ie
2x^2 - 2x + 3 = 0
x^2 + 5x - 7 = 0

giving
2x^2 - 2x = -3
x^2 + 5x = 7

multiplt the second equation by 2

2x^2 - 2x = -3
2x^2 + 10x = 14

I thought this would let me eliminate the x^2 terms and solve for x but I get the wrong solution and I don't know why?

Thanks

2. ## Re: solving quadratic system

Since they are both equal to $\displaystyle y$ then they are equal. This means

\displaystyle \begin{align*} 2x^2 - 2x + 3 &= x^2 + 5x - 7 \\ x^2 - 7x + 10 &= 0 \\ (x - 2)(x - 5) &= 0 \\ x = 2\textrm{ or }x &= 5\end{align*}

Now substitute each point into one of the equations to find the $\displaystyle y$ values.

3. ## Re: solving quadratic system

Is this the method for all quadratic systems that are equal.

4. ## Re: solving quadratic system

The problem with your method is you assume $y=0$, but that's not given so the method Prove It has given you is a general method to solve exercices like this one.

5. ## Re: solving quadratic system

Originally Posted by anthonye
Is this the method for all quadratic systems that are equal.
It works for any kinds of systems that are equal. If y= A and y= B then A= B.