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Solve for $x$ and $y$: \begin{cases} y &= 4x^2 - x - 6 \\ y &= 2 - x. \end{cases}

I have tried rearranging to get $x + y = 2$ and then substituting $y$ into it but hit a dead end. I'm pretty sure it turns into a quadratic equation. Any help would be appreciated thanks.

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What seems to be your problem? Your approach is one way of doing it.

You could also start by subtracting the two equations from another to get

$$0 = 4x^2-8.$$

This should also be what you arrive at using your method. Can you solve this now and then solve for $y$? Note that you can divide by $4$ and get:

$$x^2-2=0.$$

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    This helped me answer my question thanks! – 2017-02-02
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    I'm glad I was able to help you. – 2017-02-02
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You can substitute $y$ into the first equation to get:

$4x^2-x-6=2-x$

$4x^2-8=0$

Plus this into the quadratic formula where $a=4$, $b=0$, and $c=-8$

$x = \dfrac{-(0)\pm \sqrt{(0)^{2} - 4 (4)(-8)}}{2(4)}$

$x=\pm \sqrt {2}$

$y=2\pm \sqrt 2$

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    You don't need the quadratic formula. $4x^2 = 8 \iff x^2 = 2 \iff x = \pm \sqrt{2}$. – 2017-02-03