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I keep trying to solve this problem, but i keep on getting crazy answers, i think i am right up to a certain point and then doing something wrong, the question is to solvie this :

$$ \begin{align*} 2x^2 + 3y + z &= 8\\ x - 2y &= 4\\ 3z - x &= 7 \end{align*} $$

I've tried many things but can't work out what i'm doing wrong, any help will be appreciated.

Thanks in advance.

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You can use $x-2y=4$ to get $\frac{1}{2}x-2=y$. You can use $3z-x=7$ to get $z=\frac{7}{3}+\frac{1}{3}x$. Plug these into $2x^2+3y+z=8$ to get

$$ 2x^2+3\left(\frac{1}{2}x-2\right) +\frac{7}{3}+\frac{1}{3}x=8. $$

Simplify and use quadratic formula to find values of $x$. Then use the above equations to get $z$ and $y$.

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  • $\begingroup$ awesome that worked, i was doing it where i multiplied the top so the other 2 fitted nice, and i must of done that wrong, so i kept getting the wrong answer, Thanks for your help! $\endgroup$
    – Alex Trott
    Mar 19 '12 at 15:47

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