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Okay so I am normally good at these kinds of things but I received this problem that even the top people in my class had trouble solving. The problem is that everyone is getting different results. We check through each others logic yet we cannot find a flaw. I asked my math teacher and he said that the answer was not pretty. So how do you solve these sets of equations simultaneously?

$$\begin{align}y&=10x-3\\y&=x^2-3x\end{align}$$


What I got

$$x=\frac{13\pm\sqrt{157}}{2}$$

But I am not sure if this is right.

How would you go around solving this?

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  • $\begingroup$ Did you try feeding your solution back into the original equations? $\endgroup$ – Mark Bennet Sep 22 '13 at 19:43
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To check your answers, substitute into the original equations and check whether or not they are satisfied. Your answer is correct!

Regarding "how to proceed":

Put $y = y$: $$10x - 3 = x^2 - 3x\iff x^2 - 13x + 3=0$$

Solve the resulting quadratic equation (find the zeros), which I'm assuming you did!

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  • $\begingroup$ This needs a TU +1 $\endgroup$ – Amzoti Sep 23 '13 at 0:13
  • $\begingroup$ Thanks for the support, @Amzoti! $\endgroup$ – Namaste Sep 23 '13 at 0:13
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You would set the two equations equal to each other since each equation is equal to $y$. Then, $10x-3=x^2-3x \implies x^2-13x+3=0 \implies x=\frac{13\pm\sqrt{157}}{2}$. Your answer is indeed ugly, but correct.

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