Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


What I got


But I am not sure if this is right.

How would you go around solving this?

share|cite|improve this question
Did you try feeding your solution back into the original equations? – Mark Bennet Sep 22 '13 at 19:43
up vote 3 down vote accepted

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!

share|cite|improve this answer
This needs a TU +1 – Amzoti Sep 23 '13 at 0:13
Thanks for the support, @Amzoti! – amWhy Sep 23 '13 at 0:13

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.