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

I have this system of equations:




Is there a method of solving [x,y] without using x²=t, y²=z?

share|cite|improve this question
up vote 2 down vote accepted

Yes, since $x$ is nonzero (because of the second equation), we can eliminate $y=(60-2x^2)/(7x)$ and substitute this in the first equation, which gives $$ 25(x + 4)(x + 3)(x - 3)(x - 4)=0. $$

share|cite|improve this answer
All the answers were really informative but your answer was exactly what I needed, thank you. – Ofir Orpaz Elman Jul 14 '13 at 8:54

This is a rather ad hoc solution, and not at all pretty. I notice that if I add twice the second equation to the first, I get


or $7x^2+14xy+7y^2=175$, which readily simplifies to $x+y=\pm5$, or $y=\pm5-x$. Substituting those linear equations into either of the original equations gives you an easy quadratic in $x$.

share|cite|improve this answer

Multiply both sides of the first equation by $12$, of the second by $11$, and subtract. The idea is to obtain a homogeneous equation.

We get $14x^2-77xy+84y^2=0$. Rewrite as $(2x-3y)(x-4y)=0$.

Now express one variable, say $x$, in terms of the other, and substitute back in one of the equations.

Remark: The idea always works for systems of the shape $P(x,y)=a$, $Q(x,y)=b$, where $P(x,y)$ and $Q(x,y)$ are homogeneous quadratics in $x$ and $y$. In general after obtaining the homogeneous quadratic, we will need the Quadratic Formula.

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.