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I have this system of equations:

3x²+7y²=55

and

2x²+7xy=60

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

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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 $$ 25(x + 4)(x + 3)(x - 3)(x - 4)=0. $$

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  • $\begingroup$ All the answers were really informative but your answer was exactly what I needed, thank you. $\endgroup$ – Ofir Orpaz Elman Jul 14 '13 at 8:54
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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

$$\left(3x^2+7y^2\right)+2\left(2x^2+7xy\right)=55+2\cdot60\;,$$

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 $x$.

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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.

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