# Can the Elimination Method be used to solve a system of polynomial equations?

I just learned about this technique here, but everything I find online only mentions linear equations specifically. What about systems of polynomial equations? If it doesn't work, why not?

More info: I have solved a system of polynomial equations using the elimination method and it seems to be giving me incorrect results, so I'm wondering if this is the cause. It seems to have reduced the degree of the resulting polynomial equation from 4 to 2, which seems fishy. $$|v_n|^2 = \mathbf{v_{xn}}^2 + \mathbf{v_{yn}}^2 \tag 1$$ $$|v_e|^2 = \mathbf{v_{xe}}^2 + \mathbf{v_{ye}}^2 \tag 2$$ $$p_x = m_e \mathbf{v_{xe}} + m_n \mathbf{v_{xn}} \tag 3$$ $$p_y = m_e \mathbf{v_{ye}} + m_n \mathbf{v_{yn}} \tag 4$$ Unknowns are bolded. I solved it mainly by using substitution, but at one point I used (2) to eliminate a pair of squared terms, leaving me with a linear equation on one side and a square root on the other, which, after squaring both sides, turned into a quadratic equation. I was then able to solve it using the quadratic formula, but the roots seems wrong when I substitute them back in.

(This is a followup to another question I had, if you're curious where the equations came from).

• maybe you should type in your system. – Will Jagy Sep 8 '18 at 23:46
• sure thing, edited – Ken Sep 9 '18 at 0:15
• Elimination can be used with any system of equations in general. Why it seems to be giving me incorrect results cannot be answered unless you post your calculations. – dxiv Sep 9 '18 at 0:37
• Thanks, that's the answer I needed. Now I know that that shouldn't be the problem. As you can imagine, the calculations get really hairy. I don't think I can write it all out nor expect people to trawl through it all looking for errors. I'll just keep looking for errors in my logic/understanding, and perhaps posting new, bite-sized questions like this one if necessary. – Ken Sep 9 '18 at 0:44
• @Ken It's not all that bad and, yes, in the end you should get a quadratic in one of the variables. – dxiv Sep 9 '18 at 0:56

Let the unknowns be $\,a,b,c,d\,$ and divide the linear equations by $\,m_e\,$, then the system is of the form: \begin{align} a^2 + b^2 &= r \tag 1 \\ c^2 + d^2 &= s\tag 2 \\ c + m a &= p \tag 3 \\ d + m b &= q \tag 4 \end{align}

Substituting $\,c=p-ma\,$ and $\,d=q-mb\,$ from $\,(3)$-$(4)\,$ into $\,(2)\,$ gives:

\begin{align} s &= (p-ma)^2+(q-mb)^2 \\ &= p^2+q^2 - 2m(pa+qb)+m^2(\color{blue}{a^2+b^2}) \\ &= p^2+q^2 +m^2 \color{blue}{r} - 2m(pa+qb) \tag{5} \end{align}

Rearranging $\,(5)\,$:

$$2mq\,b = p^2+q^2 +m^2r -s - 2mp\,a \;\;\iff\;\; b = \lambda - \mu a\tag{6}$$

Substituting $\,b\,$ from $\,(6)\,$ into $\,(1)\,$ gives the quadratic in $\,a\,$:

$$a^2 + (\lambda - \mu a)^2 = r \tag{7}$$

• Where does $b = \lambda - \mu a$ come from and what are $\lambda$ and $\mu$? – Ken Sep 9 '18 at 10:20
• @Ken Divide the previous equation by $\,2mq\,$ and define: $$\require{cancel} \lambda = \frac{p^2+q^2 +m^2r -s - 2mp}{2mq}\,,\;\; \mu = \frac{\bcancel{2m}p}{\bcancel{2m}q}$$ – dxiv Sep 9 '18 at 16:03
• This worked, thanks! Continuing.... Get equation 7 in quadratic equation form: $$a^2 + \mu^2a^2 - 2\lambda\mu a + \lambda^2 - r = 0 \\ (1+\mu^2)a^2 - 2\lambda\mu a + \lambda^2 - r = 0$$ then solve using quadratic formula. – Ken Sep 20 '18 at 0:52

What you have is two unknown vectors $v,w\in\Bbb R^2$ with $$\|v\|^2=c_1^2,\quad \|w\|^2 = c_2^2,\quad m_1v+m_2w = z,$$ where $z\in\Bbb R^2$ is constant and also $c_1,c_2,m_1,m_2\in\Bbb R$ are constant. Hence, $$m_2^2c_2^2 = \|m_2w\|^2 = \|z-m_1v\|^2 = \|z\|^2-2m_1\langle z,v\rangle + m_1^2\|v\|^2,$$ i.e., $$2m_1\langle z,v\rangle = \|z\|^2+m_1^2c_1^2-m_2^2c_2^2.$$ This gives you an additional linear equation. Hence, you have a linear system with 4 unknowns and 3 equations, that is, one free scalar parameter. Solve the linear system w.r.t. this parameter and adapt it accordingly to the quadratic equations.