# Smarter way to solve a nonlinear system of two equations

I have the following nonlinear system of two equations \begin{align} 1+4x-y =\frac{49}{15}\qquad\text{ together with }\qquad \frac{x}{x-1} =\left(\frac{y}{y-1}\right)^4\\ \end{align}

I can solve it by substitution, and find the solution of the resulting cubic equation $$34 - 121 y + 144 y^2 - 46 y^3=0$$, yielding $$y=2$$, and thus $$x=16/15$$. However, the simplicity of the solution, and the form of the nonlinear equation of the two, suggests that a smarter and faster way than my brute force attempt may exist. I would appreciate any suggestion.

We have that

$$\frac{x}{x-1} =\left(\frac{y}{y-1}\right)^4 \iff \frac{x}{x-1} =\left(\frac{f(x)}{f(x)-1}\right)^4$$

with $$f(x)=4x-\frac{34}{15}$$.

Since

• $$g(x)=\frac{x}{x-1}:(1,\infty)\to (1,\infty)$$ is bijective
• $$f(x)=4x-\frac{34}{15}$$ is bijective
• $$h(x)=\sqrtx:[0,\infty)\to [0,\infty)$$ is bijective

we expect exactly one solution that is indeed $$x=\frac{16}{15}$$.

Eliminating $$y$$ from your system you will get $$2484000 x^3-6166800 x^2+5004495 x=1336336$$ with only one real solution $$x=\frac{16}{15}$$

• Yes, I know...you would need to solve a cubic equation, though. What I am asking is: is there a smarter way that avoids cubic equations? – Pierpaolo Vivo Dec 2 '19 at 18:05
• I think there isn't. – Dr. Sonnhard Graubner Dec 2 '19 at 18:07

Probably not "faster", but: suppose you guess that there is a rational solution. If $$y/(y-1) = \alpha/\beta$$ with $$\alpha$$ and $$\beta$$ coprime integers, $$x/(x-1) = \alpha^4/\beta^4$$. Then $$y = \alpha/(\alpha - \beta)$$ and $$x = \alpha^4/(\alpha^4 - \beta^4)$$, so your first equation becomes $$1 + \frac{4 \alpha^4}{\alpha^4-\beta^4} - \frac{\alpha}{\alpha- \beta} = \frac{49}{15}$$ Now $$\alpha^4 - \beta^4 = (\alpha - \beta)(\alpha^3 + \alpha^2 \beta + \alpha \beta^2 + \beta^3)$$, where $$\alpha^3 + \alpha^2 \beta + \alpha \beta^2 + \beta^3$$ is coprime to $$\alpha$$. Unless $$\alpha = 0$$ (which certainly does not work), the denominator of the left side will be divisible by $$\alpha^3 + \alpha^2 \beta + \alpha \beta^2 + \beta^3$$, and the only way to make this equation work is that this factor divides $$15$$. It's easy to check that $$\alpha = 2$$, $$\beta = 1$$ makes $$\alpha^3 + \alpha^2 \beta + \alpha \beta^2 + \beta^3 = 15$$, and this does happen to satisfy the equation.

• Would you not need to check all factors of $60$ because the expression could have a common factor of $4$ with the numerator of $4\alpha^4$? – S. Dolan Dec 2 '19 at 19:13

For a rational solution, let $$\frac{y}{y-1}=\frac{u}{v}$$ with $$u$$ and $$v$$ coprime. Then $$11u^4-15u^3v-15u^2v^2-15uv^3+34v^4=0.$$ Therefore $$u$$ is a factor of $$34$$ and $$v$$ is a factor of $$11$$.

The solution is $$u=2,v=1$$.