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My problem is the following. I want to determine if there is a solution $x\geq 0$ to equations of the following general type: $$W=\sum_{i=1}^n\frac{a_i}{b_i+c_ix},$$ where $W\geq 0$, $a_i\geq 0$, $b_i\geq 0$ and $c_i\geq 0$.

For example, the following equation $$3=\frac{2}{1+x} + \frac{5}{5+2x} + \frac{1}{7+x} + \frac{10}{3+3x}$$ has one solution for $x\geq 0$, which is around $1.38$. I know I could solve such an equation by writing the equation with a common denominator and then by using a numerical solver for the numerator. But is there, by any chance, a closed form solution to such equations?

I also noticed something special but I am not able to prove it. Is there a way to show that there is at most one solution to such equations with $x\geq 0$?

Thank you very much for your help!

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    Yes, just multiply to common denominator. – 2017-01-15
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    This question is basically the same as asking: Can I find the roots of a $n$ degree polynomial algebraically? – 2017-01-15
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    All the terms on the right are decreasing with $x$, so there will be at most one solution. The right is very close to zero when $x$ is large, so if the right is larger than the left at $x=0$ there will be exactly one solution. – 2017-01-15
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    See my answer to [this recurring question](http://math.stackexchange.com/q/1941871). – 2017-01-15

3 Answers 3

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$$f'(x):=\left(\sum_{i=1}^n\frac{a_i}{b_i+c_ix}\right)'=-\sum_{i=1}^n\frac{a_ic_i}{(b_i+c_ix)^2}<0$$ for all $x\ge0$.

As this function is strictly decreasing, the equation cannot have more than one solution. And it has one when

$$f(0)=\sum_{i=1}^n\frac{a_i}{b_i}\ge W,$$ because $f(\infty)=0$.

As said by others, the equation is an algebraic one so there is no closed formula for $n>4$.

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    I am sorry but this is **not true**: this equation has 4 real roots. In fact, f is not defined on all $\mathbb{R}$. See my answer to [this equation](http://math.stackexchange.com/q/1941871) – 2017-01-15
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    @JeanMarie What you missed (and Yves did not bother to point out to you) is the restriction $x>0$ – 2017-01-15
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    @leonbloy You are perfectly right. I will look in a more attentive way next time... – 2017-01-15
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moving $3$ to the left and clearing the denominators we get $$-1/3\,{\frac {18\,{x}^{3}+136\,{x}^{2}+41\,x-365}{ \left( 1+x \right) \left( 5+2\,x \right) \left( 7+x \right) }} =0$$

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Your equation is equivalent to the cubic polynomial equation $$ - 18x^3 - 136x^2 - 41x + 365=0, $$ together with the restriction $(1+x)(7+x)(5+2x)\neq 0$.

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    With $x\ge0$, the restriction holds. – 2017-01-15