Is it possible to split this fraction?

Question

I am puzzled with this problem.

$\cfrac{1}{b+c}=\cfrac{x_o}{b}+\cfrac{x_1}{c}$

Is it possible to define $x_0$ and $x_1$ in terms of $b$ and $c$? If so, could I at least get a hint?

Answer 1

Assuming $b$, $c$, and $b+c$ are all nonzero, then there are infinitely many choices for $x_0$ and $x_1$. For example, $x_0=0$ and $x_1=c/(b+c)$.

In general, you can choose any $x_0$ you want and then $$x_1=c\left(\frac{1}{b+c}-\frac{x_0}{b}\right).$$

Answer 2

Let $b, c \in \Bbb R^*$ such that $b+c\neq 0$. We are then looking for $x_0, x_1 \in \mathbb R$, if possible , such that $$\cfrac{1}{b+c}=\cfrac{x_o}{b}+\cfrac{x_1}{c}$$ Then, $\frac{1}{b+c}=\frac{cx_0+bx_1}{bc}$. Then, $(cx_0+bx_1)(b+c)=bc=bcx_0+b^2x_1+c^2x_0+bcx_1=bc(x_0+x_1)+c^2x_0+b^2x_1$.

Let's then take $x_1=\frac{-c^2}{b^2-c^2}$ and $x_0=1-x_1$, provided $b\neq c $, it should work. Let's try !

Let us take $b=7$ and $c=2$. You obtain $x_1=-\frac{4}{45}$ and $x_0=\frac{49}{45}$ : it works.