Solving system of equations with fraction

Question

I am having difficulties solving the following system :

$u \neq t$ and $(t, u) \in \mathbf{R} - \{-1, 1\}$

$\frac{t}{t^2-1}-\frac{u}{u^2-1}=0$

$\:\frac{t^2}{t-1}-\frac{u^2}{u-1}=0$

I tried expanding everything, but I still can't achieve anything

$\frac{tu^2-t-ut^2+u}{\left(t^2-1\right)\left(u^2-1\right)}$

$\frac{t^2u-t^2-u^2t+u^2}{\left(t-1\right)\left(u-1\right)}$

also tried the hint below, ending up finding one equation with the two variables $t(t+1) = u(u+1)$ ... and $u \neq t$

Any idea ?

Answer 1

$$ \begin{cases} \frac{t}{t^2-1}-\frac{u}{u^2-1}=0 \\ \frac{t^2}{t-1}-\frac{u^2}{u-1}=0 \\ \end{cases} \begin{array}{c} \overset{\times(t^2-1)(u^2-1)\ne0}{\longrightarrow} \\ \overset{\times(t-1)(u-1)\ne0}{\longrightarrow}\\ \end{array} \begin{cases} t(u^2-1)-u(t^2-1)=0 \\ t^2(u-1)-u^2(t-1)=0 \\ \end{cases} $$ $$ \Rightarrow \begin{cases} tu^2-t-ut^2+u=0 \\ t^2u-t^2-u^2t+u^2=0 \\ \end{cases} \Rightarrow \begin{cases} tu(u-t)+u-t=0 \\ tu(t-u)-(t-u)(t+u)=0 \\ \end{cases} $$ $$ \Rightarrow \begin{cases} (u-t)(tu+1)=0 \begin{cases} u-t=0 \Rightarrow u=t \text{ incorrect} \\ tu+1=0 \Rightarrow tu=-1 \\ \end{cases}\\ (t-u)(tu-(t+u))=0 \begin{cases} t-u=0 \Rightarrow u=t \text{ incorrect} \\ tu-(t+u)=0 \Rightarrow tu=t+u \\ \end{cases} \end{cases} $$

$$ \Rightarrow \begin{cases} tu=-1 \\ t+u=tu \\ \end{cases} \Rightarrow \begin{cases} tu=-1 \\ t+u=-1 \\ \end{cases} \Rightarrow u^2+u-1=0 \Rightarrow \text{...} $$

Answer 2

Hint

$$\dfrac{\dfrac{t^2}{t-1}}{\dfrac t{t^2-1}}=?$$

$$\implies t(t+1)=u(u+1)$$

But $u\ne t$