# Solving system of equations with fraction

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 ?

• It is a quadratic equation in $t$, which you can solve. Sep 12, 2020 at 17:52
• @nginx9101 Show please how you made this expanding. Sep 12, 2020 at 17:53

Hint

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

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

But $$u\ne t$$

• I don't get what you are trying to do?.. Details please? Sep 12, 2020 at 18:56
• @nginx9101, replace the values of the numerator and denominator Sep 12, 2020 at 19:01
• well yes, but your reasoning isn't complete yet. What to do after ? I end up having 2 variables which I can't solve Sep 12, 2020 at 19:09

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

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