Answer
$t=\dfrac{1}{2},\text{ }2$
Work Step by Step
Multiplying both sides of the given equation, $ \dfrac{2}{t}+\dfrac{1}{t-1}=2 ,$ by the $LCD=t(t-1)$ results to \begin{align*} t(t-1)\left(\dfrac{2}{t}+\dfrac{1}{t-1}\right)&=2(t)(t-1) \\\\ (t-1)(2)+t(1)&=2t^2-2t \\ 2t-2+t&=2t^2-2t \\ 3t-2&=2t^2-2t \\ 0&=2t^2-2t-3t+2 \\ 2t^2-5t+2&=0 .\end{align*} Using the factoring of quadratic trinomials, the factored form of the equation above is \begin{align*} (2t-1)(t-2)&=0 .\end{align*} Equating each factor to zero (Zero Product Property) and solving for the variable, then \begin{align*} 2t-1&=0 \\ 2t&=1 \\\\ \dfrac{2t}{2}&=\dfrac{1}{2} \\\\ t&=\dfrac{1}{2} \\\\\text{OR}\\\\ t-2&=0 \\ t&=2 .\end{align*} Hence, the solutions to the equation $ \dfrac{2}{t}+\dfrac{1}{t-1}=2 $ are $ t=\dfrac{1}{2},\text{ }2$.
