4t-3-(3t+1)=5t-4

Given, $4 t-3-(3 t+1)=5 t-4$

$\Rightarrow \quad 4 t-3-3 t-1=5 t-4$

$\Rightarrow \quad t-4=5 t-4$

$\Rightarrow \quad t-5 t=-4+4$ [transposing $5 t$ to LHS and $-4$ to RHS]

$\Rightarrow \quad-4 t=0$ 

$\Rightarrow$ $\frac{-4 t}{-4}=\frac{0}{-4}$ [dividing both sides by $-4$ ]

$\therefore$ $t=0$