Question:
4t-3-(3t+1)=5t-4
Solution:
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 $-4$ ]
$\therefore$ $t=0$
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