Question:
$\frac{3 t+5}{4}-1=\frac{4 t-3}{5}$
Solution:
Given, $\frac{3 t+5}{4}-1=\frac{4 t-3}{5}$
$\Rightarrow$ $\frac{3 t+5-4}{4}=\frac{4 t-3}{5}$
$\Rightarrow$ $5(3) t+5-4)=4(4 t-3)$ [by cross-multiplication]
$\Rightarrow \quad 5(3 t+1)=4(4 t-3)$
$\Rightarrow$ $15 t+5=16 t-12$
$\Rightarrow$ $15 t-16 t=-12-5$ [transposing 16 tto LHS and 5 to RHS]
$\Rightarrow$ $-t=-17$
$\Rightarrow$ $\frac{-t}{-1}=\frac{-17}{-1}$ [dividing both sides by $-1$ ]
$\therefore$ $t=17$
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