Question

$\text{Solve for the value of t.}\frac{3t+1}{16}-\frac{3t-3}{7}=\frac{t-4}{8}+\frac{3t+1}{14}$

Answer 1

$\text{The given equation is,}\frac{3t+1}{16}-\frac{3t-3}{7}=\frac{t-4}{8}+\frac{3t+1}{14}$

$\text{On arranging the terms according to}\text{the denominator, we get}$

$\Rightarrow \frac{3t+1}{16}-\frac{t-4}{8}=\frac{3t-3}{7}+\frac{3t+1}{14}\Rightarrow \frac{3t+1-2(t-4)}{16}=\frac{2(3t-3)+3t+1}{14}\Rightarrow \frac{3t-2t+1+8}{16}=\frac{6t+3t-6+1}{14}\Rightarrow \frac{t+9}{16}=\frac{9t-5}{14}$

On cross multiplication, we have
$\Rightarrow 14(t+9)=16(9t-5)\Rightarrow 14t+126=144t-80\Rightarrow 14t-144t=-126-80\Rightarrow -130t=-46\Rightarrow t=\frac{-46}{-130}\Rightarrow t=\frac{23}{65}$