Question

Solve the following equations:
$\frac{3t+1}{16}-\frac{2t-3}{7}=\frac{t+3}{8}+\frac{3t-1}{14}$

Answer 1

Given $\frac{3t+1}{16}-\frac{2t-3}{7}=\frac{t+3}{8}+\frac{3t-1}{14}$

$\Rightarrow \frac{3t+1}{16}-\frac{t+3}{8}=\frac{2t-3}{7}+\frac{3t-1}{14}$

$\Rightarrow \frac{3t+1}{16}-\frac{(t+3)\times 2}{8\times 2}=\frac{(2t-3)\times 2}{7\times 2}+\frac{3t-1}{14}$

$\Rightarrow \frac{3t+1}{16}-\frac{2(t+3)}{16}=\frac{2(2t-3)}{14}+\frac{3t-1}{14}$

$\Rightarrow \frac{3t+1-2(t+3)}{16}=\frac{2(2t-3)+3t-1}{14}$

$\Rightarrow \frac{3t+1-2t-6}{16}=\frac{4t-6+3t-1}{14}$

$\Rightarrow \frac{t-5}{8}=\frac{7t-7}{7}$

$\Rightarrow \frac{t-5}{8}=\frac{7(t-1)}{7}$

$\Rightarrow t-5=8t-8$

$\Rightarrow 7t=-5+8$

$\therefore t=\frac{3}{7}$