Question

Solve each of the following equation and also check your result in each case:
$\frac{(3x+1)}{16}+\frac{(2x-3)}{7}=\frac{(x+3)}{8}+\frac{(3x-1)}{14}$

Answer 1

$$\begin{gathered} \frac{3\mathrm{x}+1}{16}+\frac{2\mathrm{x}-3}{7}=\frac{\mathrm{x}+3}{8}+\frac{3\mathrm{x}-1}{14} \\ \mathrm{or}\frac{3\mathrm{x}+1}{16}-\frac{\mathrm{x}+3}{8}=\frac{3\mathrm{x}-1}{14}-\frac{2\mathrm{x}-3}{7} \\ \mathrm{or}\frac{3\mathrm{x}+1-2\mathrm{x}-6}{16}=\frac{3\mathrm{x}-1-4\mathrm{x}+6}{14} \\ \mathrm{or}\frac{\mathrm{x}-5}{8}=\frac{-\mathrm{x}+5}{7} \\ \mathrm{or}7\mathrm{x}-35=-8\mathrm{x}+40 \\ \mathrm{or}15\mathrm{x}=75 \\ \mathrm{or}\mathrm{x}=\frac{75}{15}=5 \\ \mathrm{Check}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{3\times 5+1}{16}+\frac{2\times 5-3}{7}=\frac{16}{16}+\frac{7}{7}=2 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{5+3}{8}+\frac{3\times 5-1}{14}=\frac{8}{8}+\frac{14}{14}=2 \end{gathered}$$

$\therefore$ L.H.S. = R.H.S. for x = 5