Question

Solve the following equation and also check your result:
\[\frac{(3x + 1)}{16} + \frac{(2x - 3)}{7} = \frac{(x + 3)}{8} + \frac{(3x - 1)}{14}\]

Answer 1

\[\frac{3x + 1}{16} + \frac{2x - 3}{7} = \frac{x + 3}{8} + \frac{3x - 1}{14}\]
\[\text{ or }\frac{3x + 1}{16} - \frac{x + 3}{8} = \frac{3x - 1}{14} - \frac{2x - 3}{7}\]
\[\text{ or }\frac{3x + 1 - 2x - 6}{16} = \frac{3x - 1 - 4x + 6}{14}\]
\[\text{ or }\frac{x - 5}{8} = \frac{- x + 5}{7}\]
\[\text{ or }7x - 35 = - 8x + 40\]
\[\text{ or }15x = 75\]
\[\text{ or }x = \frac{75}{15} = 5\]
\[\text{ Check: }\]
\[\text{ L . H . S . }= \frac{3 \times 5 + 1}{16} + \frac{2 \times 5 - 3}{7} = \frac{16}{16} + \frac{7}{7} = 2\]
\[\text{ R . H . S . }= \frac{5 + 3}{8} + \frac{3 \times 5 - 1}{14} = \frac{8}{8} + \frac{14}{14} = 2\]
∴ L.H.S. = R.H.S. for x = 5