# Solve the Following Equation and Also Check Your Result: ( 3 X + 1 ) 16 + ( 2 X − 3 ) 7 = ( X + 3 ) 8 + ( 3 X − 1 ) 14 - Mathematics

Sum

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

#### Solution

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

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#### APPEARS IN

RD Sharma Class 8 Maths
Chapter 9 Linear Equation in One Variable
Exercise 9.2 | Q 13 | Page 11