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

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Solution

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

$\Rightarrow \frac{3 x + 1}{16} + \frac{2 x - 3}{7} - \frac{x + 3}{8} - \frac{3 x - 1}{14} = 0$

$\Rightarrow \frac{(3 x + 1) 7 + (2 x - 3) 16 - (x + 3) 14 - (3 x - 1) 8}{112} = 0$

$\Rightarrow \frac{21 x + 7 + 32 x - 48 - 14 x - 42 - 24 x + 8}{112} = 0$

$\Rightarrow \frac{15 x - 75}{112} = 0$

$\Rightarrow 15 x - 75 = 0$

$\Rightarrow 15 x = 75$

$\Rightarrow x = 5$

To check : $\frac{3 x + 1}{16} + \frac{2 x - 3}{7} = \frac{x + 3}{8} + \frac{3 x - 1}{14}$ for $x = 5$

$L . H . S = \frac{3 x + 1}{16} + \frac{2 x - 3}{7}$

$= \frac{3 (5) + 1}{16} + \frac{2 (5) - 3}{7}$

$= \frac{15 + 1}{16} + \frac{10 - 3}{7}$

$= \frac{16}{16} + \frac{7}{7}$

$= 1 + 1$

$= 2$

$R . H . S = \frac{x + 3}{8} + \frac{3 x - 1}{14}$

$= \frac{5 + 3}{8} + \frac{3 (5) - 1}{14}$

$= \frac{8}{8} + \frac{15 - 1}{14}$

$= \frac{8}{8} + \frac{14}{14}$

$= 1 + 1$

$= 2$

$∴ L . H . S = R . H . S$

Hence, the given equation is verified for $x = 5$ .

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Similar questions

\frac{(3 x + 1)}{16} + \frac{(2 x - 3)}{7} = \frac{(x + 3)}{8} + \frac{(3 x - 1)}{14}

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