Solve each of the following equations and also check your results in each case:

(xix) $\frac{45 - 2 x}{15} - \frac{4 x + 10}{5} = \frac{15 - 14 x}{9}$

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Solution

Given : $\frac{45 - 2 x}{15} - \frac{4 x + 10}{5} = \frac{15 - 14 x}{9}$

$\frac{45 - 2 x}{15} - \frac{4 x + 10}{5} - \frac{15 - 14 x}{9} = 0$

$\frac{(45 - 2 x) 3 - (4 x + 10) 9 - (15 - 14 x) 5}{45} = 0$

$\frac{135 - 6 x - 36 x - 90 - 75 + 70 x}{45} = 0$

$\frac{28 x - 30}{45} = 0$

28x -30 = 0
28x = 30

$x = \frac{30}{28}$

$x = \frac{15}{14}$

To check : $\frac{45 - 2 x}{15} - \frac{4 x + 10}{5} = \frac{15 - 14 x}{9} f o r x = \frac{15}{14}$

$L . H . S . = \frac{45 - 2 x}{15} - \frac{4 x + 10}{5}$

$= \frac{45 - 2 (\frac{15}{14})}{15} - \frac{4 (\frac{15}{14}) + 10}{5}$

$= \frac{45 - \frac{15}{7}}{15} - \frac{\frac{30}{7} + 10}{5}$

$= \frac{300}{105} - \frac{100}{35}$

$= \frac{300 - 300}{105}$

= 0

$R . H . S = \frac{15 - 14 x}{9}$

$= \frac{15 - 14 (\frac{15}{14})}{9}$

$= \frac{15 - 15}{9}$

= 0

$∴ L . H . S . = R . H . S$

Hence, the given equation is verified for $x = \frac{15}{14} .$

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