Using the property $\frac{a}{b} = \frac{a k}{b k}$ , fill in the blanks substituting proper numbers in the following.
$\frac{9}{14} = \frac{4.5}{. . .} = \frac{. . .}{42} = \frac{. . .}{3.5}$

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Solution

Given property $\frac{a}{b} = \frac{a k}{b k}$
$\frac{9}{14} = \frac{4.5}{. . . .} = \frac{. . . .}{42} = \frac{. . . .}{3.5}$
$\frac{9}{14} = \frac{9 (0.5)}{14 (0.5)} = \frac{4.5}{7.0}$ as $k = 0.5$
$\frac{9}{14} = \frac{9 (3)}{14 (3)} = \frac{27}{42}$ as $k + 3$
$\frac{9}{14} = \frac{9 (0.25)}{14 (0.25) = \frac{2.25}{3.5}}$ as $k = 0.25$
$∴ \frac{9}{14} = \frac{4.5}{7.0} = \frac{27}{42} = \frac{2.25}{3.5}$

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\frac{a}{b} = \frac{a k}{b k}

\frac{5}{7} = \frac{. . .}{28} = \frac{35}{. . .} = \frac{. . .}{3.5}

\frac{a}{b} = \frac{a k}{b k}

\frac{5}{7} = \frac{. . . .}{28} = \frac{35}{. . . .} = \frac{. . . .}{3.5}

\frac{9}{14} = \frac{4.5}{. . . .} = \frac{. . . .}{42} = \frac{. . . .}{3.5}

\frac{a}{b} = \frac{a k}{b k}

\frac{5}{7} = \frac{. . . .}{28} = \frac{35}{. . .}

\frac{9}{14} = \frac{4.5}{. .} = \frac{. .}{42} = \frac{. .}{3.5}

\frac{5}{12} = \frac{- - - - - -}{6}

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