$$\frac{5}{5}(5f-3)=\frac{1}{20}(10f-9)$$
$f=\frac{17}{30}\approx 0.566666667$
$$1\left(5f-3\right)=\frac{1}{20}\left(10f-9\right)$$
$$5f-3=\frac{1}{20}\left(10f-9\right)$$
$$5f-3=\frac{1}{20}\times 10f+\frac{1}{20}\left(-9\right)$$
$$5f-3=\frac{10}{20}f+\frac{1}{20}\left(-9\right)$$
$$5f-3=\frac{1}{2}f+\frac{1}{20}\left(-9\right)$$
$$5f-3=\frac{1}{2}f+\frac{-9}{20}$$
$$5f-3=\frac{1}{2}f-\frac{9}{20}$$
$$5f-3-\frac{1}{2}f=-\frac{9}{20}$$
$$\frac{9}{2}f-3=-\frac{9}{20}$$
$$\frac{9}{2}f=-\frac{9}{20}+3$$
$$\frac{9}{2}f=-\frac{9}{20}+\frac{60}{20}$$
$$\frac{9}{2}f=\frac{-9+60}{20}$$
$$\frac{9}{2}f=\frac{51}{20}$$
$$f=\frac{51}{20}\times \frac{2}{9}$$
$$f=\frac{51\times 2}{20\times 9}$$
$$f=\frac{102}{180}$$
$$f=\frac{17}{30}$$
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