Question

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Answer 1

47.
$\frac{x}{2} = \frac{y}{3} = k = > x = 2k \: \: y = 3k$
a.
$\frac{2k + 3k}{2k} = \frac{5k}{2k} = \frac{5}{2}$
b.
$\frac{6k + 3k}{10k + 6k} = \frac{9k}{16k} = \frac{9}{16}$
c.
$\frac{8k - 3k}{2k + 3k} = \frac{5k}{5k} = \frac{5}{5} = 1$
d.
$\frac{10k + 9k}{10k - 9k} = \frac{19k}{1k} = \frac{19}{1}$
E.
$\frac{20k + 3k}{14k - 3k} = \frac{23k}{11k} = \frac{23}{11}$
48.
$\frac{x}{2} = \frac{y}{5} = k = > x = 2k \: \: \: \: y = 5k$
a.
$\frac{6k + 10k}{2k + 5k} = \frac{16k}{7k} = \frac{16}{7}$
b.
$\frac{14k}{20k} = \frac{14}{20}$
c.
$\frac{10k - 5k}{4k + 2k} = \frac{5k}{6k} = \frac{5}{6}$
d.
$\frac{18k + 10k}{4k} = \frac{28k}{4k} = \frac{28}{4} = \frac{7}{1} = 7$
E.
$\frac{4k + 10k}{2k + 5k} = \frac{14k}{7k} = \frac{14}{7} = \frac{2}{1} = 2$