  Selina solutions

Q12) Find x, such that:

(i) \frac{-2}{3}=\frac{6}{x}

(ii) \frac{7}{-4}=\frac{x}{8}

(iii) \frac{3}{7}=\frac{x}{-35}

(iv) \frac{-48}{x}=6

(v) \frac{36}{x}=3

(vi) \frac{-27}{x}=9

Solution 12:

(i) \frac{-2}{3}=\frac{6}{x}

-2x = 6\times3

-2x = 18 x = \frac{18}{-2}

x = - 9

\therefore\frac{-2}{3}=\frac{6}{-9}

(ii) \frac{7}{-4}=\frac{x}{8}

7\times8 = -4x

56 = - 4x x = \frac{56}{-4}

x = - 14

\therefore\frac{7}{-4}=\frac{-14}{8}

(iii) \frac{3}{7}=\frac{x}{-35}

-35\times3 = 7x

-105 = 7x

x = \frac{-105}{7}

x = -15

\therefore\frac{3}{7}=\ \frac{-15}{-35}

(iv) \frac{-48}{x}=6

48 = 6x

x = \frac{-48}{6}

x = - 8

\therefore\frac{-48}{-8}=6

(v) \frac{36}{x}=3

36 = 3x

x = \frac{36}{3}

x = 12

\therefore\frac{36}{12}=3

(vi) \frac{-27}{x}=9

27 = 9x

x = \frac{-27}{9}

x = -3

\therefore\frac{-27}{-3}=9

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