Find the value of x such that :

(i) $\dfrac{-2}{3} = \dfrac{14}{x}$

(ii) $\dfrac{8}{-3} = \dfrac{x}{6}$

(iii) $\dfrac{5}{9} = \dfrac{x}{-27}$

(iv) $\dfrac{11}{6} = \dfrac{-55}{x}$

(v) $\dfrac{15}{x} = -3$

(vi) $\dfrac{-36}{x} = 2$

(i) $\dfrac{-2}{3} = \dfrac{14}{x}$

We have,

$\dfrac{-2}{3} = \dfrac{14}{x} \\[1em] \Rightarrow (-2) \times x = 3 \times 14 \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \dfrac{3 \times 14}{-2} \\[1em] \Rightarrow x = 3 \times (-7) = -21$

Hence, x = -21.

(ii) $\dfrac{8}{-3} = \dfrac{x}{6}$

We have,

$\dfrac{8}{-3} = \dfrac{x}{6} \\[1em] \Rightarrow (-3) \times x = 8 \times 6 \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \dfrac{8 \times 6}{-3} \\[1em] \Rightarrow x = 8 \times (-2) = -16$

Hence, x = -16.

(iii) $\dfrac{5}{9} = \dfrac{x}{-27}$

We have,

$\dfrac{5}{9} = \dfrac{x}{-27} \\[1em] \Rightarrow 9 \times x = 5 \times (-27) \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \dfrac{5 \times (-27)}{9} \\[1em] \Rightarrow x = 5 \times (-3) = -15$

Hence, x = -15.

(iv) $\dfrac{11}{6} = \dfrac{-55}{x}$

We have,

$\dfrac{11}{6} = \dfrac{-55}{x} \\[1em] \Rightarrow 11 \times x = 6 \times (-55) \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \dfrac{6 \times (-55)}{11} \\[1em] \Rightarrow x = 6 \times (-5) = -30$

Hence, x = -30.

(v) $\dfrac{15}{x} = -3$

Since $-3 = \dfrac{-3}{1}$, we have

$\dfrac{15}{x} = \dfrac{-3}{1} \\[1em] \Rightarrow (-3) \times x = 15 \times 1 \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \dfrac{15}{-3} = -5$

Hence, x = -5.

(vi) $\dfrac{-36}{x} = 2$

Since $2 = \dfrac{2}{1}$, we have

$\dfrac{-36}{x} = \dfrac{2}{1} \\[1em] \Rightarrow 2 \times x = (-36) \times 1 \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \dfrac{-36}{2} = -18 \\[1em]$

Hence, x = -18.