Exercise 3.3 Practice

Two rational numbers $\dfrac{a}{b}$ and $\dfrac{c}{d}$ are equal if $a \times d = b \times c$ (cross-multiplication rule).

(i) $\dfrac{2}{3}$ and $\dfrac{4}{6}$: $$2 \times 6 = 12 \quad \text{and} \quad 3 \times 4 = 12$$ Since both cross products are equal, $\dfrac{2}{3} = \dfrac{4}{6}$. ✓

Alternatively, $\dfrac{4}{6} = \dfrac{4 \div 2}{6 \div 2} = \dfrac{2}{3}$.

(ii) $\dfrac{5}{4}$ and $\dfrac{10}{8}$: $$5 \times 8 = 40 \quad \text{and} \quad 4 \times 10 = 40$$ Since both cross products are equal, $\dfrac{5}{4} = \dfrac{10}{8}$. ✓

(iii) $-\dfrac{3}{5}$ and $-\dfrac{6}{10}$: $$(-3) \times 10 = -30 \quad \text{and} \quad 5 \times (-6) = -30$$ Since both cross products are equal, $-\dfrac{3}{5} = -\dfrac{6}{10}$. ✓

(iv) $\dfrac{9}{3}$ and $3$: $$\frac{9}{3} = 9 \div 3 = 3$$ So $\dfrac{9}{3} = 3$. ✓

To add rational numbers, first find the Lowest Common Denominator (LCD), then add the numerators.

(i) $\dfrac{2}{5} + \dfrac{3}{10}$
LCD of 5 and 10 is 10. $$\frac{2}{5} + \frac{3}{10} = \frac{4}{10} + \frac{3}{10} = \frac{7}{10}$$

(ii) $\dfrac{7}{12} + \dfrac{5}{8}$
LCD of 12 and 8: $12 = 2^2 \times 3$, $8 = 2^3$, so LCD $= 2^3 \times 3 = 24$. $$\frac{7}{12} + \frac{5}{8} = \frac{14}{24} + \frac{15}{24} = \frac{29}{24}$$

(iii) $-\dfrac{4}{7} + \dfrac{3}{14}$
LCD of 7 and 14 is 14. $$-\frac{4}{7} + \frac{3}{14} = -\frac{8}{14} + \frac{3}{14} = \frac{-8 + 3}{14} = -\frac{5}{14}$$

To subtract rational numbers, find the LCD, convert, then subtract numerators.

(i) $\dfrac{5}{6} - \dfrac{1}{4}$
LCD of 6 and 4: $6 = 2 \times 3$, $4 = 2^2$, LCD $= 12$. $$\frac{5}{6} - \frac{1}{4} = \frac{10}{12} - \frac{3}{12} = \frac{7}{12}$$

(ii) $\dfrac{11}{8} - \dfrac{3}{4}$
LCD of 8 and 4 is 8. $$\frac{11}{8} - \frac{3}{4} = \frac{11}{8} - \frac{6}{8} = \frac{5}{8}$$

(iii) $-\dfrac{7}{9} - \left(-\dfrac{2}{3}\right)$
Subtracting a negative is the same as adding a positive: $$-\frac{7}{9} + \frac{2}{3}$$ LCD of 9 and 3 is 9. $$= -\frac{7}{9} + \frac{6}{9} = \frac{-7 + 6}{9} = -\frac{1}{9}$$

To multiply rational numbers: $\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}$. Simplify where possible.

(i) $\dfrac{2}{3} \times \dfrac{3}{10}$ $$= \frac{2 \times 3}{3 \times 10} = \frac{6}{30} = \frac{1}{5}$$ (Cancelled the common factor of 6.)

(ii) $\dfrac{7}{11} \times \dfrac{5}{8}$ $$= \frac{7 \times 5}{11 \times 8} = \frac{35}{88}$$ (No common factors; result is already in lowest terms.)

(iii) $-\dfrac{4}{7} \times \dfrac{5}{14}$ $$= \frac{(-4) \times 5}{7 \times 14} = \frac{-20}{98} = -\frac{10}{49}$$ (Cancelled the common factor of 2.)

To divide rational numbers, multiply the first fraction by the reciprocal of the second: $$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc}$$

(i) $\dfrac{2}{3} \div \dfrac{3}{10}$ $$= \frac{2}{3} \times \frac{10}{3} = \frac{20}{9}$$

(ii) $\dfrac{7}{11} \div \dfrac{5}{8}$ $$= \frac{7}{11} \times \frac{8}{5} = \frac{56}{55}$$

(iii) $-\dfrac{4}{7} \div \dfrac{5}{14}$ $$= -\frac{4}{7} \times \frac{14}{5} = \frac{-4 \times 14}{7 \times 5} = \frac{-56}{35} = -\frac{8}{5}$$ (Cancelled common factor of 7.)

Left-Hand Side (LHS):
First compute the bracket: $$\frac{1}{2} + \frac{3}{4} = \frac{2}{4} + \frac{3}{4} = \frac{5}{4}$$ Now multiply: $$\text{LHS} = \frac{5}{4} \times \frac{8}{3} = \frac{40}{12} = \frac{10}{3}$$

Right-Hand Side (RHS): $$\frac{1}{2} \times \frac{8}{3} + \frac{3}{4} \times \frac{8}{3}$$ $$= \frac{8}{6} + \frac{24}{12}$$ $$= \frac{4}{3} + \frac{2}{1}$$ Let me redo this carefully: $$\frac{1}{2} \times \frac{8}{3} = \frac{8}{6} = \frac{4}{3}$$ $$\frac{3}{4} \times \frac{8}{3} = \frac{24}{12} = 2$$ $$\text{RHS} = \frac{4}{3} + 2 = \frac{4}{3} + \frac{6}{3} = \frac{10}{3}$$

Since LHS $= \dfrac{10}{3} =$ RHS, the distributive property is verified. ✓

Method 1: Distributing first
Apply the distributive law: $a(b - c) = ab - ac$ $$\frac{7}{9} \times \frac{6}{7} - \frac{7}{9} \times \frac{3}{4}$$ $$= \frac{42}{63} - \frac{21}{36}$$ $$= \frac{2}{3} - \frac{7}{12}$$
LCD of 3 and 12 is 12: $$= \frac{8}{12} - \frac{7}{12} = \frac{1}{12}$$

Method 2: Bracket first (verification)
LCD of 7 and 4 is 28: $$\frac{6}{7} - \frac{3}{4} = \frac{24}{28} - \frac{21}{28} = \frac{3}{28}$$ Now multiply: $$\frac{7}{9} \times \frac{3}{28} = \frac{21}{252} = \frac{1}{12}$$

Expand the left-hand side using the distributive property: $$\frac{5}{6} \cdot x + \frac{5}{6} \cdot \frac{3}{5} = \frac{5}{6}x + \frac{1}{2}$$ $$\frac{5}{6}x + \frac{15}{30} = \frac{5}{6}x + \frac{1}{2}$$ $$\frac{5}{6}x + \frac{1}{2} = \frac{5}{6}x + \frac{1}{2}$$

This is an identity — it is true for all rational values of $x$. Both sides are identical regardless of what $x$ is.

This demonstrates that the distributive property holds: $$\frac{5}{6}\left(x + \frac{3}{5}\right) = \frac{5}{6}x + \frac{5}{6} \times \frac{3}{5} = \frac{5}{6}x + \frac{1}{2}$$ The equation holds for every rational number $x$.