This part is interesting. In the last section we saw that we could \textbf{not} further simplify terms that were different such as \(2x + 4y\). But now with multiplication and division we can! \(2x \times 4y = 8xy\) and \(2x \div 4y = \frac{\cancel{2}x}{\cancel{4}y} = \frac{x}{2y}\). Isn't that cool? Okay, now you guys try: \begin{examplebox} \subsection{Examples} \begin{questions} \Question[6] Multiplications \begin{multicols}{3} \begin{parts} \part \(4 \times 3a = \) \begin{solutionordottedlines}[1cm] 12a \end{solutionordottedlines} \part \(2d \times 5e = \) \begin{solutionordottedlines}[1cm] 10de \end{solutionordottedlines} \part \(4m \times 5m = \) \begin{solutionordottedlines}[1cm] $20m^2$ \end{solutionordottedlines} \part \(3p \times 2pq = \) \begin{solutionordottedlines}[1cm] $6p^2q$ \end{solutionordottedlines} \part \(3x \times (-6) = \) \begin{solutionordottedlines}[1cm] $-18x$ \end{solutionordottedlines} \part \(-5ab \times -3bc = 15abc^2\) \begin{solutionordottedlines}[1cm] $15abc^2$ \end{solutionordottedlines} \end{parts} \end{multicols} \Question[6] Divisions \begin{multicols}{3} \begin{parts} \part \(24x \div 6 = \) \begin{solutionordottedlines}[1cm] 4x \end{solutionordottedlines} \part \(36a\div 4 = \) \begin{solutionordottedlines}[1cm] 9a \end{solutionordottedlines} \part \(-18x^2 \div (-3) = \) \begin{solutionordottedlines}[1cm] $6x^2$ \end{solutionordottedlines} \part \(\frac{15a}{3} = 5a\) \begin{solutionordottedlines}[1cm] $5a$ \end{solutionordottedlines} \part \(\frac{12x}{21} = \frac{4x}{7}\) \begin{solutionordottedlines}[1cm] $\frac{4x}{7}$ \end{solutionordottedlines} \part \(\frac{-24xy}{6y} = \) \begin{solutionordottedlines}[1cm] $-4x$ \end{solutionordottedlines} \end{parts} \end{multicols} \end{questions} \end{examplebox} \begin{exercisebox} \subsection{Exercises} \begin{questions} \Question[6] Rewrite as a single fraction: \begin{multicols}{3} \begin{doublespace} \begin{parts} \part \(\frac{2a}{5} \times \frac{a}{4}=\frac{2a^2}{20}=\frac{a^2}{10}\) \begin{solutionordottedlines}[1cm] $\frac{a^2}{10}$ \end{solutionordottedlines} \part \(\frac{3x}{7} \times \frac{5y}{12}=\frac{15xy}{84}=\frac{5xy}{28}\) \begin{solutionordottedlines}[1cm] $\frac{5xy}{28}$ \end{solutionordottedlines} \part \(\frac{4p}{q} \times \frac{3}{2p}=\frac{12p}{2pq}=\frac{6}{q}\) \begin{solutionordottedlines}[1cm] $\frac{6}{q}$ \end{solutionordottedlines} \part \(\frac{15}{x} \times \frac{2}{3x}=\frac{30}{3x^2}=\frac{10}{x^2}\) \begin{solutionordottedlines}[1cm] $\frac{10}{x^2}$ \end{solutionordottedlines} \part \(\frac{2x}{3} \div \frac{3x}{5}=\frac{2x}{3} \times \frac{5}{3x}=\frac{10}{9}\) \begin{solutionordottedlines}[1cm] $\frac{10}{9}$ \end{solutionordottedlines} \part \(\frac{6a}{7b} \div \frac{2ab}{3}=\frac{6a}{7b} \times \frac{3}{2ab}=\frac{18}{14b^2}=\frac{9}{7b^2}\) \begin{solutionordottedlines}[1cm] $\frac{9}{7b^2}$ \end{solutionordottedlines} \end{parts} \end{doublespace} \end{multicols} \Question[6] Simplify \begin{multicols}{3} \begin{parts} \part \(5c \times 2d = 10cd\) \begin{solutionordottedlines}[1cm] 10cd \end{solutionordottedlines} \part \(-6l \times (-5m)= 30lm\) \begin{solutionordottedlines}[1cm] 30lm \end{solutionordottedlines} \part \(-2m \times (-4m)= 8m^2\) \begin{solutionordottedlines}[1cm] $8m^2$ \end{solutionordottedlines} \part \(24a^2 \div 8 = 3a^2\) \begin{solutionordottedlines}[1cm] $3a^2$ \end{solutionordottedlines} \part \(7 \times 15p \div 21 = 5p\) \begin{solutionordottedlines}[1cm] $5p$ \end{solutionordottedlines} \part \(18y \div 6 \times 2= 6y\) \begin{solutionordottedlines}[1cm] $6y$ \end{solutionordottedlines} \end{parts} \end{multicols} \Question[8] Simplify by first cancelling out common factors: \begin{multicols}{4} \begin{doublespace} \begin{parts} \part \(\frac{14p}{21} =\frac{2p}{3}\) \begin{solutionordottedlines}[1cm] $\frac{2p}{3}$ \end{solutionordottedlines} \part \(\frac{22x^2}{33} =\frac{2x^2}{3}\) \begin{solutionordottedlines}[1cm] $\frac{2x^2}{3}$ \end{solutionordottedlines} \part \(\frac{2xy}{6xy}=\frac{1}{3}\) \begin{solutionordottedlines}[1cm] $\frac{1}{3}$ \end{solutionordottedlines} \part \(\frac{-4xy}{8x} =\frac{-y}{2}\) \begin{solutionordottedlines}[1cm] $\frac{-y}{2}$ \end{solutionordottedlines} \part \(\frac{2y}{5} \times \frac{y}{4} =\frac{2y^2}{20}=\frac{y^2}{10}\) \begin{solutionordottedlines}[1cm] $\frac{y^2}{10}$ \end{solutionordottedlines} \part \(\frac{p}{6q} \times \frac{9p}{4q} =\frac{9p^2}{24q^2}=\frac{3p^2}{8q^2}\) \begin{solutionordottedlines}[1cm] $\frac{3p^2}{8q^2}$ \end{solutionordottedlines} \part \(\frac{2yz}{5xy} \times \frac{3xy}{4yz} =\frac{6xyz}{20xyz}=\frac{3}{10}\) \begin{solutionordottedlines}[1cm] $\frac{3}{10}$ \end{solutionordottedlines} \part \(\frac{2y}{5} \div \frac{y}{4} =\frac{2y}{5} \times \frac{4}{y}=\frac{8}{5}\) \begin{solutionordottedlines}[1cm] $\frac{8}{5}$ \end{solutionordottedlines} \end{parts} \end{doublespace} \end{multicols} \end{questions} \end{exercisebox}