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Question

# Verify the following: (i) $\frac{3}{7}×\left(\frac{5}{6}+\frac{12}{13}\right)=\left(\frac{3}{7}×\frac{5}{6}\right)+\left(\frac{3}{7}×\frac{12}{13}\right)$ (ii) $\frac{-15}{4}×\left(\frac{3}{7}+\frac{-12}{5}\right)=\left(\frac{-15}{4}×\frac{3}{7}\right)+\left(\frac{-15}{4}×\frac{-12}{5}\right)$ (iii) $\left(\frac{-8}{3}+\frac{-13}{12}\right)×\frac{5}{6}=\left(\frac{-8}{3}×\frac{5}{6}\right)+\left(\frac{-13}{12}×\frac{5}{6}\right)$ (iv) $\frac{-16}{7}×\left(\frac{-8}{9}+\frac{-7}{6}\right)=\left(\frac{-16}{7}×\frac{-8}{9}\right)+\left(\frac{-16}{7}×\frac{-7}{6}\right)$

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Solution

## $\text{(i)}\phantom{\rule{0ex}{0ex}}\text{LHS=}\frac{3}{7}×\left(\frac{5}{6}+\frac{12}{13}\right)\phantom{\rule{0ex}{0ex}}=\frac{3}{7}×\left(\frac{65+72}{78}\right)\phantom{\rule{0ex}{0ex}}=\frac{3}{7}×\frac{137}{78}\phantom{\rule{0ex}{0ex}}=\frac{137}{182}\phantom{\rule{0ex}{0ex}}\text{RHS}=\left(\frac{3}{7}×\frac{5}{6}\right)+\left(\frac{12}{13}×\frac{3}{7}\right)\phantom{\rule{0ex}{0ex}}=\frac{3×5}{7×6}+\frac{12×3}{13×7}\phantom{\rule{0ex}{0ex}}=\frac{15}{42}+\frac{36}{91}\phantom{\rule{0ex}{0ex}}=\frac{195+216}{546}\phantom{\rule{0ex}{0ex}}=\frac{411}{546}\phantom{\rule{0ex}{0ex}}=\frac{137}{182}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ ∴ ​$\frac{3}{7}×\left(\frac{5}{6}+\frac{12}{13}\right)=\left(\frac{3}{7}×\frac{5}{6}\right)+\left(\frac{3}{7}×\frac{12}{13}\right)$ $\text{(ii)}\phantom{\rule{0ex}{0ex}}\mathrm{LHS}=\frac{-15}{4}×\left(\frac{3}{7}+\frac{-12}{5}\right)\phantom{\rule{0ex}{0ex}}=\frac{-15}{4}×\left(\frac{15-84}{35}\right)\phantom{\rule{0ex}{0ex}}=\frac{-15}{4}×\frac{-69}{35}\phantom{\rule{0ex}{0ex}}=\frac{\left(-15\right)×\left(-69\right)}{140}\phantom{\rule{0ex}{0ex}}=\frac{1035}{140}\phantom{\rule{0ex}{0ex}}=\frac{207}{28}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{RHS}=\left(\frac{-15}{4}×\frac{3}{7}\right)+\left(\frac{-15}{4}×\frac{-12}{5}\right)\phantom{\rule{0ex}{0ex}}=\frac{\left(-15\right)×3}{4×7}+\frac{\left(-15\right)×\left(-12\right)}{4×5}\phantom{\rule{0ex}{0ex}}=\frac{-45}{28}+\frac{180}{20}\phantom{\rule{0ex}{0ex}}=\frac{-225+1260}{140}\phantom{\rule{0ex}{0ex}}=\frac{1035}{140}\phantom{\rule{0ex}{0ex}}=\frac{207}{28}\phantom{\rule{0ex}{0ex}}\therefore \frac{-15}{4}×\left(\frac{3}{7}+\frac{-12}{5}\right)=\left(\frac{-15}{4}×\frac{3}{7}\right)+\left(\frac{-15}{4}×\frac{-12}{5}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ (iii) $\left(\frac{-8}{3}+\frac{-13}{12}\right)×\frac{5}{6}=\left(\frac{-8}{3}×\frac{5}{6}\right)+\left(\frac{-13}{12}×\frac{5}{6}\right)\phantom{\rule{0ex}{0ex}}\text{LHS}=\left(\frac{-8}{3}+\frac{-13}{12}\right)×\frac{5}{6}\phantom{\rule{0ex}{0ex}}=\frac{-32-13}{12}×\frac{5}{6}\phantom{\rule{0ex}{0ex}}=\frac{-45}{12}×\frac{5}{6}\phantom{\rule{0ex}{0ex}}=\frac{-45×5}{12×6}\phantom{\rule{0ex}{0ex}}=\frac{-225}{72}\phantom{\rule{0ex}{0ex}}=\frac{-225÷9}{72÷9}\phantom{\rule{0ex}{0ex}}=-\frac{25}{8}\phantom{\rule{0ex}{0ex}}\text{RHS}=\left(\frac{-8}{3}×\frac{5}{6}\right)+\left(\frac{-13}{12}×\frac{5}{6}\right)\phantom{\rule{0ex}{0ex}}=\frac{-8×5}{3×6}+\frac{\left(-13\right)×5}{12×6}\phantom{\rule{0ex}{0ex}}=\frac{-40}{18}+\frac{-65}{72}\phantom{\rule{0ex}{0ex}}=\frac{-160-65}{72}\phantom{\rule{0ex}{0ex}}=\frac{-225}{72}\phantom{\rule{0ex}{0ex}}=\frac{-225÷9}{72÷9}\phantom{\rule{0ex}{0ex}}=\frac{-25}{8}\phantom{\rule{0ex}{0ex}}\therefore \left(\frac{-8}{3}+\frac{-13}{12}\right)×\frac{5}{6}=\left(\frac{-8}{3}×\frac{5}{6}\right)+\left(\frac{-13}{12}×\frac{5}{6}\right)$ (iv) $\frac{-16}{7}×\left(\frac{-8}{9}+\frac{-7}{6}\right)=\left(\frac{-16}{7}×\frac{-8}{9}\right)+\left(\frac{-16}{7}×\frac{-7}{6}\right)\phantom{\rule{0ex}{0ex}}\text{LHS}=\frac{-16}{7}×\left(\frac{-8}{9}+\frac{-7}{6}\right)\phantom{\rule{0ex}{0ex}}=\frac{-16}{7}×\left(\frac{-16-21}{18}\right)\phantom{\rule{0ex}{0ex}}=\frac{-16}{7}×\frac{-37}{18}\phantom{\rule{0ex}{0ex}}=\frac{592}{126}\phantom{\rule{0ex}{0ex}}=\frac{296}{63}\phantom{\rule{0ex}{0ex}}\text{RHS}=\left(\frac{-16}{7}×\frac{-8}{9}\right)+\left(\frac{-16}{7}×\frac{-7}{6}\right)\phantom{\rule{0ex}{0ex}}=\frac{128}{63}+\frac{112}{42}\phantom{\rule{0ex}{0ex}}=\frac{256+336}{126}\phantom{\rule{0ex}{0ex}}=\frac{592}{126}\phantom{\rule{0ex}{0ex}}=\frac{296}{63}\phantom{\rule{0ex}{0ex}}\therefore \frac{-16}{7}×\left(\frac{-8}{9}+\frac{-7}{6}\right)=\left(\frac{-16}{7}×\frac{-8}{9}\right)+\left(\frac{-16}{7}×\frac{-7}{6}\right)$

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