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Question

If $\frac{3a+7b}{3a-7b}=\frac{4}{3}$ then find the value of the ratio $\frac{3{a}^{2}-7{b}^{2}}{3{a}^{2}+7{b}^{2}}$

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Solution

$\frac{3a+7b}{3a-7b}=\frac{4}{3}$ Applying componendo and dividendo, we get $\frac{\left(3a+7b\right)+\left(3a-7b\right)}{\left(3a+7b\right)-\left(3a-7b\right)}=\frac{4+3}{4-3}\phantom{\rule{0ex}{0ex}}⇒\frac{3a+7b+3a-7b}{3a+7b-3a+7b}=\frac{7}{1}\phantom{\rule{0ex}{0ex}}⇒\frac{6a}{14b}=7\phantom{\rule{0ex}{0ex}}⇒\frac{a}{b}=\frac{14×7}{6}=\frac{49}{3}.....\left(1\right)$ Now, $\frac{3{a}^{2}-7{b}^{2}}{3{a}^{2}+7{b}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\frac{3{a}^{2}}{{b}^{2}}-\frac{7{b}^{2}}{{b}^{2}}}{\frac{3{a}^{2}}{{b}^{2}}+\frac{7{b}^{2}}{{b}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{3×{\left(\frac{a}{b}\right)}^{2}-7}{3×{\left(\frac{a}{b}\right)}^{2}+7}\phantom{\rule{0ex}{0ex}}=\frac{3×{\left(\frac{49}{3}\right)}^{2}-7}{3×{\left(\frac{49}{3}\right)}^{2}+7}\left[\mathrm{Using}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{\frac{2401-21}{3}}{\frac{2401+21}{3}}$ $=\frac{2380}{2422}\phantom{\rule{0ex}{0ex}}=\frac{2380÷14}{2422÷14}\left(\mathrm{HCF}\mathrm{of}2380\mathrm{and}2422=14\right)\phantom{\rule{0ex}{0ex}}=\frac{170}{173}$

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