Question

Question:

If $\displaystyle{\sf\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}$, then find the values of the given ratios.

$\dfrac{5a\:+\:3b}{5a\:-\:3b}$

$\dfrac{2a^2\:+\:3b^2}{2a^2\:-\:3b^2}$

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Answer 1

$\: \: \: \: \:{\large{\pmb{\sf{\underline{ Here's \: your \: required \: solution!! }}}}}$

• We are given $\displaystyle{\sf \purple{\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}}$

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⠀$\displaystyle{\sf\green{:\implies\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}}$

$\displaystyle{:\implies\sf\:\dfrac{a}{b}\:\times\:\dfrac{5}{3}\:=\:\dfrac{7}{3}\:\times\:\dfrac{5}{3}\:\:\:\:-\:-\:\purple{\left [\:Multiplying\:both\:sides\:by\:\dfrac{5}{3}\:\right]}}$

$\displaystyle{:\implies\sf\:\dfrac{5a}{3b}\:=\:\dfrac{35}{9}}$

$\displaystyle{:\implies\sf\:\dfrac{5a\:+\:3b}{5a\:-\:3b}\:=\:\dfrac{35\:+\:9}{35\:-\:9}}$

$\displaystyle{:\implies\sf\:\dfrac{5a\:+\:3b}{5a\:-\:3b}\:=\:\cancel{\dfrac{44}{26}}}$

$\displaystyle{:\implies\boxed{\red{\sf\:\dfrac{5a\:+\:3b}{5a\:-\:3b}\:=\:\dfrac{22}{13}}}}$

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Again

$\displaystyle{\sf\green{:\implies{\:\dfrac{a}{b}\:=\:\dfrac{7}{3}}}}$

$\displaystyle{:\implies\sf\:\dfrac{a}{7}\:=\:\dfrac{b}{3}\:=\:k }$

$\displaystyle{\implies\sf\:a\:=\:7k\:\:\&\:\:b\:=\:3k}$

$\displaystyle{\sf:\purple {\implies \:\dfrac{2a^2\:+\:3b^2}{2a^2\:-\:3b^2}}}$

$\displaystyle{:\implies\sf\:\dfrac{2\:\times\:(\:7k\:)^2\:+\:3\:\times\:(\:3k\:)^2}{2\:\times\:(\:7k\:)^2\:-\:3\:\times\:(\:3\:k\:)^2}}$

$\displaystyle{:\implies\sf\:\dfrac{2\:\times\:49k^2\:+\:3\:\times\:9k^2}{2\:\times\:49k^2\:-\:3\:\times\:9k^2}}$

$\displaystyle{:\implies\sf\:\dfrac{98k^2\:+\:27k^2}{98k^2\:-\:27k^2}}$

$\displaystyle{:\implies\sf\:\dfrac{125\:\cancel{k^2}}{71\:\cancel{k^2}}}$

$\displaystyle{:\implies\boxed{\red{\sf\:\dfrac{2a^2\:+\:3b^2}{2a^2\:-\:3b^2}\:=\:\dfrac{125}{71}}}}$

Answer 2

Answer is in above attachment .

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