Question

plzzzzz help me out​

Answer 1

Question:

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

Answer:

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

( ii ) $\displaystyle{\boxed{\pink{\sf\:\dfrac{2a^2\:+\:3b^2}{2a^2\:-\:3b^2}\:=\:\dfrac{125}{71}}}}$

( iii ) $\displaystyle{\boxed{\orange{\sf\:\dfrac{a^3\:-\:b^3}{b^3}\:=\:\dfrac{316}{27}}}}$

( iv ) $\displaystyle{\boxed{\blue{\sf\:\dfrac{7a\:+\:9b}{7a\:-\:9b}\:=\:\dfrac{38}{11}}}}$

Step-by-step-explanation:

( i )

We have given that,

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

We have to find the value of

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

Now,

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

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

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

$\displaystyle{\implies\sf\:\dfrac{5a\:+\:3b}{5a\:-\:3b}\:=\:\dfrac{35\:+\:9}{35\:-\:9}\:\:\:\:-\:-\:[\:By\:Componendo\:-\:Dividendo\:]}$

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

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

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( ii )

We have given that,

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

Now,

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

$\displaystyle{\implies\sf\:\dfrac{a}{7}\:=\:\dfrac{b}{3}\:=\:k\:\:\:-\:-\:-\:[\:By\:cross\:multiplication\:]}$

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

We have to find the value of

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

Now,

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

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

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

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

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

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( iii )

We have given that,

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

Now,

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

$\displaystyle{\implies\sf\:\dfrac{a}{7}\:=\:\dfrac{b}{3}\:=\:k\:\:\:-\:-\:-\:[\:By\:cross\:multiplication\:]}$

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

We have to find the value of

$\displaystyle{\sf\:\dfrac{a^3\:-\:b^3}{b^3}}$

Now,

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

$\displaystyle{\implies\sf\:\dfrac{343\:k^3\:-\:27\:k^3}{27\:k^3}}$

$\displaystyle{\implies\sf\:\dfrac{316\:\cancel{k^3}}{27\:\cancel{k^3}}}$

$\displaystyle{\implies\boxed{\orange{\sf\:\dfrac{a^3\:-\:b^3}{b^3}\:=\:\dfrac{316}{27}}}}$

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( iv )

We have given that,

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

We have to find the value of

$\displaystyle{\sf\:\dfrac{7a\:+\:9b}{7a\:-\:9b}}$

Now,

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

$\displaystyle{\implies\sf\:\dfrac{a}{b}\:\times\:\dfrac{7}{9}\:=\:\dfrac{7}{3}\:\times\:\dfrac{7}{9}\:\:\:\:-\:-\:\left[\:Multiplying\:both\:sides\:by\:\dfrac{7}{9}\:\right]}$

$\displaystyle{\implies\sf\:\dfrac{7a}{9b}\:=\:\dfrac{49}{27}}$

$\displaystyle{\implies\sf\:\dfrac{7a\:+\:9b}{7a\:-\:9b}\:=\:\dfrac{49\:+\:27}{49\:-\:27}\:\:\:\:-\:-\:[\:By\:Componendo\:-\:Dividendo\:]}$

$\displaystyle{\implies\sf\:\dfrac{7a\:+\:9b}{7a\:-\:9b}\:=\:\cancel{\dfrac{76}{22}}}$

$\displaystyle{\implies\boxed{\blue{\sf\:\dfrac{7a\:+\:9b}{7a\:-\:9b}\:=\:\dfrac{38}{11}}}}$

Answer 2

solution

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We have given that,

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

Now,

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

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

We have to find the value of

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

Now,

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

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

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

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

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