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If \frac{a}{b} = \frac{3}{2} then \: find \: the \: value \: of \: \frac{4 {a}^{2} + 3 {b}^{2} }{4 {a}^{2} - 3 {b}^{2} } ​ please answer it I will follow you and also mark you brainlist

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323 cents Poluhina
Answer:
  • ➳ 2

GivEn:-

  •  \sf\dfrac{a}{b} = \dfrac{3}{2}

To Find :-

  •  \sf value \: of \: \dfrac{4 {a}^{2} + 3 {b}^{2} }{4 {a}^{2} - 3 {b}^{2} }

➠ We will find the value with two simple methods! Let's do it.

➤ Method 1 :-

 \sf\dfrac{a}{b} = \dfrac{3}{2}

 \implies \sf \: 2a = 3b

Now,

 \sf \dfrac{4 {a}^{2} + 3 {b}^{2} }{4 {a}^{2} - 3 {b}^{2} }

 \implies \sf \:\dfrac{ {(2a)}^{2} +3 {b}^{2}}{{(2a)}^{2}-3 {b}^{2}}

As, we get 2a = 3b, we will put the value of 2a there!

 \implies \sf \:\dfrac{ {(3b)}^{2} +3 {b}^{2}}{{(3b)}^{2}-3 {b}^{2}}

 \implies \sf \:\dfrac{ 9b^{2} +3 {b}^{2}}{{9b}^{2}-3 {b}^{2}}

 \implies \sf \:\dfrac{ 12b^{2}}{6 {b}^{2} }

 \boxed{\implies \sf \:2}

Now, another method!

➤ Method 2 :-

 \sf\dfrac{a}{b} = \dfrac{3}{2}

➠ a = 3k

➠ b = 2k (where k,common factor ≠ 0)

 \sf \dfrac{4 {a}^{2} + 3 {b}^{2} }{4 {a}^{2} - 3 {b}^{2} }

Now putting the values, we would get,

 \implies\sf \dfrac{4 \times({3k})^{2} + 3\times({2k})^{2} }{4\times ({3k})^{2} - 3\times ({2k})^{2} }

 \implies \sf \:\dfrac{4 \times 9 {k}^{2} + 3 \times 4 {k}^{2}}{4 \times 9 {k}^{2}-3 \times 4 {k}^{2} }

 \implies \sf \:\dfrac{36 {k}^{2} + 12 {k}^{2}}{36 {k}^{2}-12{k}^{2} }

\implies\sf \dfrac{{48k}^{2} }{ {24k}^{2} }

 \boxed{\implies \sf \:2}

Therefore, the answer would be ➾ 2

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Poluhina
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