Question

If 15a²+4b²/15a²-4b²=47/7 then find the values of a/b

Answer 1

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Answer 2

Answer:

$\rm \dfrac ab = \pm \dfrac 35.$

Step-by-step explanation:

Given:

$\rm \dfrac{15a^2+4b^2}{15a^2-4b^2}=\dfrac {47}{7}$

According to Componendo and Dividendo rule, if there is a fraction such as,

$\rm \dfrac ND=\dfrac ab$,

then,

$\rm \dfrac{N+D}{N-D}=\dfrac{a+b}{a-b}$.

Applying this rule to the given equation by putting,

$\rm N= 15a^2+4b^2\\D=15a^2-4b^2\\a=47\\b=7$

we get,

$\rm \dfrac{(15a^2+4b^2)+(15a^2-4b^2)}{(15a^2+4b^2)-(15a^2-4b^2)}=\dfrac{47+7}{47-7}\\\dfrac{15a^2+15a^2+4b^2-4b^2}{15a^2-15a^2+4b^2-(-4b^2)}=\dfrac{54}{40}\\\dfrac{30a^2}{8b^2} =\dfrac{27}{20}\\\dfrac{a^2}{b^2} = \dfrac {8}{30} \times \dfrac{27}{20} = \dfrac {9}{25}\\\Rightarrow \dfrac ab=\pm \sqrt{\dfrac{9}{25}} = \pm \dfrac 35$