Question

If 15a²+4b²/15a²-4b²=47/7 then find the values of 7a-3b/7a+3b

Answer 1

$\frac{7a - 3b}{7a + 3b?} = \frac{1}{6}$

Answer 2

Answer:

$\rm \dfrac{7a-3b}{7a+3b} =\dfrac 16\ \ \ \text{; if a}>0\\\dfrac{7a-3b}{7a+3b} =6\ \ \ \text{; if a}<0$

Step-by-step explanation:

Given:

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

The Componendo and Dividendo theorem states that if there is a fraction such as,

$\rm \dfrac {N^m}{D^m}=\dfrac pq$,

then,

$\rm \dfrac{N^m+D^m}{N^m-D^m}=\dfrac{p+q}{p-q}$.

Applying this rule to the given equation by putting,

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

we get,

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

Therefore,

$\rm a = \pm \dfrac 35 b$

Taking positive value of a.

$\rm a=\dfrac 35 b\\\\\dfrac{7a-3b}{7a+3b}=\dfrac{7\left( \dfrac 35 b\right) -3b}{7\left( \dfrac 35 b\right) +3b}\\=\dfrac{\dfrac {21b}{5}-3b}{\dfrac {21b}{5} +3b}\\=\dfrac{\dfrac{21b-15b}{5}}{\dfrac{21b+15b}{5}}\\=\dfrac{6b}{36b}\\=\dfrac 16$

Taking negative value of a.

$\rm a=-\dfrac 35 b\\\\\dfrac{7a-3b}{7a+3b}=\dfrac{7\left( -\dfrac 35 b\right) -3b}{7\left( -\dfrac 35 b\right) +3b}\\=\dfrac{-\dfrac {21b}{5}-3b}{-\dfrac {21b}{5} +3b}\\=\dfrac{\dfrac{-21b-15b}{5}}{\dfrac{-21b+15b}{5}}\\=\dfrac{-36b}{-6b}\\=6.$